 Yeah. Right. And so is the question that, so the question is that's why you'll always get these two as item potents? Yeah. I'm sorry, I think I'm missing the question of the point. There are possibly many item potents for this operation. There's only two if the ring happens to be an integral of the main. That's approved for homework. We proved way back in the group theory section that if you look at a structure that happens to be a group, then there's only two items, then there's only one item potent, right? Zero. So the only thing that has zero plus or a plus a equals a is a equals zero. So somehow not really knowing anything about the structure of this binary operation in general means that in general lots of things could happen and typically does. But in the situation where you're talking about just multiplication, in any ring at least you've got those two that work. Why those two work? I guess these two are item potents for different reasons. One is because we prove that zero times anything is zero. And this is because one times anything is itself. So somehow this one sucks up everything. That's why zero itself works. This one on the other hand behaves as an identity. So one times a is a. So zero times a is zero, and one times a is a, and that means if you plug in these two specific things you get item potents. But then there might be more. Maybe I haven't hit the answer. Let's chat after class. Alright. Alright. So what we're up to tonight is we're going to, we're really going to introduce the key player in what is going to be the focus of our attention for the remainder of the semester. It's a player that we've touched on briefly for the last couple of weeks but haven't really analyzed in detail. And the player is this. We're going to talk about what are called polynomial rings. And if you're thinking, well we've already seen those, yeah we have, but we haven't formally defined what they are. I just asked you to sort of believe here's what the symbols are and what we're going to do tonight is look a little bit deeper as to what the structure of these things are. And so the idea is this. Start with any ring. So let capital R denote any ring you want. This construction that we're about to look at works for any ring you want. Of course the rings that we're going to focus most on are rings with unity so it's not unreasonable to think that this ring has a 1. That's fine. We'll look briefly tonight at a situation that you've already encountered. In fact you looked at for a homework problem where the ring might not be an integral domain. That's going to cause certain issues. And then what we'll do after noting those issues is say look, we're only going to be interested in using rings that happen to be integral domains. In fact more specifically we're only going to be interested in using rings R that happen to be fields because that's where the most interesting structure occurs. But the point is when I talk about this construction it holds regardless of what ring you start with. What we're going to look at is this. The polynomial ring, polynomial ring with coefficients in R. So that's what I'm defining is defined to be symbols of this type. The set of symbols that look like this, a0 plus a1x plus a2x squared plus dot, dot, dot, plus a sub nx to the n where n can be anything and is some positive integer. So I'm going to write it as z plus. N could be zero. That's the degree of this thing. And each of the sub i's is in the underlying ring R. So the set that I'm asking you to look at are symbols that look like the things that you are used to calling polynomials. They're a constant quote unquote plus some element of the underlying set, often the one you're used to working with is the real numbers times some symbol called x plus a2 times x squared, etcetera, etcetera, etcetera. Well here's the set that I want you to look at. Now I have to define two binary operations on the set. One called addition and one called multiplication. And the point is those operations are exactly as you're used to in the standard situation where the ring R happens to be the real numbers. So we define addition and multiplication on this set as usual for polynomials, polynomials. And folks, I could spend ten minutes showing you specifically what those things look like, but we'll never need the formal description of what does it mean to say take this and multiply it by this or take this and just do it. Let me give you some clues to what the issues are. Presumably you've handed me a symbol like this and then you'll hand me another symbol that looks like that. Of course the second symbol that you hand me might not correspond to the same value of n here. You might hand me a polynomial of eventually, of course we're going to call this the degree, a polynomial of degree three and a polynomial of degree five. So how do you add those together? Well technically if I'm going to define the addition I somehow have to add some terms in to make sure that the two things sort of look the same. In other words I have to make the degree three polynomial sort of expand it out a little bit by adding 0x to the fourth plus 0x to the fifth. Now I have two degree n polynomials for the same n. Now I'm going to teach you how to add them just by adding the coefficients just as you'd expect. I mean that's something that you don't even worry about anymore. If I ask you to add 1 plus x squared to x cubed, well the first one's a degree two and the second one's a degree three. How do you add them in? So you just, you sort of in your head are thinking well there's a 0x cubed tacked on to the end of the 1 plus x squared. No problem. Similarly for multiplication, well multiplication could cause a little bit more grief only because here the underlying coefficients, the ring r, might not even be commutative. So that, and we've seen an example of this, if you try to multiply two things like you know a plus b or 1 plus x and not one like, let me not go there. The key is you're going to need to be a little bit careful on the order that you multiply these things in and make sure that you go ahead and keep whatever order you're handed these things in straight. Enough said I don't want to write out the formal definition. When you see two polynomials you're going to have to add them you'll be able to do that easily. If you're handed two polynomials and you have to multiply them you just multiply the two expressions together as you would expect. Alright we call this ring notationally this r bracket x. We'll do some examples in a minute but it's important to keep in mind what the role of this symbol x is. You're used to thinking of x as some sort of indeterminate, as some sort of expression that you're going to somehow plug something into and that the polynomials viewed as a function that's going to spit out a value. Okay that's been a good description of polynomials and that will typically serve as well but I want you to view the symbols in this set purely as symbols. Whatever this x means alright it might be the case that we might plug something into it later but for now just manipulate it as if you would a polynomial and don't try to ascribe any sort of meaning to what that is. In particular there's really nothing special about calling this thing x. We could just easily call it t and that's the indeterminate call y or something like that. The ring that you'll get even though the letters look different of course will be the same ring. It's polynomials in whatever the specified variable name is. So we get a ring and it turns out then r of x but I'm not going to prove much if any of this is a ring. So here's going to be the question if I know something about the coefficients that I'm using to build the polynomials can I then conclude something about this collection as a ring in and of itself. Well there's lots of easy observations, proof omitted, it all works out just fine. Everything is straightforward. Well one thing that we might be interested in doing is checking that for instance if the original ring r, if the coefficient ring has unity that r bracket x also has unity and that's pretty easy to show. I mean if I hand you the number one I can view it as an element of r bracket x. You just look at one plus, well that's it, just one because that's a completely legitimate term. Here's a symbol inside r bracket x just let n equal zero and let a sub zero be one. So you have the number one inside the ring r bracket x. In fact you have any symbol that lives inside the coefficient ring viewable as an element sitting inside our bracket x. Question? Just any subset you want. And ai come from the set capital r. So for example let's look at how about z2 in bracket x actually let's look at something a little bit more interesting how about z5 bracket x boy and say that z2 of x is totally interesting but z5 x will play up the ideas that I'm trying to get across a little bit more clearly. So here is the ring, the coefficients happen to come from that field. So here's something in there maybe two plus three x squared plus x to the seventh. So here's an element in z, z5 bracket x. So I've simply written down expressions where the coefficients in front of each of the x to the power terms comes from z sub five. Well just as we're used to I don't have to put a one in I'm going to interpret a one being there. If you need to of course you can interpret this as a zero x that you haven't written in and a zero x cube that you haven't written in and a zero x to the fourth that you haven't written in and you know etc. a zero x to the sixth that you haven't written in. In addition if you wanted to you could write it as zero x to the eighth as well. But just make sure that you've written it in a form that looks like this because I mean technically the way I've written it out you're supposed to list out things from a constant term and then an x to the first and x to the second all the way up to some number n. I don't care what that number is. So technically I've got all these terms that I would otherwise write in but of course that gets pretty cumbersome. I'll just leave those out. What do you want to call that element? Well I mean typically we call polynomials functions. We use the letters f or g or something like that so maybe this is a good name for this element or sometimes we might call it f of x just to indicate what the underlying variable happens to look like. Here's another one. How about g of x? g of x is 1 plus x plus 2x squared. That's also in z5x and let's go ahead and well if I add them in z5x. Alright well let's see the definition of addition is you just add like terms here. So the constant term turns out to be 3. Let's see the linear term here is 0 and the linear term here is 1 so I just get x. The quadratic term here is 3 and the quadratic term here is 2 and so when I add those together I get 5x squared which because the coefficients come from z5 the interpretation of this is inside that ring and of course 5 is the same as 0 inside z5. So it turns out when you add these two things together the quadratic terms disappear plus 0 plus let's see the last term that I'm going to get is x to the 7th. So adding them turns out to not be an issue and you'll notice when you add some of the terms might disappear. If I try to multiply them now f of x times g of x, alright well let's list them next to each other 2 plus 3x squared plus x to the 7th. I'm going to multiply that times 1 plus x plus 2x squared equals let's see what we get. Well folks we're just foiling these things out right so you got to do this times this. Now because I'm in a community of ring I sort of don't have to worry whether I'm doing that times that or that times that but that's alright. 2 times 1 is 2. Let's see are there any more terms that could contribute to the constant term? No because all these other terms have powers x on them. So there's the constant term how about the x term? Well if I have x times 2 there's the only x term all the other products are going to give me higher powers. How about an x squared term? Well here's an x squared term 2x squared times 2 so I've got 2 times 2 which is 4. I've then got 3 times 1 which is 3 and that'll be the x squared term so again I've done 2 times 2x squared and I've done 3x squared times 1 and I've chosen to write the coefficient just by factoring the x squared term out. So there's the x squared term. How about x cubed terms? Well let's see yeah I have an x squared times an x here so I'll get 3x cubed. How about an x to the fourth term? Yeah 3x squared times 2 is 6x squared. I'll put that in prens for now because we'll have to trade this in for something that looks like it's an element in the coefficient ring inside z5. Let's see any x to the fifth terms? Nope because the only way I'd get an x to the fifth term is by multiplying one of these by let's see I'd have to multiply this by an x to the fifth and I don't have one. I'd have to multiply this by an x to the fourth and I don't have one. I'd have to multiply this by an x cubed and I don't have one. How about x to the sixth terms? No good same reason. How about x to the seventh terms? Yeah I got one of those so x to the seventh. x to the eighth terms yeah I got one of those and how about x to the ninth terms yeah I got one of those x to the seventh times 2x squared is 2x to of the ninth. Now let's just do the quick simplification. Four plus three is seven but inside z five that's the same as two. Six of course inside z five is the same as one. So the product of these two polynomials you just knock it out. That's what you get. Okay so there's really nothing different here other than you have to keep track of what the underlying coefficients are doing. Specifically you have to keep track of what the ring r is that's being used as the coefficients and simply do all of the ring operations with that system in mind. Alright. Questions, comments? Yeah question Tracy. So let's see. So is the question how many elements are in this ring? The yeah so let's write that down. Exactly right. So the proposition is for any ring r, I don't care if r is finite or infinite, r, the ring r bracket x always contains infinitely many elements, infinitely elements. And there's lots of ways to prove it. Let me just convince you that inside there there's always infinitely many different things regardless of what the underlying ring is. Proof here's infinitely many things in there. Well I'm gonna assume okay with unity sort of a non assumption. So here is something that's a symbol of the correct form. It's the situation where a sub zero is zero and n equals zero. So that's perfectly legit. Here's another one. Technically I've sort of left the one out but that's alright. It's the situation where a sub zero is zero, a sub one is one and n is one. You stop there. There's another legit symbol. Another legit symbol. Now maybe Tracey it's a little clearer if I point back to this example here. You'll notice folks that the coefficients are taken from z sub five. So if you see a five you get to replace it with a zero. If you see a six you get to replace it with one. But that's only happening in the coefficients. It's certainly not happening in the exponents. The fact that this is x to the seventh is x to the seventh. Just because my coefficients are coming from z five you don't treat this like x squared or anything like that. So be careful there's a significant difference between what's going on in the exponents and what's going on in the coefficients. So even though on the surface you look at this and say oh there's finally many things in here. Yeah that's fine but the powers of x sort of propagate as big as they want. So here is I mean here's the sort of question that we need to address early on which is if somebody hands you a ring that you know something about. Like maybe somebody hands you a ring that's a field or that is commutative or that doesn't have zero divisors or is an integral of a main or has unity. Whatever it is whatever properties we can talk about a ring are we can then ask the question does that same property carry over to the polynomial ring r bracket x. Well one thing we've already noted let's say properties of r do properties of r carry over to properties of r bracket x. And the answer in general is definitely no. But this is sort of two questions in one. One is if I hand you a property does that same property hold in r bracket x. The question too is if I hand you a property for r is there some property that holds in r bracket x based on the property of the appropriate coefficients. Let me give you a good example. Example if r has unity. I hate that notation I've mentioned that a few times then r bracket x has unity. Yeah we already talked about tonight. You know if you start with something that behaves like the number one inside r then you can view that thing as in effect a constant polynomial and it will behave exactly the way it's supposed to. So here's one very minimalist remark but it has the right flavor. If you can tell me something about r then in fact you can tell me that that same thing happens in r bracket x. Here's another example. Example if r is commutative then so is r bracket x is also commutative. Because when you do in the multiplication of two polynomials if in each of the situations that you're multiplying the corresponding coefficients together you can switch the order then basically you just take whatever the product is switch all of the individual pieces in each of the coefficients and you realize that's the same as the product of two things having been taken as polynomials in r bracket x. You know I'm not going to write the technical proofs of these things because they're I mean they're they're somewhat tedious and distasteful. I'm just trying to give you a feel as to the sorts of questions that we can ask here. Similarly if r is not commutative then r bracket x is not commutative. If there's two things inside r that you can't change the order on then you can sort of build things inside r bracket x that you can't change the order of multiplication on. Here is a much bigger example example if I'm going to list that as a proposition. So example three will be phrased as a proposition. Example three if r I'll phrase it as an integral domain is an integral domain into a integral domain and so is r bracket x. I hesitate just for a minute there I could have phrased this if r has zero divisors then r bracket x has zero divisors and I can phrase this as if r doesn't have zero divisors then our bracket x doesn't have zero divisors so the point is this if you give me coefficients and you can't multiply two non-zero things to get zero in the coefficients then it's the case that you can't multiply two non-zero polynomials and get zero and the reason is reason all right well let's see I'm going to avoid defining as many terms as I can tonight but eventually I have to get to this description I'm going to tell you in the ring of coefficients in the collection of things that you're multiplying together is the coefficient I'm going to tell you that it's impossible to find two non-zero things so that when you multiply them together you get zero and that's the given information that's what it means say r is an integral domain technically it means a little bit more it means r is commutative and has unity okay that's fine but we've just noted that if r has unity then our bracket x does and we've noted that if r is commutative then our bracket x is so okay so those two pieces of being an integral domain are already satisfied in our bracket x now we just have to show that our bracket x has no zero divisors so I have to convince you that if I take two non-zero polynomials and I multiply them together that I get something not zero you scratch your hand thinking I don't know how to multiply two non-zero polynomials together ever and get something zero so why is this an issue I'll show you an example of that in a minute but here's the point if I multiply two polynomials together notice what happens in these highest terms if I'm multiplying stuff corresponding to lower powers and then combining it's possible that I might have two similar power terms that add together to somehow contribute to the coefficient on one of these but the only way that you can get something contributing to the coefficient on the highest power term and what is the highest power term it's going to be x to whatever the highest power term is here plus whatever that number is here just like you used to so the degree of the product is going to be nine the point is that the coefficient in front of the highest power term in the product the coefficient here is simply the product of the highest power coefficient here and the highest power coefficient here that's what appears here and the point is to even talk about the highest power coefficient means that the coefficient that you've got sitting in front of here is not zero because if this was zero then it wouldn't be degree seven it'd be degree two similarly the highest power term in the other polynomial has a coefficient in front of it that's not zero and the point is if the underlying ring is assumed to be an integral domain it means if you take two non-zero elements and you multiply them together you get something not zero so the coefficient in front of the highest power term here is going to be not zero because of the hypothesis in our and if this thing is not zero then the polynomial is not the zero polynomial and that's all we need to show so the reason here is this let let f of x have degree oh let's call it n one where degree means what it means tell me the highest power of x that appears and let g of x have degree n sub two and I'm going to assume that I'm going to start with two non-zero polynomials now technically folks it's possible for a polynomial to have degree zero that just means it's a constant term so degree zero is fine degree zero is fine but the point is the product f of x times g of x has as the coefficient coefficient on the x to the n one plus n two term the product of the leading coefficients of each of the individual polynomial coefficients to original one f of x and g of x and the point is these leading coefficients are assumed to be non-zero that's what allows me to write down the statement that the first one has degree n one and the second one has degree n two in particular so since ours in our domain and integral domain we get this coefficient is not zero it's non-zero and that's all we need so the product is not so that's all we need to do to show something is an integral domain specifically here to show our bracket x is an integral domain all we have to do is show that if we take two non-zero elements in there and multiply them together that we get something non-zero and that's exactly what we've done here in fact we get a non-zero expression and I can actually identify at least one term that turns out to be non-zero x to the n one plus n two term where you simply add the two degrees of the polynomials that you're multiplying together all right all right now if you say well an integral domain really is a commutative ring with unity that blah blah blah you're right but remember these first two examples say as long as you start with a ring with unity that is commutative and that's part of the definition of our being an integral domain and in fact our bracket x has both those properties as well so it's sort of interesting it says as long as you start with a commutative ring with unity that has no zero divisors then you don't impose any new zero divisors inside the collection of polynomials now I'm going to offer just for contrast but we're not going to look at we won't look at any of these examples with much interest if at all for the remainder of the semester contrast that with if I start with something that's not an integral domain like z six and I look at polynomials over what might happen well here's a polynomial half of x equals 2x to the seventh and here's another one g of x equals 3x to the ninth what happens if I multiply half of x times g of x what do I get I get six times x to the sixteenth but six inside z six is zero see it's possible to get what seemed to be honest to goodness polynomials multiplying together to give zero if somehow the underlying system has zero divisors to begin with and so we're going to be interested in avoiding this sort of situation the reason this could wind up being zero is look I took the two highest power terms of course the highest power terms are the entire polynomial haven't added anything down here but that's fine this is a polynomial of degree seven is a polynomial of degree nine and I multiplied them together and I got a polynomial of degree well I got zero questions what we're about to do is look more formally at the notion of the degree of two polynomials and then we'll write down a result that's true for integral domains and that's this so more formally formally we need to define what the degree of a polynomial is the degree of an element of our bracket x is one of two things is either well it's two possibilities it's either the exponent or the maximum exponent appearing in the polynomial in the polynomial I say that's what it is right polynomial that only happens in case the polynomial that you've handed me is not the zero polynomial polynomial that you start with is not the zero polynomial so somebody hands you a polynomial and ask you what's the degree of that polynomial just look at the thing pick out the highest power of x that appears and say the polynomial has degree blah but now the question is all right if somebody hands you the polynomial that is zero not degree zero but zero question what number do we assign to the degree of that polynomial the answer is we don't assign it any number or undefined if the polynomial is zero I've written it out in words because this seems to cause a little bit of grief for students let me write it out now more formally in other words i e the degree of a polynomial f of x and here's the notation for degree is either the highest or maximum maximum exponent if the polynomial is not zero and is undefined if f of x is the zero everything well why make such a big deal about the degree of the zero polynomial the answer is we're going to write down some results that just aren't true if you assign the zero polynomial degree zero in fact as we look at the results that will prove about the degree of various polynomials it turns out what some authors wind up doing is very naturally assigning the value negative infinity to the degree of the zero polynomial seems totally strange other authors for perfectly good reasons assign the zero polynomial degree negative one but whatever you do don't assign it the value zero because then none of our formulas are going to work out proposition proposition if r is an integral domain then f of x g of x are non-zero not degree non-zero are non-zero polynomials in other words i haven't just given you the big goose egg here take a polynomial i don't care if it's a constant polynomial the constant polynomial of two that's perfectly legit just don't take zero for either of these then the punch line is then the degree of the product is exactly what you'd expect it to be look if you multiply two polynomials together what's the degree you tell me what the highest power of x is here you tell me what the highest power of x is here you multiply those terms together that gives you the highest power of x over here when you multiply the highest powers together you're adding the exponent is the degree of f of x plus the degree of g of x plus here now that makes sense if you hand me a polynomial of degree three and a polynomial polynomial of degree two and a polynomial of degree seven and you multiply them together then you get a degree 9 polynomial again next to the ninth term. Okay, two quick comments. The first comment is we need this sort of hypothesis. Because if you start with a system that's not an integral domain, heck, it's possible to have polynomials that when you multiply them together don't give you a polynomial of degree 16 here. Heck, in fact, the thing can totally crap out and give you zero. Or as another example, if I would have just added something onto here, I would have written something like 1 plus 2x to the seventh and multiplied it times this. Alright, then the product wouldn't have been zero. But it certainly wouldn't have been a degree 16 polynomial because that degree 16 term is still going to wipe out. Alright, you get a degree 7, you get something less than it. But as long as the underlying system is an integral domain, technically all you need is dot zero divisors. You don't need to worry about commutativity or existence of unity or something like that, but those are always going to be true in the systems that we're going to look at. Then it turns out there's a formula for degree. Now notice, folks, if I would have allowed one or the other or both of these to be the zero polynomial, then this formula would not have been true. Because, hey, if one of these is zero, then the result is zero. And so, you know, if you start with a degree 6 polynomial and you multiply it by zero, you're not going to get a polynomial of degree 6 plus anything. So for what it's worth, here's at least somewhat reasonable or somewhat logical reason why it might not be unreasonable to define the degree of the zero polynomial to be negative infinity. If you define it to be negative infinity, at least this formula holds true. Because then, if you take something like a degree 6 times the zero polynomial, then this is the zero polynomial, so its degree is negative infinity. Over here you get 6 plus negative infinity, which is negative infinity, and your formula works out. But that, to me, seems a little bit artificial. So for us, just don't worry about the zero polynomial. That won't prove it. I just proved it for you in words. This is the property of polynomials that we're going to use. So what we've focused on for the last 15 minutes is this. In the very nice situation where you happen to start with an integral domain for the coefficients, we've shown that the resulting polynomial ring is also an integral domain. But now here's where things get a little bit different. So now what you might be expecting is, look, we showed that if r has unity, then so does r bracket x. We've shown that if r is commutative, then so is r bracket x. We've shown that if r is an integral domain, then so is r bracket x. Well, the sort of next candidate in line is what if r happens to be a field? And you may be expecting, well then, r bracket x is a field, and that is never true. So the punch line turns out to be, I don't care what system you start with for the coefficients, when you form the polynomial rings, you never get a field, not just sometimes. So the next thing that you might be expecting in line, the sort of next natural statement, what can you say about r bracket x based on what you know about r, fails miserably if you try starting with r equals a field. So, note, the situation of most interest, of most interest, interest, will be when r, the ring of coefficients, is a field. All right, well, of course, we proved that every field is an integral domain. So in particular, in particular, this last example, r is an integral domain when r is a field, that's nice. So we just proved, so in this case, this case, at least the polynomial ring is an integral domain. Just proved. So that's reassuring. If I start with elements coming from a field, at least I know that the polynomials don't contain any zero divisors. Questioning, though, is, though, is that the case that every polynomial, every non-zero polynomial has a multiplicative inverse, and the answer is absolutely not. But, and this is something we've actually stated already twice, proposition, I don't care what r is, r bracket x is never a field. Not a field. I don't care what r is. Make r good, make r bad, make r anything, you never get a field. And the proof of this is, I can write down a non-zero element, let's call it this. I mean, I can write down a lot of examples. I can write down x to the fourth if you want. Here is a non-zero element of r bracket x. Does it have a multiplicative inverse? Can you find something in r bracket x? Can you find a polynomial so that when you multiply that polynomial times x that you get the multiplicative identity. Of course, the answer is no. Has no multiplicative inverse. Why? Because the unity of r bracket x is one. You might say, well, I thought that was the unity of r. Yeah, but it's also the unity of r bracket x viewed as the polynomial of degree zero. And so the point is you can't solve i.e. x times g of x is never equal to one for every g of x in r bracket x. Now, just to avoid the silly situations, but it turns out the silly situations can't come up anyway. It's just I'm not worried about focusing on those. Even if you start with r as a field, alright, what are you trying to do? You're trying to cook up a polynomial so that when you multiply it times x, you get one. And folks, that can't happen. Look, because if I take x, if you're working over a field, it's an integral domain, there's no zero divisors. The degree of a product is always the sum of the two degrees. So as long as you've got something of degree bigger than zero, I've just happened to have chosen x. I could have chosen a whole lot of other things. It's impossible to multiply it and get a result that has lower degree than the thing you started with. This has degree one, this has degree zero, and there's x to the zero. And that's when it's impossible to cook up a polynomial for which that happens because there are no polynomials of negative degree. And if you're asking, well, why don't you just invent some polynomials of negative degree, you can. That's a completely different beast, though. For us, polynomial means constant plus something times x plus something times x squared, etc. So here's the punch line. These things are bracket x that we're going to focus on for the remainder of the semester are typically pretty nice because if you start with the coefficients being from an integral domain, at least our bracket x is an integral domain. In particular, the case that'll be of most interest to us, if you start with a field for the coefficients, x is an integral domain, but it's never a field. There's always things in there that don't have multiplicative inverse. Now, we've been looking at, for the last three weeks, an important example of a commutative ring with unity that is an integral domain that isn't a field. That's the ring of integers. The integer is an integral domain. If you multiply two non-zero integers together, you certainly get not zero. The coefficients in the integers don't have multiplicative inverses. The only two that do is one and minus one. The intuition that I'm going to have you start developing is that these things, our bracket x, at least when you start with r being a field, are integral domains that aren't fields. And it'll turn out that a lot of the things that you know about properties of the integers are also going to be true about properties of polynomials. The sort of analogy that I want to start playing up in here, properties of the integers and properties of the ring of polynomials with coefficients taken from a field turn out to be extremely similar in many situations. Not identical. We can't say everything that's true in z is also true in r bracket x, but many things are and we will get to those. Actually, we'll get to a couple of examples tonight. Okay. Questions, comments? I think we'll do this now. So, yeah, because that'll be a good lead into the following comment. So, the intuition, intuition is this, that if r is a field, is a field, then the ring, r bracket x behaves a lot like z. At least at this stage, what we can say is, for example, both are integral domains, integral domains and both are not fields. So, there's two things that these particular rings have in common. I'm going to give you a third one. For those of you that saw the number theory course, this will be familiar. For those of you that maybe haven't, the idea is completely familiar. It's just third grade arithmetic. If I hand you a couple of whole numbers, integers, it makes sense to divide one into the other. It makes sense to divide 7 into 23 or something like that. Divide 7 into 23. What that's supposed to mean is that you sort of take as many 7s out of 23 as you can and then tell me what the remainder is and the remainder is understood to be something that, well, is bigger than or equal to 0 and somehow less than the thing that you're dividing by. I mean, I can divide 7 into 23 and get 1 with a remainder of 16. I mean, 7 goes into 23 in time with 16 left over. That's a technically correct statement but what you're thinking, well, that's not really what I'm trying to do. What I'm trying to do is sort of, you know, take more out until I can't take a full one out again. That process, for those of you that saw the number theory course, it's called the division algorithm. So the division algorithm algorithm in the integers, this is the third grade long division. Where do you do that? Fourth grade maybe? I don't know. It's the moral equivalent of third grade long division. You know, give me any, ooh, gotta be a little bit careful. You can't ask me to divide 0 into something because division by 0 is meaningless but give me something that's not 0. Then I can ask you to write it in a certain way and the way that plays out, folks, is as follows for any, yeah, for any for any two integers, a and b in the integers but don't pick the thing that you're going to divide by to be 0. There is a unique way way to write. a is q times b plus a remainder where the thing that you've left over in absolute value is bigger than or equal to 0 and r is less than the absolute value of the thing that you're dividing by. You can always take a non-zero integer and divide it into any other integer and what we ask that to mean is you can write the second integer or some multiple of the first one but this remainder has the property that its value is always less than the absolute value of the thing that you're dividing by and now if you haven't seen the division algorithm in this form, just think about it for a minute. The interesting case, of course, is when you're dividing by something bigger than 0, that's typical and all it says is, hey, you just keep taking as many b's out of a's as you can how many, you take that many out until the thing that you've left over well, you can certainly allow a zero remainder that's perfectly legit that you've left over is less than the thing that you've been dividing by. That's the division algorithm in Z. What we're about to write down is its corresponding statement in r bracket x. There is a division algorithm for r bracket x. It says if you hand me two polynomials but make sure the second one isn't just the zero polynomial that it makes sense to divide this polynomial into the other one. So what we're doing is long division of polynomials and it turns out that a result quite similar to this one holds there. That there's only one way to do it if you specify that the thing that you want me to leave out here, the thing that you want to act as the remainder is intuitively smaller than the thing that you've divided by. The difference is going to be the measure for how big the original divisor was and the corresponding remainder was simply its size. The quantity that plays the corresponding role when we're dividing polynomials is the degree of the polynomial. If I hand you two polynomials and ask which is bigger that question doesn't make any sense. I can ask which is the bigger degree polynomial. That makes sense. So what we use for some sort of standard or basis for comparison of polynomials isn't the polynomial itself but the degree of the polynomial. So here it is. This is the corresponding division algorithm in r bracket x and this is theorem what number theorem 23.1 23.1 the text says this let r be a field and that's key here folks. You can't get away with this division algorithm in r bracket x unless you're starting with coefficients from a field. And then the punchline is let f of x and g of x be elements of r bracket x where the degree of g of x I think we can get away with at least zero but let g of x be a polynomial that actually has some guts to it. That's got some x's in it. Don't just tell me to divide by two. Tell me to divide by x squared plus three or something like that. Then here's the punchline. Then there exist unique polynomials. We're going to call one of them suspicious and the other one r of x that's suspicious with the property that two things are true first that you can write the original polynomial f of x as q of x times g of x plus r of x in other words you divide f into g you get something that we call q of x q for quotient and you get some remainder and secondly where the degree of r bracket x I'm sorry of r of x is less than the degree of g of x or r of x is zero. Remember we've got to sort of tease out as a special case the situation where this remainder is zero which certainly can happen if you divide one polynomial into another it might go in I don't like to use the word evenly but there might not be a remainder in other words the remainder is technically zero might say well then isn't the degree of the remainder less than the degree of the thing you've started with well no if this thing is zero then remember we haven't defined this expression degree of r of x makes no sense so I have to separate out the remainder being zero as simply a separate case here's the division algorithm in a higher bracket x so quiz four happens on Monday and quiz four is this just tell me this I'm going to ask you to come in Monday right at the beginning of class and I'll hand you a sheet of paper that says state theorem 23.1 state the division algorithm r bracket x where r is a field and you can just either spit out what I've just written on the board if you want or if you'd like to memorize the books version verbatim they might state things minimally differently but it's exactly the same idea things that you need to be careful of are first of all the division algorithm for r bracket x requires that the underlying coefficients come from a field so that's here let me point out another couple of things that students often forget secondly the division algorithm only holds when you're dividing by polynomials with some guts when you're dividing by something that looks like g of x equals x squared or something like that the division algorithm isn't true if you start with something from the original field in other words a degree zero polynomial let's see most students do fine here make sure you put this word in there exists unique polynomials in other words if you found two polynomials q of x and r of x that make both of these conditions true then there's only one way to do that you and your friend will have come up with the same answer and finally most people write down this piece but you have to remember to include the possibility that the remainder polynomial might be the zero polynomial not a degree zero but the zero polynomial let's do an example this will just be a refresher on long division of polynomials which some of you haven't done for a while it's alright alright this one might be interesting because remember the at least the statement I'm not going to prove this for you it's essentially the long division of polynomials works but the statement is this holds regardless of what field you start with you're used to doing it in r bracket x where r is the reals or maybe complexes but we are a little more sophisticated than that now so I'll let f of x be x to the fifth plus one and g of x be x squared plus one in this polynomial ring z to x now here's the phrase divide g into f as in the division divide g into f of x so that request means complete the division algorithm by writing f of x is something times g of x plus the remainder I mean you do the long division as you're used to it's just you have to keep in mind where the coefficients are coming from so what do you do you write x squared plus one and then you're going to divide it into well x to the fifth plus one so you typically put some placeholders here plus zero plus zero x plus one this is standard when you're doing long division of polynomial you divide highest coefficient into highest coefficient x squared into x to the fifth goes x cubed times and now you do the multiplication x cubed times x squared is x to the fifth x cubed times one is x cubed and then we do you subtract so all this is golden so far so the x to the fifth minus x to the fifth cancels out and you're all none your heads that's good here's where things get a little bit different x cubed now I'm sorry that's zero x cubed minus x cubed so you'd want to write down that but remember where the coefficients are coming from so the coefficient here is minus one but inside the ring z2 minus one is one so that's the same as that hmm minus one equals one in z2 we'll make some sort of simplification based on the coefficients that we're working with alright that's fine so now I have an x cubed here so what's the next step divide x squared into x cubed and what are we going to get an x term right because x squared into x cubed gives x so now I have x times x squared gives x cubed and then I have x times one gives x and subtract again x cubed minus x cubed is zero I get x minus x gives well minus x which is the same as plus x because minus one is plus one alright now what let's see oh now I bring down the hmm and now I'm done why because the degree of the thing that I'm dividing by is less than the degree of the thing that I'm trying to divide into so here's the punch line so the original polynomial which was x to the fifth plus one is q of x that's what appears on the top x cubed plus x times the thing that we're dividing by g of x x squared plus one plus the remainder and the remainder here is x plus one the key being that the remainder the degree r of x is one and that's less than the degree of g of x which in this case happens to be two a quick suggestion especially when you're working over z to x just multiply these things out and make sure you get back to your original one and on the surface that looks a little bit weird I mean it looks like you're adding everything and somehow all you're going to be left with is x to the fifth plus one so it's a little bit counter to but yeah you are let's see if we multiply this out we get x to the fifth plus x cubed plus x cubed plus x plus so when I multiply that out I get this plus x plus one and what's the beauty of working in z to x look if I have x cubed plus x cubed zero because if you want to think of it as two x cubed that's fine or because we proved last time that if we have the coefficient ring being characteristic d then the polynomials also have characteristic d here I started with a coefficient ring of characteristic two if I add two things together then I get zero here I've added the same thing together you get zero so this is just x to the fifth plus one and it checks questions there comments let me make a quick observation example if you're working in something a little bit more complicated than z to but still a field let's look at a couple of polynomials maybe in z seven if I hand you something like three x squared plus one that's a g of x and I hand you five x cubed plus two x squared plus four that's f of x and I want this to happen divide g of x into f of x but I want all these being viewed inside z seven x in other words I want you to interpret the coefficients as living inside z seven then we set things up exactly the same way we do a long division so you'll be dividing three x squared plus one into five x cubed plus two x squared and I'll write it out as zero x plus four but now the game gets a little bit more interesting I mean if you're going to do this into this in general what you're figuring to put here is what five thirds x but I don't know what five thirds means so here's what you have to do you're interested in somehow putting a term here so that when you multiply this times this that you get five x cubed well I know what power of x to use because it's going to be x times x squared to give me an x cubed term what I need to figure out though is what the coefficient is so I need to figure out some way of writing down a coefficient here so that when I multiply it by three I get five and I'm working in the system z sub seven there's a couple ways of doing that one is just list everything out and just try out possibilities three times one is three no good, three times two is six no good, three times three is nine which is two, no good three times four is twelve which is five so the idea would be put a four there that was three times four okay not as systematic as you would hope here's a better way to go about doing it think about what this means folks five thirds it's five times three inverse and at least in this notation this looks like the notation that we use in rings as long as this thing happens to be a unit so what you really want to do is take the thing that you're dividing by find its inverse inside the given ring so how the heck do I know this thing has an inverse answer because the thing that we're starting with is a field find what its inverse is then simply multiply by whatever this coefficient is so for example another way to do it is what is three inverse in z seven well that we well it turns out there's some algorithmic ways to do it but for us all we're interested in doing is finding it we know it exists because seven is prime and we've proved that z sub p is a field whenever p is prime so we know that we've got to be able to find it three times one is three three times two is six not one three times three is nine not one three times four is not one three times five is 15 which is one so three inverse is five so just by sheer coincidence three inverse happens to be five so this is 25 and 25 in z seven is four so here's another way of viewing how I got the four I like playing up this example because it shows that if you're going to hope to have a chance of doing the division algorithm the things that you've got here and here have to come from a field the reason being the coefficient that you're going to put on the highest power of x in this thing that gets kicked out as the quotient as q has to have the property that this times this is that and the only systems you're guaranteed for that to happen in are underlying systems for fields question John yeah the first time I got four I just went through the list I need something so that when I multiply four times three I get five four times one no I'm sorry three times one is three three times two is six not five three times three is nine which in z seven is two but it's not five three times four is twelve which in z seven happens to be five and that's how we get the answer alright so I'll let you sort of finish this up etc etc you'd multiply this out you get what five x cube because we've rigged it that way and you get four x times three is plus four x and then you yeah plus four x right column here plus four x and then you go ahead and etc questions comments let's see if I need to give you any more info about this that's a big place to quit oh yeah let me give you one last remark because this will help you on one of the homework problems that you were given folks if I hand you one of these I'm going to call it f bracket x where capital F stands for a field now so these rings are always infinite there's always infinitely many different polynomials regardless of how big the coefficients are even if the coefficients only have two elements zero and one you can still produce infinitely many polynomials here's a question that you can ask tell me all of the polynomials of degree three well if I hand you the field as the real numbers and you ask to list out all the polynomials having degree three forget it x cubed to x cubed, 3x cubed, 4x cubed, 5x cubed and I haven't even gotten to x cubed plus x squared so the question find all of the or how many are there in general is not a legit question but if you're working in a system like z to x it makes perfect sense to ask the question tell me all the polynomials of degree 3. Well, yeah, here are some of them. All polynomials or elements of this ring of given degree make sense if f is finite. So, for example, all polynomials in z2x of degree 3, well, here's one, here's another one, here's another one, here's another one, here's another one, etc. At least in this case, because the coefficients are finite, you have a chance of being able to list out all polynomials of a given degree and that's what you'll be asked to do in one of the homework problems that I handed you on Monday. So if you're going to look at these over the weekend, then that's a good place to look for information about. But don't, you know, don't go overboard here. The question list out all of or tell me how many there are, polynomials of a given degree is only a reasonable question if the coefficient field happens to be. All right, sorry to have run over a little bit. So if you've got the homework due tonight, I'll let you leave it on the table here. I will have that graded by Monday. Let's see, I'm in tomorrow from 11.30 till noon if you want to come by and if I don't see before then, have a good weekend and I'll see you on Monday. Lindsay, questions? Well, if you have negative four and Z7, sure negative four, just, you know, add seven to it, it's positive three. Yeah? Or if you have negative two, it's the same as positive five, you're just, you're working.