 that technically you need to tell me that they're in f bracket x. It's highly possible that I forgot to write that down when I wrote up the division algorithm but the point is that the two polynomials you get us the quotient and the remainder are actually polynomials in f bracket x as well. The uniqueness of those two polynomials meaning if you and your friend each start with the same f and g and you're put in separate rooms and asked to divide g into f then you'll both come up with the same polynomials kruv x and r of x. There's only one way to solve the given equation with the stipulations on the degrees and second or and the last comment is some of you missed the subtlety here you can do it in such a way that either the degree of r of x is less than the degree of g of x. That's one possibility or r of x is zero. What a couple of you did was you put or the degree of r of x is zero. The point is if you're going to require the degree of r of x to be less than the degree of g of x obviously this already includes the case where the degree of r of x is zero because you've assumed the degree of g of x is bigger than or equal to one. This situation is the situation that we need to get around because we haven't given the zero polynomial a degree. We don't define the degree of it because we want our other formulas to work out so we simply have to separate out as a special case the polynomial zero not the degree zero polynomial so those are perfectly fine and all of the non-zero ones are taken up with the requirement that the degree of r of x is less than the degree of g of x. All right let's see I have a couple of administrative announcements. What are they? Yeah two administrative announcements. The first is so next week there's no class on Wednesday because school is closed for Thanksgiving holiday Wednesday Thursday Friday. The homework that I gave you on Monday is due next Monday. Two comments about that. One is there won't be an SI session on Tuesday so no SI Tuesday what's that the 20th? There's obviously no SI on Wednesday the 21st because school is closed but there will be an SI Monday November 19th instead of the Wednesday one that's nine to ten thirty so if you need some extra advice on how to get the homework done that's due Monday do that Monday morning with Jen. The homework that's due on Monday if folks you really think you need an extra 24 hours or so and you want to turn it in Tuesday afternoon or Tuesday evening either send it to me electronically or slide it under my doors that's fine so the Monday due date is I mean it's really Tuesday at six it's just that the campus will be closed on Wednesday so I can't make it due on Wednesday so you know feel free to take an extra 24 hours if you want it and this is looking ahead I'll definitely remind you of this and try to remember to send an email out two weeks from tonight no two weeks from Monday Monday November 26th we'll meet in Dwyer Hall 121 I don't know if you'll be on that night but okay I'll have to remind myself to send an email to the appropriate video people um there's some I don't know there's something going on that is a little bit unclear to me but they've asked whether we could meet in there that night so keep that in mind I'll try to remind you of that a few more times before the night just for that night just for that one night the new old building the one that they just gutted and it's actually really nice and they've been in there twice and whoo like it okay at some point I'm gonna tip my hand as to why we're doing all this stuff there's sort of a natural place to do that we may get there tonight but more likely we'll get there next Monday there's sort of this overarching theme that the author calls the basic goal but for now what I want to do because that'll take me 10 or 15 minutes to describe is continue to play up this sort of analogy between the integers the ring of whole numbers and these polynomial rings f bracket x where f is a field and we've already seen certain properties I'm gonna click through the properties that we've already seen just so that you continue to hopefully develop an intuition about the similarities between them so ring of integers and f bracket x both are integral domains neither is a field both have some sort of division algorithm associated with them and what we started doing on Monday night was looking at things inside f bracket x looking at polynomials that somehow play the role that the prime numbers play inside the integers there the the polynomials that somehow you can't break down any further that's the intuition behind prime numbers so just as a reminder we're working in f bracket x where f is a field and an element let's call it little f of x in capital f of x is called irreducible in case let's see the degree of f of x is at least one is at least one what this means folks is in particular the zero polynomial f of x equals zero is not viewed as an irreducible polynomial irreducible because its degree isn't bigger than or equal to one because it doesn't have a degree assigned to it so obviously doesn't have a degree sign to it then it's degrees isn't bigger than or equal to one and secondly the only way to factor f of x as g of x times h of x where each of these is an f bracket x where g of x h of x are in the underlying polynomial ring is when the degree of g of x is zero or the degree of h of x is zero and I'm going to add because I I didn't make a strong enough issue of this on Monday I'll add it in again here is irreducible in f bracket x let me go ahead and online those three irreducible intuitively means it's prime it doesn't break down anymore and a way to describe what not breaking down anymore is is look it's not that prime numbers don't break down it's not that you can't factor prime numbers it's just the only way to factor prime numbers are trivial ways if I hand you the number seven I'm about to factor it it's one time seven so I've technically factored it it's just one of the things that I've used as a factors number one I can also factor in the number seven is negative one times negative seven it's just I've used the number negative one and inside the integers the numbers one and negative one play a special role they are the units and what we saw inside f bracket x is that the units in f bracket x are the things of degree zero two six I mean the constants are the things that have degree zero other than the zero polynomial itself and so the irreducibility idea is all right if you hand me the polynomial maybe it factors but if it does the only way it factors is sort of trivially like this that it factors only as a unit times a polynomial now another way to phrase it is a numbers prime if you can't factor it as the product of two things each of which is less than absolute value than the thing itself and every time you factor seven one of the factors has absolute value seven it's one time seven or negative one times negative seven another way to view irreducibility of polynomials is the only way to factor it is if one of the factors has the same degree as the original polynomial so to say for instance a polynomial of degree three is irreducible means the only way you can factor it is as a degree three times a degree zero polynomial and degree zero polynomials are nothing more than the constant so what this says is the irreducible polynomial can only be factored in sort of the silly way like I can factor x plus one watch it's one half times quantity to x plus two but then do anything for you all right that's what your due stability in f bracket x means and the example that I left with left you with on Monday is that it turns out the notion of irreducibility and this is where things don't really have counterparts in the integers they have counterparts in extensions of the integers we're not going to talk about those in here they have there's sort of no counterparts of the following idea if I hand you polynomial by x squared minus two and the question is is it irreducible the answer always depends on tell me what f is tell me what the field is that you're using is the coefficients because what we showed last time is if you start with something like x squared minus two then it turns out and we'll show this will actually prove it in detail tonight turns out that so give you this would I give you x squared I give you x square plus one x square plus one will do x squared minus two tonight turns out that f is irreducible if you view it in q bracket x but it's not irreducible in c bracket x because x squared plus one can be written as x minus I times x plus I and I'm going to throw a third example on here I hand you the polynomial x square plus one turns out it's also not irreducible inside z two x turns out not irreducible in z two x watch I'm about to factor it x square plus one is x plus one times x plus one because remembering z px you prove that a plus b squared is a square plus b squared so a plus b squared if you want you just multiply it out so what happened oh well you got a middle term of two x but remember the coefficients are turns out it is irreducible irreducible in z three x in words it's impossible to factor x squared plus one as the product of two polynomials each of which have some guts in other words not just by pulling out a constant here inside z three x so here's the big question for the night tonight what we're going to ask is the question we ask and sometimes are actually able to answer not always because it's a really hard question given a polynomial little f of x in capital f of x when is f of x irreducible I guess we can in some situations ask the opposite question when is little f of x not irreducible I mean not irreducible sounds like double negatives than it is I mean you can say that the polynomial is reducible but just because I speak quickly often it starts sounding like reducible irreducible start sounding the same so I prefer to use the phrase not irreducible rather than the phrase reducible and the thing is well intuitively not prime when you can actually break the thing down all right well what we showed on Monday this thing we called the factor theorem is if you hand me a polynomial in f bracket x and there's some element in f with the property that when you plug it in for x that zero comes out in other words if you have a zero of the polynomial then you always have a factor that looks like x minus alpha and if you have a factor it looks like x minus alpha well there's a factor of degree one if you've started with some polynomial that has degree at least one polynomial that's not just a linear term then that will necessarily give a non-trivial factorization that'll imply that the corresponding polynomial is not irreducible so we've already seen a situation where depending on what the degree of the original polynomial is where there's a relationship between whether or not the polynomial has a zero and whether or not the polynomial has a factor of certain type so we already know we have seen this is the factor theorem I'll remind you of it I won't write it out in all its detail says this little f of x in capital f of x take a polynomial if alpha is some element in the same field has phi sub alpha of f of x equal to zero then f of x is x minus alpha times g of x for some g of x in half bracket x so the point is if you happen to start with a polynomial with coefficients in f and if you can find some element in that field notice this thing matches this thing here with the property that well that this expression equals zero intuitively all that means folks with the property that when you plug alpha in everywhere you see an x that zero comes out then necessarily f factors this way so you're thinking alright then it's the situation that f is not irreducible because you've just factored it well there is one somewhat silly possibility which is if the original polynomial happened to have a degree one then what this is going to amount to is nothing more than saying that x minus alpha is a factor in which you might have factored out as a two or six or something like that which really doesn't give a factorization so the point is so if the degree of f of x is bigger than or equal to two and if there is some element alpha in f for which when you plug alpha into f zero comes out the alpha of f of x equals zero then a little f of x is not irreducible irreducible in half bracket x so what this means is if you can find a zero inside the field if you can find some element so that when you plug it into f that zero comes out then necessarily the polynomial reduces I'll give you a quick hint I was able to immediately know that in z2 this polynomial was not irreducible why because I had in the back of my head all I need to do is make sure that I found a situation where if I plug something in from the field that zero comes out if you plug one in for x you get one plus one but you're working inside z2 and that zero so in fact I can conclude from that that if I hand you like x cube plus one that's also not irreducible in z2 because x cube plus one plug in one gets zero x to fourth plus one not irreducible plug in one get zero x to the fourth plus x cube plus x plus one plug in one yeah one plus one plus one plus one which is zero in z2 so that polynomial factors as well in fact not only do you know that it's somehow factors that it's not irreducible I can actually tell you all the factors is it's x minus the thing that you've realized is the zero right so at least this first example is a situation where you can reduce a polynomial but let's see the factor theorem said more than that yeah if I can find a zero then I can find a non-trivial factorization so here's the question alright is it the case then that if you can't find a zero that the polynomial doesn't factor in other words is the converse true and the answer well fortunately or unfortunately the reality situation is that the converse isn't true if the converse was true it would make life significantly easier and would be able to do essentially all of the stuff that we're going to do tonight and next Monday in about 10 minutes but it turns out the converse is not true converse is not true in other words IE if f of x in capital f of x does not have any zeros have any zeros in f I'll rephrase that in terms of these evaluation homomorphisms IE there are no elements we call them alpha in f for which v sub alpha of f of x is zero then it turns out then little f of x may or may not be irreducible we don't know irreducible it depends on lots of things folks I don't want to discount or play down the connection between whether or not a polynomial has a zero inside f in other words whether or not there's something that you can plug into the polynomial and have zero come out and whether or not the polynomial is irreducible there's obviously an important relationship but the relationship only goes one way and the relationship is only as good as your ability to say more about the given polynomial let me give you an example here's a polynomial excuse me f of x is let me give you this one x to the fourth plus three x squared plus two in let's see where we will do this thing how about our yeah our bracket x our bracket x of the reels here question are there any elements inside the reels with the property that when you plug them in for x that zero comes out not a chance whatever you plug in for acts here I'm asking you to raise it to the fourth power here I'm asking you to square it so the expressions that are getting kicked out regardless of what real number you plug in here always going to be positive I'm multiplying this one by one that one by three so these things are always bigger than or equal to zero if I had to I always get an expression bigger than or equal to two so there's no way that I can find a real number so that when I plug it in I get zero so the punchline is there are no zeros for f of x in the reels I'm not suggesting there aren't zeros for this thing somewhere else maybe there's some zeros for it in the complex numbers or something like that but at least in this particular field there are no zeros but you're thinking oh no zeros maybe the thing doesn't factor I'm about to factor it for you it's easy f of x is x squared plus one times x squared plus two in our bracket x if I have a zero I can factor it if I have no zeros maybe I can factor it maybe I can't example let's see f of x equal to x squared plus one in our bracket x r of x question is f irreducible does it factor does it not factor well I've told you not well I've told you at least not in q bracket x not in our bracket x well it turns out it doesn't in other words f is irreducible you can't break it down any further the answer is yes and here's why here's why by contradiction I'm going to show you that f is irreducible by showing you that if f did factor that somehow there would be a real number that when you squared it equals negative one and we know there are no real numbers with property whose squares negative one here's why by contradiction contradiction is this let's see suppose suppose what oh suppose that this polynomial x squared plus one did factor in other words that I could write it is g of x times h of x where what does it mean to be not irreducible it means that you can actually write down an honest to goodness factorization of the thing in other words the factorization where each of the polynomials is in the given polynomial ring in other words where the coefficients are taken from whatever the field is that you're working over and where the degrees of each of them are less than the degree of the polynomial that you started with well folks if you start with a polynomial degree two and you've actually factored is the product of two things each with smaller degree there's only one choice that has to be degree one and that has to be degree one so we're sort of lucky here so where the degree of g of x is less than two and the degree of h of x is also less than two that's what a legitimate factorization looks like a non irreducible factorization of x squared plus one look like yeah but in this case since the degree of x squared plus one is two this necessarily means there's only one possibility that the degree of g of x is one and the degree of h of x is one and what we're about to do is show that that can't happen I'm showing you that yet but we'll show that that can't happen folks if I had handed you the polynomial x to the six plus one or something like that I wouldn't be able to get so nicely to a situation where I'd be able to analyze things because hey if I'm going to then ask you to factor it as something times something else maybe that has degree one that has degree five or maybe that has degree two and that is degree four that is degree three and that is degree three but when I'm starting with something with such a small power small degree it turns out I can analyze the possible factorizations relatively easily in other words well let's see what does it mean to say g of x is a polynomial degree one it looks like g of x is then let's call it alpha x plus beta you want to call it ax plus b I don't care whatever you want and h of x is you know gamma x plus delta where each of these things alpha beta gamma delta are reels because that's where I'm requiring the factorization to take place and folks to say that the degrees of g and h are each one means that the coefficients on the x term have to be nonzero because if alpha was zero then it wouldn't be a polynomial degree one be a polynomial degree zero and alpha not zero and gamma not zero I'm still about to show that this can't happen in other words I've got let's see x squared plus one is alpha x plus beta times gamma x plus delta but alpha isn't zero but let's see alpha is not zero so what I'm about to do is I'm going to multiply everything through by one over alpha and that makes perfect sense inside the reels if you want to call it alpha inverse that's not a bad idea either so I'm going to write this as x squared plus one equals oh yeah so alpha inverse exists in the reels and I can write this as follows I'm going to write it as x plus beta over alpha times well I'm just factoring an alpha out but I'm going to absorb it into here alpha gamma x plus alpha delta okay so let's see that's a real number that's a real number dividing by alpha is legit and now I'm going to write this just this is the elementary arithmetic again this is x minus negative beta over alpha times alpha gamma x plus alpha gamma alpha delta it looks like a mess but here's finally where we've arrived I've taken this polynomial and I factored it is this times this and all of the expressions that you're looking at here alpha one over alpha beta gamma delta etc are real numbers so what I've just done is I've taken half of x and I've written it as x minus something times g of x where this thing and this thing are both polynomials in our bracket x but what does the factor theorem say the factor theorem says if you've got x minus something as a factor then this thing is a zero of that polynomial so by the factor theorem because I've got a factor of the correct form it necessarily says that minus beta over alpha is a zero of the given polynomial x squared plus one in other words this thing has the property that when you plug it in everywhere you see an x that you get zero folks beta divided by alpha is a real number so what I've just presumably come up with is a real number with the property that when you plug it into x squared plus one that you get zero and you can't have that you can't have a real number who's squared add it to one to zero because that would give you a real number who squares negative one and there is no such thing so that implies that minus beta over alpha squared equals minus one and that is a contradiction in the real contradiction of the fact that there are no squares there are no square roots of negative one in the real numbers and now that was sort of long and painful but at least it shows you one possible way of proving that this particular polynomial x squared plus one is irreducible in our bracket x questions there comments now seem totally painful and long-winded and it was but I wanted to show you what the details were so that when I write down the more general result at least you'll have an idea of how things play out so it turns out this sort of argument leads to the following to the following now remember I mentioned that the converse to this factor theorem the converse to the statement if there are no zeros does the thing factor isn't true in general because I gave you a situation of a polynomial that has no zeros inside this field but certainly factors the example I gave you that was a polynomial of degree four and what we're about to write down is all of the counter examples to the converse of that previous statement have to be with polynomials of degree four or more in other words if you hand me a polynomial of degree two and you want to know whether or not it's irreducible in f bracket x just tell me whether or not you can find a zero for it in f you can't find zeros it's irreducible that's what we just did here we couldn't find a zero for x square plus one and the punch line is irreducible why because if the degree was to it had the factor into a degree one times a degree one and we showed that if we had a degree one factor that that led to a zero and there are no zeroes similarly folks if I had handed you something like if I had me not do it in the reals if I had started with a polynomial of degree three don't think of x cube plus one because that's not a good example in this particular context but if I had handed you a polynomial of degree three and I was trying to convince you that that polynomial was irreducible I could run through the same sort of proof by contradiction proof by contradiction would look like well suppose it's not irreducible in other words suppose the thing actually factored well let's see when I had a degree two polynomial and honest to goodness factorization meant I could write in this degree one times degree one I have a degree three polynomial and honest to goodness factorization would mean that I have either well there's only one possibility I'd have to have degree one times degree two but the point is in that case I'd still have to have a degree one factor in other words I'd have to have a factor that looks something like this and whether or not the other factor was also degree one turns out to be irrelevant as soon as I have a degree one factor I can play exactly this game where instead of calling it divided by alpha I just call it out the inverse in general that's fine and then I can play exactly this game which then leads me to a zero for the polynomial in the field so if the thing factors and the thing had degree two I could get a zero if the thing factors and has degree three I could get a zero if the thing factors and has degree four there's no guarantees that there's a zero because it might factor down into a product of two quadratic terms and neither one of these necessarily gives us zero the factor so this argument leads to the following theorem and let me not call a theorem called proposition proposition says this if you start with the degree of a polynomial either being two or three or three and if F has no zeros in the field F then you've actually got your hands on an irreducible polynomial then F of X is irreducible and what historically happens to students is they look at this result and say cool you mean I can figure out whether or not polynomials irreducible just by deciding whether or not it has zeros in the field yes but only if you start with a polynomial of degree two or three and the reason again that those are special is if you have a degree two or a degree three polynomial if there was a factorization it would lead necessarily to a degree one factor and anytime you have a degree one factor you can play exactly the same game that we just played and pull a zero out of the field so for example if I hand you this polynomial F of X equals X squared minus two and it turns out this is irreducible in Q bracket X why well it's a polynomial of degree two so I can conclude that the polynomial is irreducible in Q bracket X as long as I can conclude that it's impossible to find an element in the field a rational number with the property that when you plug it in for X the zero comes out folks the point is I know what both of the zeros of this polynomial look like root two and minus root two and the point is none of those neither of those are rational so reason okay use the previous result previous proposition because the degree of F is two and the fact that F of X has no zeros in the rationals and I'll put parenthetically IE the zeros of this thing squared of two and minus squared of two are not in Q so this thing is your oh but wait a minute let's see if I ask you a different question like is this thing irreducible in our bracket X you know it's not irreducible in our bracket X it actually factors down in our bracket X why because it has a zero in our bracket X there's a real number that I can plug in here to get zero out squared two so in our bracket X it factors as follows note F of X equals X squared minus two is not irreducible in our bracket X irreducible in our bracket X since well I'll factor it for you F of X equals X minus root two X plus root two and if the question is well how come you can do that over here the answer is irreducibility depends on what field you're allowing me to use as coefficients when I write down this factorization it's a pretty it's a perfectly good factorization if you allow me to take coefficients from the reals because squared of two is a real number no problem squared of two is not a rational number so this isn't a factorization inside cube bracket X so you're starting to get some feel hopefully for the importance that the underlying field plays in the question of whether or not polynomial is irreducible and I'm going to start throwing out little hints about what this main idea is or what the author calls the basic goal so I started with X squared minus two and if I just looked in sort of the smaller field of rational numbers the field of fractions of integers it turns out this thing had no zero in there or was irreducible in there but if I went to a larger field then it turns out this polynomial actually did factor had a zero and the intuitive question that we're going to sort of lay out for you on next Monday and then start specifically trying to answer you know sort of rigorously is all right if you hand me polynomial and you view the polynomial with coefficients in some field it may or may not be the case that you can find something in that field that acts as a zero here I handed you this polynomial there's no zero for this polynomial in Q question can you somehow find a bigger field where that polynomial does have a zero well yeah I can find a bigger field where that polynomial has a zero our works of course before we leave that this example it's not necessarily the case that r is the only field that works like the complex numbers would work here but more interestingly remember two weeks ago this problem number 12 let s be the collection of real numbers with the property that they look like a plus b times the square of two where a and b are rational which is spent well some of you didn't spend enough time doing it but what you what you propose to do was to prove for me that s is a field yeah so the collection of things that look like a plus b times the square root of two where a and b are rational form a field field certainly contains the rational a plus zero times square root of two so every rational number is in there but it also contains the square root of two zero plus one times the square root of two is the square root of two so square root of two is in there and it's a field so there's a field with the property that this thing now does have a zero inside that field so in some sense I don't even have to go as big as the reals to find a field where this polynomial has a zero I really didn't have to look much beyond Q I just had to sort of throw the square root of two in and then whatever other stuff I needed to make the thing into a field and then in fact I came up with a field for which this polynomial has zero so the note is I forget what section that is turns out that x squared minus two also is not irreducible in s bracket x where s is the field from problem number 12 and section was it 18 or 19 or something like that let's call it 18 and keep fingers crossed that we got it right okay all right so again I haven't phrased for you in its entirety what the basic goal is but that hopefully gives you some glimpse as to where we're headed hand me a polynomial if the polynomial happens to not have a zero in the field that you're interested in try to make the field a little bit bigger and see whether or not you can find a zero for that polynomial all right questions comments and did that did that okay all right the rest there yeah the the remaining 25 or 30 minutes or so of tonight will not look like something that you're familiar with in any sort of math class and here's why typically in a math class you're looking at results that are nice and polished and say if you have this situation then you can always conclude the following that's a big theorem right whenever you have this you have this as it turns out folks this question of if you hand me a polynomial in f bracket x is it irreducible there's just no nice big all-encompassing results that simply say take whatever the polynomial is and do something and that'll give you a yes no answer there's no sort of determinant idea for polynomials where you simply hand me whatever polynomial you happen to come along and you know stumble over and hand me whatever the field is and put it in this machine and out will spit a yes no answer there's just not anything like that so what that means is whatever we can say about irreducibility of polynomials is really just sort of a it's sort of this miscellaneous grab bag here are some things that we can say and we'll say those but in general there's just no big result that always answers the yes no question it's hard question maybe well hopefully that shouldn't surprise you because remember this notion of irreducibility is sort of analogous to the question about the prime numbers in the integers because that's really what they are and questions about prime numbers in the integers are really hard so it's sort of the same idea okay so this is like I know the grab bag results grab bag of results regarding irreducibility of polynomials here's sort of the first flavor the first type first type of result folks it's often the case that what we will be looking at is the question of whether or not a given polynomial is irreducible over the field q so here's a very important field that comes up a lot the rational numbers and the question won't simply be here's a polynomial is it irreducible run it through machine the question will often be here's a polynomial the polynomial just by some miracle happens to have the property that even though it's a polynomial whose coefficients could be rational numbers the coefficients of the given polynomial happen to actually be integers here's a good example of that the coefficients on this polynomial happen to be integers even though I've been asking you to treat this polynomial as an only cube bracket x if the question is well why don't you ask me to treat it as a polynomial in z bracket x because we're only interested in viewing polynomials as polynomials in polynomial rings where the coefficients come from a field because if I don't then I don't have the division algorithm anymore losing the division algorithm just won't make things well won't make things work out nicely but it turns out there's some irreducibility results in those situations and here they are sometimes we have we have f of x in cube bracket x so focus on the field here folks you can't just sort of blow off and say oh f of x in some field it's that particular field where in fact actually all the coefficients of f are in z a good example is example f of x equals x cubed for example x cubed minus 3x squared plus a folks here's a perfectly good polynomial in cube bracket x but it happens to be the case that each of its coefficients are whole numbers it turns out that the following is true proposition if you can factor such an f and f of x in cube bracket x in other words if somebody hands you a polynomial and it happens to be a polynomial where all of its coefficients happen to be whole numbers and you ask the question all right is the thing irreducible or not viewed as a polynomial cube bracket x it turns out if you can factor it in cube bracket x you could actually have factored it in such a way that both of the pieces also have whole number coefficients so a factorization of an integer coefficient polynomial into two polynomials where each of the polynomials has rational coefficients if there was such a decomposition then it turns out it may not be the one you're looking at but you could then slide things around and actually rig it so that the two factors you're looking at actually both have whole number coefficients as well so it's a mouthful to say but the point is if you can factor in cube bracket x then in fact it could have been factored it f of x f of x could have been factored factor in z bracket x you will so what well here's the so what what this means is in this case the factorization result in two polynomials integer coefficient polynomials coefficient polynomials whose constant term constant terms divide the constant term of f and you're still not impressed by that but let me show you why you should be okay here's the question folks if I hand you this polynomial x cubed minus 3x squared plus 8 and ask the question does it factor in cube bracket x or is it irreducible in cube bracket x well one way to decide that is to try to break things down and say well if it factors it looks like a degree one polynomial times a degree two polynomial in cube bracket x as soon as I have a degree one polynomial well then I know I'd have to have a zero in cube bracket x question are there any rational numbers so that when you plug the rational number into here that zero comes out I don't know we're going to try all the rational numbers and see if any of them work it's a bad idea I mean we locked out when we looked at x squared minus two because I already knew where the zeros were and I knew that both of the zeros were not rational but if I hand you something like that and ask are there any rational zeros hard you don't know the answer to that you're thinking well you know how do I go about deciding whether or not this thing factors okay I can't find zeros why because I can't simply plug in all infinitely many rational numbers and decide whether or not any of them spit out zero but here's the point this result which I'm not going to prove for you this result says if this thing actually does factor then there would have to be a factorization into two polynomials where both of the coefficients on the polynomials were from the integers and moreover where the constant terms of the two polynomials that you factor it into have to divide eight so the point is here we can show whether x cubed minus three x squared plus eight is irreducible in cube bracket x in cube bracket x by showing that there are no no decompositions that look like x cubed minus three x squared plus eight equals ax plus b times cx squared plus dx plus e with b divides eight and e divides eight what this proposition allows you to do folks is take a task that is impossible because it requires you to plug in the infinite different things ask the question whether or not this thing has a zero on q for the question is it possible to find integers with the property that you can split that thing up as a product of two things that look like this but I'm gonna significantly limit what that number is and what that number is and the point is as soon as you determine what that number is and what that number is well wait a minute let's see I also know something about what a and c have to be because if I'm going to multiply these two things together to get x cubed that means that a times c has to be one so either a is one or a is negative one and if a is one then c has to be one oh now all of a sudden you've sort of limited significantly what the possible decompositions look like and then you simply go through and say yes I can do it or no I can't do it so what you've done by using this result is you've reduced the task of determining irreducibility to something that's just a finite sequence of steps can you solve the following equations where the appropriate coefficients live in z and it turns out you can't do it plus you can't do it can't solve for a b c d z I'll leave out the work there's a good example in the text but the point is look you already know something if somehow this product is going to equal that you already know that a times c is one you know that b times e is eight or b and e are integers that in effect only leaves you to say something intelligent about d but you should be able to say something about that if you multiply these two things out and then just equate coefficients all right as a consequence of this proposition it says this corollary if let's see do I want to do this yeah just this way if the polynomial that you start with is I'll introduce this term now because you'll see in the homework is monic monic folks just means the leading coefficient equals one in other words the coefficient on the highest power of x is one coefficient equals one then if you hand me f of x in q bracket x having integer coefficients integer coefficients factors are has a zero in q if and only if it has a zero zero in z so if somebody hands you a monic polynomial like this one x cube minus 3 x squared plus a monic means that leading coefficient is one and ask the question are there any zeros for that polynomial in the rational numbers a consequence of this factorization over z is if you could find a zero in the rational it turns out you'd also be able to find a zero in the integers and the zero necessarily divides divides the constant term of the polynomial what this corollary does is in certain situations if you're interested in whether or not a polynomial is irreducible well hey it turns out if you start with a polynomial of degree two or three here's a polynomial of degree three irreducibility is the same thing as whether or not the polynomial has a zero in the given field if I'm asking about irreducibility in this particular field then I'm asking whether or not this thing has a zero in q does it I can't check all the elements of q but it turns out again as a corollary of this thing that I'm admittedly not proven for you were just sort of writing down the grab bag here the corollary is if you're in a situation like that and you're interested in whether or not the polynomial has a zero in q it'll have a zero in q precisely when it has a zero in z the thing well that doesn't buy me much there were infinitely many things in here I couldn't plug them all in there's infinitely many things in here I obviously can't plug all them in either but the punchline is if there is a zero in here necessarily it has to divide the constant term of the given polynomial so if you're going to check for zeros in q you only have to check for zeros in z and you only have to check among the well you've got to be careful among the eight possible things that divide eight they are one two four eight minus one minus two minus four minus eight so you can't forget the fact that factorization inside z actually involves not only positive factors but negative factors as well so does it work well let's just plug those eight things in I won't do it for you but so plug in one does zero come out no plug in two does there come out no plug in four does there come out no plug in eight does there come out no plug in minus one is there the answer is no so since there's no zeros in z there are necessarily no zeros in q and since this thing has degree less than or equal to three since there's no zeros then necessarily it's irreducible in q I mean if this sounds like a lot of hard work to get to the answer yeah this thing is irreducible it is a lot of hard work but it's it's as good as we get it's as good as we know just no general rules at work now you thought that was sort of squishy here's another totally squishy one yeah question Richard yeah the corollary is true for any degree yeah right so yeah I want you to give the wrong impression the corollary is true for any degree using the corollary to try to determine whether or not a polynomial is irreducible will only be useful if I would say only be useful will typically be useful mostly in situations where the polynomial happens to be a degree two or degree three to begin with all right although if you find a zero I mean if you happen to hit on one then automatically the polynomial reduces is not irreducible because as soon as you have a zero then you have x minus that thing is a factor but if you don't find any in general might be irreducible might not be reduced so that's the first type of result here's the second type of result second type result this result folks again only holds for the specific field q of x f of x in q bracket x and the question is is it irreducible in q bracket x so here's what I want you to do write your polynomial f of x is and someone asked after class two times ago when we were looking at polynomials in detail do you have to write it a zero plus a one x plus a two x squared plus a and x n or can you write out a and x n plus a and minus one I mean doesn't matter what order you write about in the answers no it doesn't really matter you want to write out the constant term then a linear term then the x the n term or reverse it makes a difference because the plus operation is a billion right so let's write it this way a this is the typical way to write out this result a and minus one x and minus one plus that that that that plus a two x squared plus a one x plus a zero so here is a polynomial in q bracket x and what we're going to do is we assume again just like in the first type of result that all the a sub i are in fact in z and the question is going to be is this thing irreducible in q bracket x and the answer turns out to be this if you can find a prime number you can find a prime let's call it p so that the following three things are true first of all the prime doesn't divide a sub n in other words a sub n isn't a multiple of that prime secondly that the prime does divide all of the rest of the coefficients that divides that one that one that one that one and the constant term for all zero less than or equal to i less than or equal to n minus one we're not done and third that the square of the prime doesn't divide the constant term enough hypotheses then here's the conclusion then then half of x is irreducible in q bracket x give me a polynomial with integer coefficients it's the same question is it irreducible in q bracket x answer if you can hunt around and find a prime i don't care what prime two three five makes a difference and if that prime has these three properties then you can conclude that the polynomial is iridescent back at x i mean this gives you the flavor of how sort of to what lengths you have to go to actually get a result that says if you have blah blah blah blah then the polynomial is irreducible and heck not only would go in great lengths this result only holds for this one particular field i can prove this one for you this one's given a name it's called eisenstein's criterion for irreducibility criterion so what does it say it says let's see so for example if we look at half of x equal to oh x squared minus two in q bracket x we already know this thing is irreducible it's degree two and we know because we know what the zeros of this thing look like we know that it has no zeros in q so we already know it's irreducible but it turns out eisenstein's criterion for p equals two works let's see it's the case that two doesn't divide that coefficient it does divide all of the other coefficients notice even though i haven't written it in i have a zero x term but any integer divide zero that's not an issue and two certainly divides to so we get second criterion third criterion is have to make sure that the square of this thing doesn't divide the constant term and it doesn't so check i mean that's sort of a big hammer it's too big a hammer to hold on this one you already know this one's irreducible let's try another one how about f of x equals this thing x to the fourth minus nine x squared plus six x plus fifteen turns out this is also irreducible irreducible in q bracket x eisenstein's criterion and the prime to use is p equals three let's see does that yeah p doesn't divide one it does divide zero nine six and fifteen but p squared doesn't divide 15 so this thing is irreducible in q bracket x this is a nice machine here you can't go overboard if i hand you a polynomial in z x it might be irreducible in q bracket x even though there's no prime that works for eisenstein's criterion so it's another sort of ad hoc thing if you luck out you happen to be able to find a prime that works good for you but just because you can't find a prime that works doesn't tell you squat about whether or not the polynomials irreducible not this thing is irreducible in q bracket x but there's no prime that can be used in the context of eisenstein's criterion to get you to that conclusion yeah p equals three doesn't divide that does divide that does divide the p equals three doesn't divide the last term so p equals three is a non-player p equals two is also a non-player because it doesn't and p equals any other prime is clearly a non-player because it's not going to divide three or eight so you know here's another deal where if it works you know you're happy but if it doesn't work typically you can't say therefore it's not typically all you can do is walk away and say well i need to haul out another tool and see if i could use questions comments i mean if it seems messy it is messy but you know at least we know a few things it's not like all hope is lost third type of result we're doing here you know for a third type of result the field is finite like you know z2 or z3 or something like that there's other finite fields folks other than just the z sub p's but we'll get to those a little bit later on if the question is does this polynomial factor is it irreducible or not in z2x the advantage you have in those situations is it makes sense to actually just list out what all the possible factors are and see if any of them work i mean for example then use i'll say you have to you can use brute force quite honestly this is i mean this phrase often appears even in the mathematical literature here's the proof brute force just do it see if anything works for example here's a polynomial in f of x equals x cube plus x squared plus one in z2 bracket x is it irreducible or not irreducible in z2 bracket x oh well let's see because it's a polynomial of degree less or equal to three i can test irreducibility by asking whether or not there's a zero okay so i'm lucky here because i happen to be looking at polynomial of degree less or equal to three i can test irreducibility by testing for zeros and testing for zeros folks is like way easy because you only have two things to check and i'm guaranteed zero doesn't work because i know what i'm going to get when i plug in zero you're one that was easy does one work well that's one plus one plus one is one it's not zero so irreducible irreducible check all possible elements of z2 that doesn't scare you there's only two of them zero one neither is a zero none is a zero so since the polynomial degree of f of x is less or equal to three conclude that it's irreducible irreducible all right so that was easy there's an irreducible polynomial in z3x the question would be ridiculously hard if the field that i asked you to view this thing in was q bracket x we have to decide whether or not had zeros in q let's try another one how about g of x equals let's do a degree for one here yep x to the fourth plus let's see plus x plus one in z2x question is this reducible in other words not irreducible or is it irreducible in z2x well the first thing you might try is to see whether or not there are zeros if there happen to be zeros then you're done necessarily it's not irreducible because x minus that thing is a factor so check for zeros zero spits out a one one spits out one so there's no zeros okay so are we done no we're not done at all we happen to be done over here because we lucked out because we happen to be looking at a polynomial degree less or equal to three but if we're looking at a polynomial of degree four or more just because there aren't any zeros doesn't tell you whether or not the thing is irreducible although at least I know that there's no factor of degree one because if you have a factor of degree one then by the game we played over there you could write down a zero so that's helpful since no zeros in z2 just check them both so no factor of degree one of degree one by the argument that we gave before so if it factors x to the fourth plus x plus one is not irreducible then it would have to look like this it would factor factor as some degree two polynomial times some other degree two polynomial let's see what you did for homework no yeah what you're doing for homework is looking at situations where you asked to write down all the degree two or the degree three or the degree something polynomials with coefficients coming from a finite field and that's doable folks there's only a certain number of things that could possibly happen here x squared x square plus one x square plus x plus one x squared plus x I think I got them all there's all the polynomials of degree two here's all the polynomials of degree two you just listen them out x squared x square plus one x square plus x x square plus x plus one so presumably what you got to do is just decide whether or not if you take those four choices and those same four choices whether or not there's any possible way of multiplying two things that look like that together to kick out this thing maybe there is maybe there isn't but in fact you can reduce the workload even further I mean you should be happy if you get to the situation where all you got to do is multiply 16 things together that's not too bad but look I can guarantee that x squared can't be a factor because if you multiply x squared times anything you're going to get an x squared certainly not going to get a one out so it can't be x squared can't be x square plus x either because that's got a factor of x on it and this thing doesn't have a factor so in fact you can significantly limit the possible factors that the thing looks like and so turns out the only possible factor this could be is x squared plus one or x squared plus x plus one because whatever polynomial of degree two you write down can't be able to factor an x out folks if you could factor an x out to one of one of the factors you could factor an x out of this thing you can't factor in so there you go now multiply those four possible products and see whether or not this thing gets kicked out if it does then you've factored it if it doesn't then the point is you've eliminated all the possible factors and you can conclude that it's here it seems like a mess somewhat of a mess but it turns out at least when the field is finite when you're starting with coefficients coming from z2 or z3 or z sub p or other possible finite fields that we'll look at before the end of the semester you can often reduce the question of irreducibility to hey if there is a factorization would have to look like this I can actually list out all the possible things that look like that in fact I can limit the search significantly more because I don't have to worry about things like x squared or x squared plus x because the corresponding polynomial that I'm trying to look for can't have either of those as a factor and then just say alright here's what's left now just hammer them out see if it works I'll let you finish that one up at home does it work maybe it does maybe it doesn't and that doesn't work oh hey even in the end look folks it's commutative so that times that is the same as that times that so if you check that that times that doesn't work and that times that doesn't work either so in the end we sort of reduce the workload to just multiplying three different things together and I'll give you hint neither of the possibilities will work and the person is there's some symmetry here you know if you multiply that times that certainly unlikely that you'll get an x term without an x term alright sorry I ran over a little bit here