 OK, so then I start. Good afternoon. Well, we already saw each other this morning. 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 a 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 element 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 in the second condition, it's 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 the z of E. So these are the 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 1. Then, and so we want to say that these are called the Gaussian integers. They claim it's a Euclidean ring with, well, if I take t 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 0 if a plus i b is not the 0 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 divided by w, which I write as a plus bi, be the 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, so the absolute value of this difference is smaller equal to one half and the absolute value of b minus m is smaller equal to one half. We can certainly do that. Just shows the nearest one, 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. And so this is one half squared, one quarter, one quarter. This is smaller equal to one. No, no, 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, here I have added it wrong. n plus im times w. So this is supposed to be the rest. So let's see whether it 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 write n plus im w plus r, so this is our division of the 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, it's 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 thing is actually one half. So this is certainly smaller than d of w. We know that this is smaller equal to one half. Okay, and so we find this. And the second statement is, obviously 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 principle ideal domains. So that means every ideal is principle. So that means every ideal is generated by just one element. And so in particular, 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 principle ideal domain. And as this is a bit long, one usually always just writes PID. If every ideal, i in r is principle. 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 it theorem, although it's not really that every Euclidean ring is a principle 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 principle ideal domain, a Euclidean ring, and we chose ourselves an ideal. So now we want to see that this ideal is principle. So if i is just a zero ideal, then it is obviously principle because then i is equal to the ideal generated by zero. So that's principle. So thus we can 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 have this wonderful d. We choose an element which has the minimal d of a, minimal 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 idea. Well, so let b 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 qa plus r with q. 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, but if you look at it. But by this equation, we have r is equal to b minus qa. These are two elements of i, so this is an element in i. And 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. 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 of a. So that's b in a. If I take an element b and i, it's an element in 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 principal ideal domains. I don't know whether it has an apostrophe or not. Okay, so we can also talk about principal, about great and greatest common divisors in principal ideal domains. So I had defined great common divisors in integral domains. And I said it's not always clear they exist. They only give a definition, but it doesn't mean they have to exist. But in principal ideal domains, they do exist and are actually quite simple. So this is a proposition. So let r be a PID and let take some elements a1 to ar, some elements in r. Then there is a greatest common divisor of a1 to ar. It exists, so maybe I call it d, exists and is of the form of the form. Can write it as a linear combination of these d equal to a1 x1 plus ar xr with x1 to xr is r elements in r. And actually quite, I mean, it's the obvious thing. These a1 to ar generate an ideal. This ideal is a principal 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 greatest common divisor. So proof. So we choose an element d in r such that if I take the ideal a1 to ar, d r generated by these, so all linear combinations of these, that this is equal to the ideal generated by our, I mean, we know that every ideal is a principal ideal, so also this one. So we can write like this. And the claim 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 ai lies in the ideal generated by d for all i, 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 a1 to ar. 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 ai 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 divides ai. It divides say xi times ai. And if it divides all xi times ai, it divides the sum of all of them. So e divides a1 x1 plus a r xr. Because it divides all the ai and therefore, which is after all d. And so that's fine. So it's a greatest common divisor. So for instance in z, we have that if 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 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, 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 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 being integral domain. So an element A and R will be called irreducible if it cannot be written in a non-trivial way as a product of two other elements. 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 0, is called irreducible. Well, so 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 AB 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 the stupid. So for instance, say, so maybe first I finish. So it is called reducible if it's not irreducible. So if Q is again is not a unit and not irreducible, it is called reducible. So this means in other words, if one wants doesn't want to say it in such a direct way, obviously just in indirect way, it obviously just means that. So Q is reducible if it exists AB in R which are not units. None of, so without units such that Q is equal to A times B. So for instance, 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. 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. Of course, 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, 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, an element in K is irreducible. Well, this thing is not a unit and it's 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 B is equal to F times G for F and G polynomials, then we know that 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 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 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. Namely when it's degree is 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 PID. 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 ideal, you get a field for position. So let R be a PID, 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 a reminder 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 strictly, it's contained in I. So this ideal is a principal ideal. So you can write I equal to A. So that P is contained in the ideal generated by A. It 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 B 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 this A by a unit, it gives you the same ideal. So that's one possibility or B is not a unit. 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. Yeah, yeah, sorry, obviously 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 the assumption was that it's true if I have an irreducible element. So yeah, obviously I have to use my assumption. 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 then because P is prime is irreducible. So then obviously 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 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 that's precisely the definition of a maximal ideal. Yeah, amazing. Okay, so we want to use this to construct new fields from old by 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, 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 a field isomorphic to what I just say K as a suffering. 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 we have the constant polynomials. So if I take the map from the constant polynomials to this quotient, this is an isomorphism onto the image. So the map, so on, so for, let's see, A and K, we have the class of A, no, so the map, restriction Kx to F to K, which are the constant polynomials, 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. So if I restrict it to K, then the map is injective, because the, it would have to be either, so as K is a field, 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 lie in 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 subfield 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 start it. 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 we'll look at later, so we'll come back to this later. Now as we have now found this, so 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 irreducibility over q and z. So we have polynomials in qx and in zx. And we want to find out about irreducibility by comparing what happens in q and z. And the first result about this is due to Gauss. So we have first the lemma and then, so first we have a lemma which is the first step towards this. So we first make the definition. So obviously 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 a unit. So therefore, we want to first look at polynomials which are primitive, which means that their coefficients have no common divisor. So the polynomial equal to sum i equals 0 to n ai z, and 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 0 to a n are relatively prime. 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 is 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 tried to see whether 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 maybe we write f equal to sum i equals what I want 0 to n ai. X to the i, g equal to sum i equal to 0 to m. Whatever, b i, x to the i. And so we assume that fg is not primitive. So then if it's not primitive, then all the coefficients of fg have a common divisor. So there's 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 of f times g. 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 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. 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 know it doesn't divide all, so there will be some. It doesn't, but it can take the smallest one. So 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? 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. So this is just the sum over all albm such that l plus m is equal to i plus j by definition. And so 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 in 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 so this contradicts our assumption 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 it as a polynomial of two polynomials in qx and vice versa. And so let's see. So let f in kx be a non-constant non-zx, 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 a non-trivial way as a product of polynomials with rational coefficients. And that doesn't seem to be so obvious. So there are, I mean, it's equivalent. So there are two directions. First, we do the easy direction. So if f is reducible in zx, so it's by counterposition 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 0, that means g is constant, then it means that g is a non-unit in z. 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 bigger than 0. And the same argument applies if the degree of h is equal to 0. Now it's symmetric. So thus we also have the degree of h is bigger than 0. 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 as being reducible in qx. So we have f h bigger than 0. 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. If I have a polynomial, OK. So this was the easy direction. Because we just have to take the same. Now we have to use the other way. So this is the most surprising part. You have to see that if you cannot write it as a product of two polynomials with z coefficient, you cannot write it as such a product with r coefficients. What? Yeah, yeah, yeah, certainly. Yeah, yeah, no, I finished. I mean, this is the argument. But what is the, yeah. Well, that's what I proved here, no? This is the direction I proved. No, no, no, but the units are different, no? So if you cannot, so let me see whether your argument makes sense, well, I mean, in some sense. But you have to realize that there are, in zx, there are, if I all right, more irreducible elements. If you have a constant in zx, which is an integer, which is not plus or minus 1, 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 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 just have to see that they have no common factor. 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. But obviously, you are right that 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 a contraposition. 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 that it's also reducible in zx. So we have this. So now we can, so these are, so g has some coefficients. So some ai x to the i, they are some rational numbers. So we can multiply with some integer so to clear all the denominators. So we multiply with all the denominators 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 g prime is g multiplied by some rational number so that it is so that with g, g prime, and h prime primitive in zx. So I mean just as an example, whatever, if you have, if f is equal to 7 fifth plus 3 divided by 8. So x plus this, then you would multiply here by 5. So in this case, you would get, you have to multiply 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 7x plus 3. So it's a polynomial which is the same up to multiplying by a, no. Hopefully you do not believe that. So yeah, yeah, yeah, 7 times, so this is 56x plus 15. I think that's more likely to be true. And yeah, obviously, should not be caught forgetting what one learned before one went to high school. Anyway, OK. So anyway, I just replace this by this multiplied by some rational number, this by this multiplied by some rational number so that these two are two primitive polynomials. And 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. That's how it works. And then you can do it in such a way that they are primitive. And so now let's look at it. And I can also assume, obviously, here we have a rational number so I can clear the common factors so that we have a and b are relatively prime. Relatively prime. 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? 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 all the coefficients is 1. So a greatest common divisor of all the coefficients in this product is b. Of coefficients of pf is b. 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 1. 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 prime. In any case, we have found the factorization of f as a product of polynomials as integer coefficients, not irreducible. So this was this statement. It is reducible in zx. 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 cross over z. So the first result is the Eisenstein criterion, the theorem, Eisenstein, so some job of mathematician, criterion for irreducibility. So it's 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 0 to say n, ai, x to the i, be a primitive polynomial in zx. And we assume so of positive degree, 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 a0, p, maybe I just say, divides a1, and so on. p divides an minus 1. But we have that p does not divide az an. It does not divide the leading coefficient of degree n. And the square of p does not divide a0. 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 equivalent. 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. And we have to show that either f or g is constant. So let's write down, not f or g, g or h. So let's write down the coefficients of g of h and h. So write g, so sum i from 0 to what we want, k, bi, x to the i, and h, sum i from 0 to, say, l, ci, 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 0, and cl is different from 0. So we know that the constant coefficient, a0, is divisible by p, but not by p squared. So as p divides a0, but p squared does not divide a0. We know that, and we have that a0 is equal to c0, p0, c0. 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 p0 and c0. So thus we can assume, obviously, the role of, which we call g and h is up to us. So we can assume that the one, so we can assume that p divides p0 and p does not divide c0. So let's see. Now we look instead at the highest coefficient. So an is not divisible by p. And obviously, if the degree we have that we have an, so obviously we have n is equal to k plus l. And an, again by the product formula, is ak times bk, is it k, times cf. And we know that p does not divide an. So thus, p does not divide bk. So we know p divides p0, but it does not divide bk. So there must be kind of a last one which it divides. So thus, there exists a maximal element j between 1 and k, such that p divides bi for all i smaller than j, and p does not divide bj. So we can find the first one which is, it does not divide. Now we want to compute the coefficient aj. So aj is, again, by this formula for the product. This is equal to pj c0 plus pj minus 1 c1 plus p0 cj. So if the number l is smaller than j, we can add some zeros. The later cj is equal to 0. So it's this. So and now it's a usual thing. So by definition, so by our choice, we have that, we know that p does not divide c0, and p does not divide bj. So p does not divide bj c0, but it divides all the smaller bi's. So it divides each summand here. Well, the conclusion is that p does not divide aj. But now we have to remember what our assumptions here were. The assumption was that p was supposed to divide all of the aj's except for the last one. So it follows that this number j must be equal to n. But what does it mean? It means j is a non-zero coefficient of the polynomial 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 0. So we have found, we have actually shown that h is constant. Actually, we was not there. And so this shows that our polynomial was indeed reduced. So now I have gone a bit over time. So maybe I stop here. I can just, so you can, for instance, if you just take any polynomial x to the 5 plus 16x plus 4, 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 lower degree, that would be difficult. That actually is, now you should have contradicted me. Maybe I put 2 here rather. I was assuming that 4 is a prime, which is a kind of rather rash assumption. Thank you very much.