 Okay, good afternoon. So I will now have the finish with the part about drinks by finishing the section about irreducibility of polynomials and then we start with fields and field extensions. So the last thing that we proved was the Eisenstein criterion. I think you will still remember it. And now we want to give some other simpler criteria to check that polynomials are irreducible. So a few more, this time a few other methods or simple methods to check irreducibility of polynomials. So the first one is just a talk. So this is the following remark. So if you have a polynomial of degree two or three, if it is reducible, then it must have a zero. So if f in kx, now moment k is any field, is a polynomial of degree two or three, then it follows, then f is reducible, so not irreducible, if and only if f has a zero in k. That's quite obvious because if f is reducible, so obviously if f has, I mean this direction is trivial, if f has a zero in k, then you can take, you can divide it by x minus that zero. We have seen that before. And so it's reducible. So if f is reducible, then this means we can write f is equal to g times h with f and g with g and h polynomials of positive degree, so degree at least one. So as the degree of f is two or three, one of these two degrees must be one. g is equal to one or the degree of h is equal to one. And so obviously this is, and then that means, for instance, obviously it's symmetric whether it's g or h. So if h has degree one, that means that h can be written as a x plus b for a and b elements in k, and then b divided by a is a zero of k. So this is very simple. Now, it's not so obvious to see that a polynomial does not have a zero. I mean, at least the polynomial of degree two, maybe you might be able to do that, but probably polynomial of degree three might be some difficulty. But one thing you can see on the other hand, we had this exercise. So if f is a polynomial with integer coefficients is monic, we can write f equal to sum as say x to the n plus sum i equals 0 to n minus 1, say ai x to the i. So monic means precisely that the leading coefficient is 1. So then it was an exercise. I think I said that first in that case, if a in q is a zero of f, then a is actually an integer. And furthermore, we have that a divides the coefficient a zero in z. You can find this by division with rest. Well, maybe I shouldn't say, but OK. So this was an exercise. And so thus, for instance, if you have a monic, so thus if f in zx is monic of degree smaller equal to 3, well, say of degree 3, degree 2, I think you can deal with. Anyway, then f is irreducible if and only if none of the devices of the constant term, which I may still call a zero of a zero, is a zero of f. And obviously, you can directly check. Your a zero is just one integer, so it will have a small number of devices. You can just check for every one of them. If you put it into polynomial, whether you get zero or not. And if you don't, then the polynomial is irreducible. None, OK? So if we take, for instance, f equal to x to the 3, so this would also work with the Isenstein criterion, but doesn't matter, plus 2x minus 2 if I'm not mistaken. So the divisors of the constant term, what is it? 1 minus 1, 2, and minus 2. And if you put each of them into the polynomial, you don't get zero. And so it follows f is irreducible. So this is a kind of trivial way to check irreducible in some cases. Now I want to have a slightly more interesting way to get some example which we will use later. So this is the following remark. It's also simple. So let again, now we have a general field. So let f, say, the polynomial sum i equals 0 to n ai x to the i, be some polynomial with coefficients in a field k. So we give ourselves an element a in k. So we can just make a new polynomial out of f by shifting it by a. So we put, so we say that then f is irreducible over k. So in kx, if and only if, if I replace the variable of f by x plus a, if that is irreducible over k. And this is, so f of x plus a is the obvious thing. You just take this and replace x by x plus a. So this is defined to be i equals 0 to n ai x plus a to the i. Obviously, this is also a polynomial. And this is clear because if I take the map, say, I call it sigma a from kx to kx, which sends any polynomial g to g of x plus a. So the variable x is replaced by x plus a. This is, as you can easily see, an isomorphism of rings. So you can easily see that it's a homomorphism of rings. If you take the sum of two polynomials, you get the sum. And if you apply it to the product, you get the product. This is basically by definition. That's obviously a homomorphism, and it's an isomorphism because the inverse is what one can immediately imagine. You shift it by a, and you can shift it back by minus a. Inverse is sigma minus a. And so if you have an isomorphism of rings, then obviously it sends irreducible elements to irreducible elements and to reducible ones. Because if something is a product of two elements, then the image will be the product of the image element. So thus it preserves irreducibility. Thus f is irreducible, if and only if, f of x plus a. Or I could now call it sigma a of f is irreducible. I want to apply this in one example, which later we have to do with if you want to deal with it with dealing with roots of unity. So example, so let p be a prime number in z. So then we can look at the polynomial f, which I just take x to the p minus 1 plus x to the p minus 2 plus, and so on, plus 1, so plus x plus 1. So just take all the powers of x from p minus 1 to 1. There's a plus here. So I claim that this is irreducible in q of x. So it's an irreducible polynomial over q. And we want to do this by using this mark. So obviously, as you have learned either in high school or in the first year of university, you will find that x minus 1 times f is equal to x to the p minus 1, beginning of some geometric series. So now I can apply to this sigma 1 to this. So if I take x times sigma 1 of f, so sigma 1 was this map. x goes to x plus 1. This will be x plus 1 to the p minus 1, which you can write down in terms of binomial coefficients. So this is x times this. Now I divide it by x. So I get sigma 1 of f is equal to this divided by x. So this is sum i equals 1 to p. So I get binomial coefficient p choose i times x to the i minus 1. So if you take this one, this would be the sum from i equals 0 to p, p choose i times x to the i. Now we subtract the constant term and divide by x. So we get this. So I hope you remember. I mean, I'm not quite sure whether in all countries the notation for binomial coefficients is the same. Yeah. But as long as you know what it is. So p choose i. You can write this as p times p minus 1 times p minus i plus 1 divided by i factorial. So i times i minus 1 times 1. OK. So it's a standard exercise that such a thing if you take p choose i for prime number, this will, unless i is equal to 0 or p, this will always be the result of p. So it's easy to check that p divides p choose i for 1 smaller equal to i, smaller equal to p minus 1. And obviously, p squared does not divide p choose 1, which is just p. So we are in the situation. And obviously, p does not divide p choose p, which is 1. So we are precisely in the situation where the Eisenstein criterion applies. So the Eisenstein criterion gives us that sigma 1 of f is irreducible and thus this polynomial here is, I mean, f is irreducible. And this somehow means if you want to take a number which is not 1, such that x to the p is equal to 1, then this will, let me see. Yeah. I mean, if p is bigger than 1, then this will not be a rational number. If p is equal to 1, this polynomial is also irreducible, but in that case, it's of degree 1, so it still has a 0. OK, now let's, so that was as much as I wanted to say for the moment about these polynomials. So let us now start to talk about fields. And what we really will be talking about is field extensions. So we want to talk about field extensions. So I first say the field extension is just that you have two fields. So you have k is a field contained in a field L, such that k is a subfield of L. So a subfield of L just means a field, which is a subring of L. So k and L are fields. So we are interested in studying such field extensions, in particular somehow field extensions which has to do with polynomials. For instance, one question you could ask oneself, assume we have a polynomial in kx, say an irreducible polynomial in kx. And you want to find the field extension of k, such that this field extension might be called L, such that in L, the polynomial has a 0. And do you want to study whether this is possible or what one can say about such fields and so on? OK, so first now I want to give the proper definitions. And OK, I don't think I need this now. So let's talk about field extensions and the degree theorem. Later we come to the degree theorem. So first I have already kind of said in words what the field extension is, but let me see it correctly. I mean explicitly as a definition. So definition is subring k of a field L is called a subfield. If it is a subring, it's called a subfield of L. So a subfield of a field is just a subring which happens to be a field. And on the other hand, we usually want to view the other round. We want to view not k as a sub thing of L, but L as an extension of k. So in this case, we call L a field extension of k. And we write also that L over k is a field extension. OK, this is this field extension. And so if, I assume we have such a thing, so if L over k and large k over k are field extensions, we will later be interested in homomorphisms between L and k, which are the identity between L and large k, which are the identity on small k. So a homomorphism of rings and thus of fields, phi from k to L is called a k homomorphism if it's the identity on small k is equal to the identity. So if I take any element a in small k, then phi of a is equal to a. And in the same way, we have k isomorphism. So it's an isomorphism, which is the identity on k and k automorphism, which would be an automorphism of say, large k to itself, which is the identity on small k. Later we will find that these k automorphisms form a nice group, which we will have to study in order to study the field extensions. So that's one thing. Now there's one thing. If we have such a field extension, one obvious very easy remark is that if L over k is a field extension, then L is a k vector space. And then we can be interested in the dimension of this k vector space. So remark. Let L over k be a field extension. So then let's just look at the axioms that we have. So we have that L with the addition is an a b-ring group because a field with addition is an a b-ring group. And we can look for the multiplication. We can just restrict the multiplication of L times L to L to a multiplication from k times L to L. And restriction of the multiplication on L to k times L gives this multiplication times from k times L to L, which is somehow we want to see that L is a k vector space. So this would be the scalar multiplication such that a few axioms are fulfilled just because we have a field and we have a few axioms and we can just specialize them. So such that we have the following rules. We have such that we have the distributive laws. We have the distributive laws. This is just if I take, say, a plus b, which are now a and b elements in k, times an element x and L. So we have a b in k, x, y in L, such a plus is equal to ax plus bx. This is just a special case of the distributive law in L. And we have if I have a times x plus y is equal to ax plus ay. And we have the associative law. If we take a times b, a times b, a and b are elements in k, times an element x, this is the same as taking a times b times x. This is just because in L we have the associative law and we just restrict it. And finally, we know if we take 1 in k times any element x, this is equal to x. Because after all, 1 is the neutral element of L. And so if you look at the, if you try to remember what you learned in your first years of university, you find out that these are precisely the axioms for a vector space, L, a k vector space. So thus we have that L is a k vector space. And so then the dimension of L as a k vector space we will call the degree of the field extension. So let me write this down. Definition L over k, a field extension. So the degree, which is denoted very similar to what we did for rings, L over k, for groups that was called index. The degree L over k, but it's slightly different. Anyway, the degree L over k of L over k is the dimension of L as k vector space, where we, where obviously this degree can therefore also be infinite if the dimension is infinite. Over k is equal to infinity if L is not a finite dimensional vector space. So this degree, which is a number, which is possibly infinite, is an invariant which you can associate to every field extension. So let's see what we can see about this number. The first obvious, so and we call L over k a finite extension if this dimension, so this degree is finite. So not L over k, so the extension we call a finite extension if the index, so if L is a finite dimensional k vector space. A one stupid remark is that if L over k is equal to 1, then L is equal to k. It's kind of essentially, I mean it's in some sense completely obvious because if L over k is equal to 1, then it means that this is a, that L is a one-dimensional k vector space and as a basis we can take the element 1 in L. So the elements of L are just 1 multiplied with any element of k and this is k. OK, so now we want to ask ourselves what happens to this degree if we kind of make several field extensions one after another. So we take L an extension of k and large k an extension of L. So what is the relation between the different degrees? And again this is similar to what we had for the index of subgroups. So first I just introduce one more word which will often be used. So it's just a different way of saying something again. So let say large k over k be a field extension and let L over k be a field extension with L is actually a subfield is contained in k. Then we call L an intermediate field of the field extension large k over k. Now obviously it lies between large k and small k. And so now we can, and so the degree theorem will tell us in this situation if you have k which contains L which contains small k, how the degrees of these things are related to each other. So it's not very, so this theorem is quite simple. Degree theorem, so let L be an intermediate field of an extension k, large k over small k. Then the degree of large k over small k is equal to the degree of large k over L multiplied with the degree of L over small k. Here obviously it might be that these field extensions are infinite. So I always say that for n in z some positive integer I have infinity times infinity is equal, infinity times n is equal to n times infinity is equal to infinity. Just that's the convention so that this holds. So if any of these numbers is infinite then I mean, if any on this side is infinite then also this is infinite and if this is infinite then one of these two must be infinite. So in particular we have that the extension k over k is a finite extension if and only if both of these are finite if and only if L. So large k over L and L over small k are both finite extensions. OK, so this is almost more difficult to state and prove. So first I'm actually not really interested very much in the case that any of these numbers is infinite so it's mostly interesting the case for finite extensions. And it's kind of obvious also that if k over L is not finite, L over k is not finite. Then it's clear k over small k is not because you find an infinite set of elements in large k which are linearly independent over L then they are also linearly independent over k. Or you find an infinite set of elements in L which are linearly independent over small k then this is also an infinite set of elements of large k which are linearly independent over small k. So in any case you find if any of these two degrees is infinite then this degree is infinite. So thus we can assume that both these numbers are finite. Maybe I should have done it with the index though because that's what I meant. So as we assume that these are both finite so let's say now I write a term that L over k small k and k over L are both finite extensions. Then we need to show that k over k is a finite extension and that the dimension is the product of these two indices, these two degrees. Well, we just do this by exhibiting a basis. So let, so we assume, so let, so the degree of L over k to be n and the degree of k over L to be m. And so we choose a basis first of this over this. So let x1 to xn be a basis of L over k. So as of L as a k vector space. And y1 to ym be a basis of k as an L vector space. And we want to show that by taking all possible products of these elements we get a basis of k over small k. If we take the set of all xi, yj, i goes from 1 to n and j goes from 1 to m is a basis of large k as a small k vector space. Well, if you want to show that something is a basis we have to show that it generates the vector space and that the elements are linear independent. So we just do it. So first I show they generate as k vector space. So I must be able to write any element in large k as a linear combination of elements of these things with coefficients in small k. So if I take, so if say y is an element in k we can write y equal say sum i equal 1 to m bj yj for some elements bj in L. This is as the yj's basis I can write any element in k as a linear combination with coefficients in L of these. Now for each of these bj's I can do the same and express them in terms of elements of small k for all j. We can write bj as sum, so this was a j, i equals 1 to n aij xi. This is again just the xi from a basis so where the aij obviously are elements in k. This is again because the xi from a basis of L over small k and so any element of L in particular bj can be written as a linear combination of them. So now we just put this into each other. So thus we have that our given element y can be written by right putting this description into this. So this is the sum of all i equal, so say i from 0 to n, sum of all j from 1 to m aij xi times yj. And so we have indeed written y as a linear combination of these products xi times yj. So they generate. And now we have to show that they are linearly independent. It's kind of basically a similar thing. It will be only easier if it just does it step by step. So they are linearly independent. So if sum aij xi times yj is equal to 0, 1. So i and j go again from 0 from 1 to n and from 1 to m. If I have such a linear combination which is 0 with elements aij in k, then we can first, we know that this yj are linearly independent. So if I have any combination of the yj's which is 0, then the coefficients of each yj must be 0. So as the yj are linearly independent over l, we find that and we have some combination of them which is 0, we find that all the coefficients are 0. It follows sum i aij xi is equal to 0 for all j. And now we apply this again. These xi are linearly independent over k. And we have a linear combination of them which is 0. We have that it follows that sum that aij is equal to 0 for all i and j. OK, so that was quite simple. In some sense, I think it's the obvious thing one would think of doing if one would be given this as an exercise. But anyway, that's it. As a corollary, I just want to mention like we did for the groups. So corollary, if k over k is a finite field extension and l is an intermediate field, then clearly we have that, I don't know which one I wanted, that say l over k divides k over k in the same way also large k over k. So anyway, that's clear because of the degree theorem. And so in particular, if k over k is a prime number, there are no intermediate fields except for large k and small k itself. So there's nothing I have to say to this. It's kind of completely obvious. So anyway, so this degree theorem is not very deep result, but it's very useful, obviously, because it gives us some way how to see whether certain field extensions can be there and to compare them. So now I want to talk about algebraic field extensions and then simple algebraic extensions. So now a field extension l over k will be called algebraic if every element in l is a 0 of a non-zero polynomial with coefficients in k. And then we will say later what this is. So let me set this up. So for the moment, we fix the field extension. I mean, for the argument now, we fix the field extension k over small k. So we say an element a in the bigger field will be called algebraic over the smaller field if it's the 0 of a non-zero polynomial. So if there exists a non-zero polynomial in so which would be sum i equals 0 n a i x to the i with coefficients in the smaller field with f of a is equal to 0. So that means sum i from 0 to n a i a to the i. We should have called that b because it's a bit confusing. P i a to the i is equal to 0. And OK, so we will be interested in such algebraic elements. So otherwise, a is called transcendent if I can spell that over k. And k is called an algebraic extension of k if all the elements of k are algebraic over small k and large k are algebraic over k. So just as examples, we have to take the square root of 3. This is algebraic over q because it's a 0 of a. So if I take the square root over 3 in the real numbers, it's a 0 of the polynomial x squared minus 3. And if I take i, so the complex number i and c is algebraic say over q because it's a 0 of the polynomial x squared plus 1. And a very celebrated theorem of about 100 years ago is that, for instance, pi, the circumflex of a circle is transcendent. I think it's called transcendental, I think, is transcendental over q. This is however much too difficult for us. I mean, I just mentioned it. I think this was proven a bit more than 100 years ago. OK, so now we want to look at, I mean, here we'll be looking also at simple algebraic extensions. So we want to look at extension which are obtained by taking a field and adding to it one element, taking the smallest field which contains also this element. And so I want to introduce this first for more than one element and then for one element, and then we want to study them which are the simplest field extensions which we can imagine. So definition. So we still remember that we have some field extension k over k somehow. So let a1 to a n be some elements of the bigger field. So the extension of the smaller field generated by a1 to an. So first I say it loosely is the smallest field l which contains the field k. And so the smallest subfield l of k which contains k and the elements a1 to an. So this kind of saying the smallest subfields which contains them is just a different way of, it's just supposed to mean. It's the intersection of all subfields of l which contain these. It follows from the definition directly that the intersection of all such subfields l will again be a subfield which contains these elements. And it is then the smallest one which does it. And this is denoted, it is denoted k with round brackets a1 to an. So by definition, this is a subfield of k of large k which contains small k and these elements and every other subfield of k which contains small k and all these elements contains also this. So now let's look at the special case that we have only one element here. So we want to look at the subfield generated by just one element. And we want this to be an algebraic element. So, definition. So let a in our element in the bigger field k be algebraic over k over small k. Then this field generated by this one element is called a simple algebraic extension. So we will actually very much study these simple algebraic extensions. It will be very important for us for two reasons. The first one is that they're really simple. So they're easy to study and one can understand them. And the second reason is that in under reasonable assumptions all finite field extensions are simple algebraic extensions. So that these things are simple and they are also everything. So therefore we certainly should study them. So the last statement is not completely true but almost. So they are important because first easy to study and second almost algebraic extension, algebraic extensions. I will explain in a moment what almost means for instance. So this is the theorem of the primitive element which we will prove later. So if one makes some suitable assumptions, then every algebraic extension is a simple algebraic extension. And now we will see that one can completely understand them. So we want to describe them explicitly. So now I want to describe explicitly a simple algebraic extension generated by one element A. And it will be described explicitly in terms of the so-called minimal polynomial. So this is an irreducible polynomial with coefficients in the smaller field, which is monic. So the constant, the leading coefficient is one. And you can completely describe the extension in terms of this polynomial. So let me first introduce this polynomial. So the polynomial, so we already had this but I say it once more. A polynomial say f with coefficients in a field k is monic. It's called monic if it's leading coefficient is one. And so if f is a non-zero polynomial, then we can turn it into a monic polynomial by dividing by the leading coefficient, obviously. So then there is a unique polynomial say g in kx, which is monic, which generates the same idea in kx. That's kind of obvious if I, we know that two polynomials, I mean two elements in an integral domain generate the same ideal if they differ only by multiplying by unit. So that means two polynomials generate the same ideal if and only if they differ by multiplying by a constant. And so in this case, I can just obtain g by taking f and dividing it by its leading coefficient. And that's obviously unique. So now let us take an element a in our bigger field k, which is algebraic over the smaller field. And we can look at the evaluation homomorphism at a. So the evaluation homomorphism was I take a coefficient, a polynomial with coefficients in the smaller field k. And we send it to k by sending f to f, I mean, yeah, of a. Every polynomial f is sent to the evaluation of the polynomial at a. We know that the evaluation is a ring homomorphism. So this is a ring homomorphism. So we have assumed that a is algebraic. So it means there's some polynomial, some non-zero polynomial, such that if I evaluate it at a, I get zero. So the kernel of this homomorphism is non-zero, kernel. And this kernel is an ideal. And so as this kx is, so this is an ideal. So skx is a principal ideal domain. We know that this ideal is principal. So it's the ideal generated by unique element. So there is a unique polynomial, a unique non-zero polynomial, such that it's generated by this. And if there's unique non-zero polynomial, there's unique monic polynomial, we record fA, such that this kernel of the evaluation morphism is equal to the ideal generated by fA. And fA is called the minimal polynomial of a over k. Okay, so this is this definition of the minimal polynomial. It's a bit maybe roundabout. So therefore, we immediately will give another characterization, which is easier to remember and often to use. But anyway, so if we take the, we have such an algebraic element over k, then the evaluation at a has a kernel, an ideal which is generated by unique monic polynomial, and this is the minimal polynomial. And now let me give another characterization, which is what I just said before, that it's a unique irreducible monic polynomial with coefficients in small k, which vanishes there. Proposition, so let, we are in the same situation. We have an element in the bigger field, which is algebraic for the smaller field. So the minimal polynomial, polynomial fA of a over k, is the unique irreducible monic polynomial, so f with coefficients in the smaller field, such which vanishes at a. So the minimal polynomial is, if I have an element, which is algebraic over some k, then the minimal polynomial is a polynomial with coefficients in this smaller field, which is irreducible and monic, and which vanishes at this element. And there is a unique such element, which is the same as what I defined here. This is the claim. The proof is not particularly difficult. I have to see that. So the first thing, so we take as fA the thing that we have defined here. So in order to show, we have to show that these two things are the same. So for this, I have to show that if I have this element, then this is irreducible. So let fA be the minimal polynomial of a. So we have to show fA is irreducible. Then we have found that this is an irreducible monic polynomial. And then on the other hand, we have to show if we have an irreducible monic polynomial with f of A equal to 0, we will have to show that it is this generator of this idea. So let's first show this thing is irreducible. So assume fA is a product of two polynomials. You have to show that one of them has to be constant. So we have 0. So fA lies in the kernel of this evaluation homomorphism. So fA of A is equal to 0. So if we have that it's this product, we get this. So here we are in some field. So if the product of two things is 0, then one of them must be 0. Obviously it doesn't matter which one we call g and h. So we can assume that g of A is equal to 0. I hope you remember the definition of the minimal polynomial. So say equal to 0. So then it means that g of A lies in the kernel of this evaluation homomorphism. So g lies, thus, g is in the kernel of the evaluation at A, which was generated by fA. So it means that g is equal to L times fA for some polynomial L in kx, no? But now we have fA is equal to g times h. And g is equal to L times fA. So we have that fA is equal to h times L times fA. So that means h times L is equal to 1. And in particular, it means that this h was a unit. So h is a non-zero element in k. So k star, I don't know whether I introduced it. It's just a unit. So this shows that fA was irreducible. So conversely, in order to show this result, we have to show that if we have an irreducible monic polynomial which vanishes, it's a generator of this idea of the kernel of the evaluation. So conversely, let say g be a monic irreducible polynomial, I mean in kx, with g of A is equal to 0. Then we have to see that g is equal to our fA. So it's the unique monic generator of the kernel of the evaluation map. Well, it's quite simple. g of A is equal to 0. So it follows that g is in the kernel of the evaluation, which is the ideal generated by fA. So there exists again, say, an L in kx, such that g is equal to L times fA. But g was supposed to be irreducible. If g is irreducible, it cannot be written in a non-trivial way as a product of these things unless this is a constant. So it follows, so as g is irreducible, L is an element in k star. And now, however, both g and fA are monic. So the leading coefficient is 1. So if you want to find an element in k star so that you multiply 1 by it to get 1, the only possibility is 1. So fA, monic. OK, so this is this characterization of the minimal polynomial. How much time do I actually have? I'm a bit slower than I thought. Actually, consider is slower. So as an example, so let p be a prime number. We take some positive integer. Then if you take the polynomial x to the n minus p, this is the minimal polynomial of the nth root of p over q, say, because this polynomial is irreducible by the Eisenstein criterion. And so it is, and it's monic, and so it's the minimal polynomial. So now I have not so much time, so which page is this? I'm actually not, oh, it's not so bad. Anyway, so now I will not be able to prove this theorem completely, but at least I can state it and maybe see to start with a proof. So this is the description of the simple algebraic extensions. So it's the formula for the description of simple algebraic extension in terms of the minimal polynomial. So the theorem is as follows. So we have, again, element a k, algebraic over k. And so with minimal polynomial f a, and we maybe fix the degree of the minimal polynomial. Then we have the following. First, then first, we have that k a. So the simple algebraic extension generated by a is isomorphic to k x divided the ideal generated by the minimal polynomial. Something that we have kind of seen before. And the second statement is the degree of the field extension is equal to the degree of this polynomial. And in fact, we can give a basis of this field over this and the elements 1, a, and so on, until a to the m minus 1 are a basis of k a over k. OK, what can I do with this now? So this is the statement. So one should note that this, in a suitable sense, completely describes this field. We know what the degree of the field extension is. And we know precisely what the ring structure is. We have this polynomial ring divided by this ideal. And we can also, OK, I don't have this. So maybe I can prove the first part and then go on the next time. So we have this evaluation homomorphism from k x to k of a. So it goes first to k, but the image is obviously of a because it's all expressed in terms of a, which sends g to g of a for any polynomial g. So this is a ring homomorphism. So it's a ring homomorphism whose kernel is the ideal generated by f a. We know that the kernel of the variation of a is the ideal generated by the minimum polynomial. So let now say l be the image of this map. So this is the image applied in relation of a to the whole of kx. This will be a suffering of k. So a suffering of k. So by definition, so the kernel is this and this is the image. So we find that this l is isomorphic to kx divided by this ideal. By the homomorphism theorem, we have that l is isomorphic as a ring to kx, modulo, this ideal generated by f a. So this is what we have on this side. So we will want to prove that l is equal to k of a. And also that it's a field, obviously. So as f a is irreducible, we have seen that it generates a maximal ideal in the principal ideal domain kx. So it follows that this l is a field. And if I identify the image of the constants under this map with k, because the map on the constant has, there's no kernel if I look at the constants. No, it's injective. So which contains k is a subfield. No, that's the image of the constant. Contains something isomorphic to kx, but doesn't matter. And obviously, so it contains k is a subfield. And obviously, we also have that a, which is equal to the polynomial x applied to a, is in this image, which I could call is the evaluation at a of the polynomial x is in l. So that means l is a field which contains k and a. And so thus l contains k and a. And also, by definition, l is contained in k of a, because all the elements in l are just polynomials in a. You apply a polynomial to a, so this gives you an expression of a with coefficients in k. So you stay in a. And l is contained in k of a, but thus l is k of a. If you take the evaluation morphism at a, it means you have some kind of linear combination with coefficients in k of powers of a. This certainly lies in the subfield generated by k and a. Because you have just used field operations to elements of k and a. So certainly l is contained in k of a, and this does this. And so l is equal to k of a. So this proves one. And then we will next time do number two, which is actually maybe, well, it's not more difficult. And maybe also review the statement again. I think we see each other on Wednesday or something. Anyway, so thank you.