 OK, so then I start. Good afternoon. Well, we already saw each other this morning, but anyway. So today, we essentially have the last lecture about rings, maybe not completely finished, but almost. And so we start out talking about Euclidean rings and principal ideal domains. There's some sort of marks on this, principal ideal domains. So last time we had seen the division with rest for polynomials with coefficients in the field. And everybody knows the division with rest for integers. So Euclidean ring will be an integral domain where you have division with rest. So an integral domain division with rest. So if we have an element a and b in this ring r, then we can write a is qb plus the rest r. And in a suitable sense, the rest should be smaller than b, than the thing we divide by. You like if you divide a number by another, the rest is smaller than the number you divide by. And so we have to find a way to measure this. And this is done by a certain function from r to z bigger than or equal to 0. So let me define it. So an integral domain r is called Euclidean ring. If we have the following, so if there's a function which is what measures how big the rest is, d from r without 0 to, say, the negative integers. Such that the first thing is that we can write whenever I'm given elements a, b, and r, and can make division of a by b with rest. So I can write, so then there exists an m and q in r, such that a is equal to qb plus r. And r should be, as I said, smaller than b. In a suitable sense, so that either r is equal to 0, or d of r, d of small r, is smaller than d of b. The second condition, which is somehow less important, says that this d is compatible with the multiplication. So the second condition is that if b is an element in r, which is not 0, and a is an element in r. So maybe like this. Then we have that d of a, b is bigger than d of a, for all a in r without 0. That's the same statement, but it's different. So whenever you multiply, this d gets bigger. So if that's the thing which measures how big our element in r is, somehow the elements get bigger whenever I multiply by a non-zero element. Yeah, this must be true, because this would otherwise be obviously wrong, is not a unit. Thank you. Because obviously otherwise, for instance, if b is equal to 1, then this can never be fulfilled, as you can see. Yeah, I copied it wrong, anyways. Clearly, it has to be that. So we have easy examples. So for instance, if we take z, and we take d of a number n, is for instance equal to the absolute value of n. This is a Euclidean ring. And up to the convention of how you define, you want the sign of the rest. This gives you the usual division with rest. And also, we had seen that if k is a field, and so then kx with d of a polynomial f is equal to the degree of f is a Euclidean ring. So this was essentially what we did the last time. Now, when we had the division with rest was precisely so that the degree of the rest is smaller than the degree of what we divide by. And the second condition is also obvious, because a polynomial is a unit if and only if its degree is zero. And the degree of the product is the sum of the degrees. And I want at least one other that we do not know. We look at z of E. So this set of all E, so n plus m, i, complex numbers such that n and m are integers. So these are just, if you look at the complex plane, these are just the points which have coordinates integers. So this is obviously a subring of the complex numbers. i is the usual i, square root of minus one. Then, and so we want to say that these are called the Gaussian integers. And I claim it's a Euclidean ring with, well if I take d of n plus i. Which one do I want it? i m equal to n squared plus m squared. So we have to see why this is the case. Well, so this, for complex numbers, this is just the restriction to the Gaussian integers of the square of the complex absolute value of the complex number. So we can extend d to the whole of c by, obviously d of a plus i b, where a and b are real numbers, is equal to a squared plus b squared. Which would be the square of the complex absolute value. As you have no doubt learned in the complex analysis. So in particular, we have that d of a plus i b is different from zero if a plus i b is not the zero in the complex numbers. And it is well known and easy to check for say z and w in c. We have that d of z times w is equal to d of z times d of w. You can immediately check this anyway from the formula if you remember the product of complex numbers and what you hopefully do. And so we have this, you can easily check this. So now we want to make our vision with rest. So let's say now z and w the elements in z of i. So this is whatever n1 plus m1, i, n2 plus m2, i, anyway it doesn't matter. So we take two elements. So we can certainly, we want to divide z by w with rest. So we first just divide it in the complex numbers. So let z divide by w which I write as a plus b i with a quotient in c. So obviously w is different from zero. So what does it, and we then just want to find as the kind of quotient in z i something which lies close enough to this. So we choose say nm in z such that say a minus n. So the absolute value of this difference is smaller equal to 1 half. And the absolute value of b minus m is smaller equal to 1 half. I can certainly do that. Just shows the nearest integer in the corresponding direction. And we put q equal to n plus im. So we have our element in z of i and this is supposed to be the quotient. So can first compute this difference between if I take z minus w minus m plus in. What is it? Well according to the definition this is a minus n squared plus b minus m squared n. Yes, and so this is one half squared one quarter one quarter this is smaller equal to one. No no I mean I think in the notes we have the other notation but I think this is consistent with itself. Okay so we have this and so we can so we put the rest as z minus n plus i here I have added wrong no n plus im times w so this is supposed to be the rest. So let's see where that works. So this is the rest so we can write now we can write z obviously is equal to we just put on the other we route n plus im w plus r. So this is our division with rest. And what we have to see that d of r is smaller than d of w. So d of r is equal well we can so if you just multiply this here is just this divided by this. So this is d of z divided by w minus n plus im times d of w because this is just this r it's just this multiplied by w. And we know that this is this thing is actually one half so this is certainly smaller than d of w and we know that this is smaller equal to one half. Okay and so we find this and the second statement is is obvious so we we find that this is indeed a Euclidean ring. Okay so now I wanted to so one of the things that Euclidean rings are is principal ideal domains so that means every ideal is principal so that means every ideal is generated by just one element and so in particular the for instance every ideal in z is just the multiples of one integer. So definition ring so I think I want an integral domain it's called the principal ideal domain and as this is a bit long one usually always just Pid if every ideal i in r is principal so that means by definition as we have already seen that i is equal to a for some which is the same as a times r for some a in i. Okay and the remark here is I mean I call the theorem although it's not really that every Euclidean ring is a principal ideal domain well it's actually quite simple so we just have to use this division with rest to do it. So let's take a principal ideal domain a Euclidean ring and we chose ourselves an ideal so now we want to see that this ideal is principal so if i is just a zero ideal then it is obviously principal because an i is equal to the ideal generated by zero so that's principal so thus we can assume assume that i is different from the zero ideal and let a be an element in i which is different from zero but we choose a particular one we choose we have this wonderful d we choose an element which has the minimal d of a minimal for all a in i whatever so we choose an element in i for which this d of a is the smallest possible one we want to claim that this generates the ideal well so so let me be an element in i we have to show that b is a multiple of a so we can do division with rest so then b is equal to q a plus r with q r and the small r is equal to zero in which case we are done because b is we have found that b lies in or d of b d of r is smaller than d of a but we know you know but if you look at it but by this equation we have r is equal to b minus q a these are two elements of i so this is an element in i and a under this assumption it would be non-zero this so this d of r smaller than d of a is a contradiction to the choice of a so this is a contradiction to the choice a because d of a was supposed to be the smallest d of something for any element in i which is different from zero so it follows r is equal to zero so that means any element b and a is a multiple any element b and i is a multiple of a so that's b in a if I take an element b and i it's an element i so i is equal to a okay so in particular we find that these euclidean rings that we have just introduced are pids so in particular we have that z z of i and k of x for k field are principle ideal domains I don't know whether it has an apostrophe or not okay okay so we can also talk about principle about great and greatest common divisors in principle ideal domains so I had defined greatest common divisors in in integral domains and I said it's not always clear they exist I only give a definition but it doesn't mean they have to exist but in principle ideal domains they do exist and actually quite simple so this is a proposition so at r p a p a d and let I take some elements a 1 2 a r some elements in r then there is a greatest common divisor of a 1 2 a r exists so maybe I call it d exist and is of the form of the form can write it as a linear combination of these d equal to a 1 x 1 plus a r x r with x 1 to x r is our elements in r and actually quite I mean it's the obvious thing these a 1 2 a r generate an ideal this ideal is a principle ideal it will be generated by one element which I call d and this element d will be the greatest common divisor or one candidate for the gates comb divisor so so proof so we choose an element d in r such that if I take the ideal a 1 2 a r generated by these to all linear combinations of these that this is equal to the ideal generated by d by our I mean we know that this is every ideal is a principle ideal so also this one so we can write like this and the claim d d the claim is obviously that claim d is a greatest common divisor this is basically obvious just look check the definitions so by definition we have obviously that each a i lies in the ideal generated by d and but what does it mean that it lies in the ideal generated by d means besides that's divisible by d whatever okay that's just what it means to lie in the ideal and so it is a common divisor is a common divisor of a 1 to a r and to be the greatest common divisor it has to be it has to be that every other common divisor divides it so let e and r be a common divisor so satisfy e divides a i for all i from one to r then I have to see that e divides d but if it divides all of this then e divide if the e divides a i it divides say x i times a i no and then if it's divides all x i times a i it divides the sum of all of them so e divides a 1 x 1 plus a r x because it divides all the a i and therefore which is after all d and so that's fine so it's the greatest common divisor so for instance in that we have that we take the ideal generated by four and six this is the same as the ideal generated by two and two is the greatest common divisor of four and six okay so so that was a rather simple remarks about this is all wondering whether I missed something no it was all so now I want to start to talk about irreducibility of polynomials we will later you know we talk about field extensions so you have a field and you want to construct a bigger field which contains the original field and you do this somehow by looking at you know with the help of an irreducible polynomial and so we want to have a way to construct them and to check that polynomials are irreducible so first I talk about irreducibility in general so definition so this I'm talking about irreducibility of polynomials and so first I define what irreducibility means in general so definition let's say r be an integral domain so an an element a and r would be called irreducible if it cannot be written in a non-trivial way as a product of two other elements so non-trivial so the trivial way would always be that you write it as a product of something with a unit so therefore an element say q in r which I assume it's not zero is called irreducible well say maybe first if q is not a unit because that would be a bit too stupid and if q is equal to a times b with a b elements in r then either a or b is a unit obviously we can always multiply with the inverse of a unit in order to get such a decomposition so that would be too stupid so for instance um say so maybe first I finish so it's it is called reducible if it's not irreducible so so if q is again is not a unit and not irreducible it is called reducible so this means in other words you know if an once doesn't want to say it in such a direct way obviously just in indirect way obviously just means that so q is reducible if it exists a b in r which are not units none of so without units such that q is equal to a times b so for instance in in z you find that the prime numbers are irreducible and in fact you can easily check that all the irreducible elements in z are plus minus p where p is a prime number so I want to give you a few more so okay so I can even write it again example so an element q in z is irreducible if and only if q is equal to plus minus p where p is a prime number that's almost by definition then if you have a field if k is a field it has no irreducible elements because after all all elements in k without zero are units and so no and zero also was excluded so there's no chance and then let's look so again in kx so we have that I claim that so ax plus b is polynomials with a an element in k which is not zero and b so k is always a field and b in any element in k is irreducible well this thing is not a unit it is not zero so it is irreducible if it cannot be written as a product of two elements which are not units and so if ax plus p is equal to f times g for f and g polynomials then we know that the in the product the degree is additive so the degree of ax plus b is equal to the degree of f plus the degree of g now the degree of ax plus b is evidently one so and the degree is at least zero so one is the sum of two non-negative numbers so at least one of the numbers must be zero so thus the degree of f is equal to zero or the degree of g is equal to zero and now as this is a non-zero element it follows that f and g are not zero so it means you know a constant polynomial is a unit in kx the constant polynomials the non-zero constant polynomials are units so it means that either f is a unit or g is a unit mainly when it's zero or g is a unit and so this means precisely that this thing is irreducible so I come back for one moment to this p id then we use this for the polynomials so if you have a principal ideal domain and you have an irreducible element then the ideal generated by it is a maximal ideal and therefore if you divide by this maximum by this ideal you get a field so proposition so let r be a p id principal ideal domain and p in r an irreducible element then the ideal generated by p is a maximal ideal in r and we know that being a maximal ideal is equivalent to the quotient by this idea being a field so the less second part is obvious and r mod p is a field so this is just to remind of it the statement is this one okay again this is quite simple but we will later use it so almost by definition so let's see so let's do the proof so we take our irreducible element and so we want to show that the ideal so let i in an ideal with say p contained in i so this contains means not necessarily it's contained so this ideal is a principal ideal so we can write i equal to a so that p is contained in the ideal generated by a means that p itself is contained in a so that i can write p is equal to a times b for some b in r so now there are two possibilities either this element p is a unit or it's not so either p is a unit we have seen that if i have a unit then an ideal generated by one element and the ideal generated by the same element multiplied by unit gives you the same ideal so if p is a unit then p is equal to a because if i just multiply you know this a by a unit it gives you the same ideal so that's one possibility or b is not a unit ah but this was actually not what i wanted to do let me see yeah yeah but i'm kind of i understand so i think i want to let's try again so well maybe you can repeat what you said ah yeah yeah sorry obviously i have to use my i have to use my assumption somewhere no i have assumed that p is a prime element obviously i'm not trying to prove that this is true for any element but i the assumption was that it's true if i have a an irreducible element no so yeah obviously i have to use my assumption so so um so p is a prime element i have written it as a product of two things so if either b is a unit or a is a unit because p is a prime element or b is not a unit then a is a unit yeah sometimes but you know then because p is prime is you know irreducible so then obviously if if a is a unit the ideal generated by a unit is the whole of the ring and so we have seen that if our prime if our irreducible element is contained in an ideal then either that ideal is equal to the ideal generated by p or it's the whole of the ring so this is precisely the definition of a maximal ideal amazing okay so we want to use this to construct new fields from old spy use of polynomials so basically if we are given an irreducible polynomial we get in this way a field so corollary let k be a field and f the polynomial in kx an irreducible polynomial then we know that after all the ideal generated by f is a maximal ideal and so then kx modulo the ideal generated by f is a field and I can view it as a field containing k as a subfield so which contains strictly speaking um a field isomorphic to what I just say k as a suffering so I will identify so because then if I so by what we said before if f is an irreducible polynomial then kx modulo f is a field and because this is a maximal ideal in kx and so the so we have the the constant polynomials so the if I take the map from the constant polynomials to this quotient this is an isomorphism onto the image so the map uh so so on so for say a and k we have the class of a no so the map the restriction kx for f to k which are the constant polynomials uh is an isomorphism onto the image so we know that this so restriction of kx pi so we have the canonical map to the quotient if I restrict it to k then the map is injective because the um it would have to be either so as k is a field and I restrict the map to k it would either have to be the zero map which is obviously isn't because the constant polynomial because k does not in line the ideal generated by f for an irreducible polynomial and so it is an injective this map from k to the image here is an injective homomorphism of fields and so therefore it is you know so it's an isomorphism onto its image which is a sub field of this and so later we will be concerned with the question we are given a field and we want to study larger fields which contain this given field and here we have kind of found a way to always construct such so if we are given ourselves an irreducible polynomial in our field then we get a larger field which contains the field where we started and for instance we will see later that we might want to find a field in which a polynomial has a zero and we will find that somehow surprisingly in this field this polynomial f will have a zero so somehow in a anyway but this will look at later so we come back to this later now as we have now found this so we this somehow is one of the reasons why we're interested in irreducible polynomials because they allow us to make fields so bigger fields from smaller ones and so now it would be useful to find some criterion for a polynomial to be irreducible and so we will do this only for the case that k is equal to q but we also have to do it then at the same time for k equal to z because that is what makes it work so we want to study the irreducibility over q and z so we have polynomials in qx and in zx and we want to find out about the irreducibility by comparing what happens in q and z and the first result about this is due to Gauss so we have first lemma and then so first of we have a lemma which is the first step towards this so we first make the definition so obviously for them for polynomials in z if you take an irreducible polynomial in z and you multiply it by a number by an integer which is not a unit it will not be irreducible anymore so because you can after all divide it by this number and this number was not unit so therefore we want to first look at polynomials which are primitive which means that their coefficients have no common divisor the polynomial equal to sum i equal to zero to n ai z x to the i is called primitive if the greatest common divisor so this is a polynomial in z of x so the ai are integers so if the greatest common divisor of the ai is one or so if the a zero to an are relatively prime which by definition just means that a greatest common divisor of the coefficients is one so we don't have any common factors for the coefficients in that case it's called primitive and the first lemma of Gauss it's actually quite simple it says that if you have two primitive polynomials then their product is also primitive seems kind of obvious but maybe it isn't so let f and g be two polynomials with coefficients in z which are primitive then f times g is also primitive so what we just try to see whether we can have a we assume it's not primitive and then we will quite easily bring this to a contradiction let's see assume f times g is not primitive well i can first but i need to do some notation so so maybe we write f equal to sum i equals whatever one zero to n ai x to the i g equal to sum here i equal zero to m whatever bi x to the i and so we assume that fg is not primitive so so then if it's not primitive then the the all the coefficients of fg have a common divisor so there's a in non a number which is not plus minus one which divides all the coefficients and so we can take this number to be a prime number there's a prime number which divides all the coefficients and of f times g no if there's some number which divides all the coefficients and this number is not a unit then i take a prime number which divides this number which divides all the coefficients and this will divide all the coefficients so then there is a prime number p which divides all coefficients of f times g but on the other hand it does not divide all coefficients of it does not divide all the ai and it does not divide all the bi and not all so so therefore we can take the smallest i for which it does not divide it so let i be minimal such that p does not divide ai and let j be minimal such that p does not divide bj so it takes the smallest such index of which it does not divide it i don't i know it doesn't divide all so there will be some it doesn't like it it can take the smallest one so so if i call maybe i call c of i plus j the coefficient of x to the i plus j in f times g so what can i say about it you know by the formula for the product we can express it in terms of the ai and the bj so you have that ci plus j is equal to ai times bj plus the sum over all k which are bigger than i ak bi plus j minus k and plus the sum over all k which are smaller than i ak bi plus j minus k no so just obviously this is just the sum over all albm such that l plus m is equal to i plus j by definition and so it's i can do it in this but now if i look at this here when k is bigger than i then this number here i plus j minus k is smaller than j so this index is smaller than j so that means p divides bi plus j minus k so it divides every sum and here when k is smaller than i then k obviously k is smaller than i and therefore p divides ak so we find that p divides this whole sum p divides both sums but it does not divide ai and it does not divide bj so it does not as it's a prime number it does not divide ai times bj it's not divide ai times bj so therefore it does not divide this one and you know so this contradicts our assumption no that that f times g was not primitive and therefore we could choose such a prime number so this is a contradiction and so we find that indeed f times g is primitive so this is then mostly used to prove a theorem which is also called gauss lemma so it's a lemma for gauss lemma which says that if i have a primitive polynomial with coefficients in z then it's irreducible in zx if and only if it is irreducible in qx so if i can write it as a product of two polynomials in zx i can also write as a polynomial of two polynomials in qx and vice versa okay and so let's see so let f in kx be a non-constant known zx be a non-constant primitive polynomial then f is irreducible in zx if and only if f is irreducible in qx so if i cannot write it in a non-trivial way as a product of polynomials with integer coefficients i also cannot write it in non-trivial way as a product of polynomials with rational coefficients and that doesn't seem to be so obvious so there i mean it's equivalent so there are two directions first we do the easy direction so if f is irreducible in zx so it's by contribution in zx then you can write f is equal to g times h where g and h are both non-units in zx yes so if the degree of g is equal to zero well that means g is constant then it means that g is a non-unit in z and and we have that this number g is a common factor of all coefficients of f equal to g times h because after all g times h is just obtained by multiplying all the coefficients of h with this number g and so f is not primitive so it follows it's bigger than zero and the same same argument applies if the degree of h is equal to zero no it's a symmetric so so thus we also have the degree of h is bigger than zero so we have written f as a product of two polynomials of positive degree so it means it is this also describes it this polynomial has been reducible in qx so we have f h bigger than zero that means f is reducible in qx because this is a way of writing it as a product of two units two non-units in qx no if I have a polynomial okay so this was the easy direction because you know we just have to take the same now we have to use the going the other way so this is the more surprising part you have to see that if you cannot write it as a product of two polynomials with z coefficient you cannot write us as such a product with r coefficients what yeah yeah yeah certainly yeah yeah no I finished I mean this is the argument I but what is the yeah well that's what I proved here no this is the direction I proved um no no no but the the units are different no so if you cannot so ah let me see whether your argument makes sense um well I mean in some sense but you know you have to realize that they are you know in zx there are a priori more irreducible elements if you have a non if you have a constant in zx which is an integer which is not plus or minus one then this is irreducible so yeah you need a tiny argument you have to use that your polynomial is primitive because otherwise it's actually false if you take the polynomial 2x for instance if you take the polynomial 2x this polynomial is irreducible in qx but not you know but not in in zx because it's 2 times x and these are both non units so I mean the argument here is very simple but you know you just have to see that they have no common factor you know if you assume it's primitive it's you know I also haven't given very much of an argument here the argument is precisely more or less what you say but you have to just exclude the possibility that they have a common factor so but you know obviously you are right that this is this is supposed to be the trivial direction and it is almost as trivial as you say but anyway so now let's go to the other one so again we do by contraposition so we so suppose that f is equal to gh where g and h are now polynomials in qx and we assume these are polynomials of positive degree so then we have to so it means the polynomial would be reducible in qx we have to see there is also reducible in zx so so we have this so now we can so so these are so g has some coefficients so some ai x to the i there are some rational numbers so we can multiply with the with some integer so to clear all the denominators so we multiply with all the denominators of of the coefficients here we multiply with all the denominators here and we divide by the greatest common denominator so let me say it like this so clear the denominators of the coefficients of f and g and divide by the greatest common divisor of the new coefficients so then we will get that f is equal to somehow a divided by b g prime h prime where you know g prime is g multiplied by some rational number so that it is so that with g g prime and h prime primitive uh in zx so I mean just you know as example whatever if you have if my if f is equal to seven fifth plus three divided by eight so x plus this then you would multiply here by five so the in this case you would get you have to multiply by by 40 and you would get that you can instead just take f prime so this was maybe g would take g prime to be seven x plus three so it's a polynomial which is the same up to multiplying by a no hopefully you do not believe that so uh so uh yeah yeah seven times so this is x plus 15 I think that's more likely to be true and uh yeah you know obviously should not be caught for getting what one learned in before one went to high school anyway okay so anyway I just replace this by this multiplied by by some rational number this by this multiplied by some rational number so that these two are two primitive polynomials and uh you know the product is like this so maybe just try it again so we have g prime is equal to alpha times g and h prime is equal to beta times h with alpha and beta are some rational numbers now that's how it works and then you can do it in such a way that they're primitive and uh so now um now let's look at it and I can also assume obviously here we have a rational number so I can kind of clear the common factors so that we have a and b are relatively prime are relatively prime no you can just put it like this so in particular we can also write this by a very complicated transformation we can write this as bf is equal to a times g prime h prime so but now what do we see you know both f and g prime h prime are primitive f is primitive because we have assumed it to be primitive and g prime and h prime were both primitive and by the lemma we had before the product is primitive that's why we proved that lemma so if we look at this equation we see here a greatest common divisor of these of all the coefficients is one so a greatest common divisor of all the coefficients in this product is b of coefficients of pf is b and the great and the greatest common divisor of the coefficients of a g prime h prime is a but how is it possible this is the same polynomial so the same polynomial the same coefficients have as greatest common divisor b and a so we know that greatest common divisors are uniquely determined up to multiplying by a unit the units in z are only plus and minus one so it follows that b is equal to a or b is equal to minus a so in other words f is equal to g prime h prime or f is equal to minus g prime h in any case we have found a factorization of f as a product of polynomials with integer coefficients not irreducible okay so this was this statement um or it is reducible in z x okay so we want to now make a few criteria for irreducibility of polynomials so one of the things we usually will be interested in the irreducibility of polynomials with coefficients in q but we do it by checking it for course over z so the first result is the Eisenstein criterion so Eisenstein so some job mathematician criterion for irreducibility so it's a something which one can sometimes apply to show that the polynomial with coefficients in z is irreducible over z and therefore over q so let f be a polynomial so sum i equals zero to say n ai x to the i be a primitive polynomial in z x and we assume so of positive degree okay well whatever otherwise it's too stupid so it's not constant and now we assume some strange condition assume there's a prime number which divides almost all the coefficients except for the leading one so with p divides a zero p maybe i just say divides a one and so on p divides a n minus one but we have that p does not divide a z a n it does not divide the leading coefficient of degree n and the square of p does not divide a zero so some strange condition about this divisibility then f is irreducible first in z of x and therefore in q of x we know that these are good okay this is a well-known criterion to check that the polynomial is irreducible it's a bit special so one cannot apply it all that often but sometimes one can so let's see so we assume again that it is reducible and then we want to find the contradiction proof assume f is equal to g times h with g and h some integer polynomials we have to show that either f or g is constant so let's write down not f or g g or h okay so let's write down the coefficients of g of h and h so write g so sum i from zero to what we want k b i x to the i and h sum i from zero to say l c i x to the i so we have fixed the coefficients and we assume in each case that this really is the degree so b of k is different from zero and cl is different from zero so we know that the constant coefficient a zero is divisible by p but not by p squared so as p divides a zero but p squared does not divide a zero we know that and we have that a zero is equal to c zero p zero c zero we must have that p can only must divide one of these two but it cannot divide both because if it would divide both then the square would divide it so we have that so and this we have that p divides precisely one of b zero and c zero so thus we can assume obviously the role of you know which we call g and h is up to us so we can assume that the one so we can assume that p divides b zero and p does not divide c zero so let's see now we want to we look instead at the highest coefficient so a n is not divisible by by p and obviously if the degree we have that we have a n so obviously we have n is equal to k plus l and a n again by the product formula is a k times a b k is it k times c l and we know that p does not divide a n so thus p does not divide b k so we know p divides b zero but it does not divide b k so there must be kind of a last one which it divides so thus there exists a maximal element j between one and k such that p divides b i for all i smaller than j and p does not divide b j so we can find the the first one which is with it it does not divide now we want to compute the coefficient a j so a j is again by this formula for the product this is equal to b j c zero plus b j minus one c one plus plus b zero c j so if you know if the number l is smaller than than j we can add some zeros the the the later c j is equal to zero so it's this so and now it's the usual thing um so by definition so by our choice we have that uh you know we know that p does not divide c zero and p does not divide b j so p does not divide b j c zero but it divides all the smaller b i so it divides each summons here well the conclusion is that p does not divide a j but now we have to remember what our assumptions here were the assumption was that p was supposed to divide all of the a j's except for the except for the last one so it follows that this number j must be equal to n okay but what does it mean it means you know j is the is a non-zero coefficient of the polynomial uh uh of the polynomial uh g so it means that the polynomial g has degree n the same as f so thus j is equal to n and therefore also k is equal to n that means the degree of g is equal to n and the degree of h must be equal to zero so we have found we have actually shown that h is constant actually we was not uh and so this shows that um our polynomial was indeed irreducible so now I have gone a bit over time so maybe I stop here I can just so you can if you just take any polynomial x to the five plus 16x plus four this would be irreducible by this criterion and obviously if you wanted to check directly that it cannot be written as a product of polynomials of degree that would be uh difficult that actually is now you should have contradicted me know maybe I put two here rather I was assuming that four is a prime which is a kind of rather rush assumption okay thank you very much