 Okay, so good afternoon. Let's start. So I will just proceed. So now I will first talk about prime ideals and maximal ideals. So these are two very simple concepts. So I will from now on for simplicity assume that our ring when I talk about ring and I don't say anything else. Then a ring by saying a ring, I mean commutative ring with one. So I mean that means usually from now on, we only look at commutative rings with one and I don't specifically say it. If a ring is not commutative with one, then I will say it. Okay. So it doesn't mean I've changed the definition of a ring. No, it's just how I simplify this. So prime ideals and maximal ideals are maybe the nicest ideals you can have. So you can see this by looking at the quotient by the ideal. So prime ideal will be an ideal such that if you take the quotient of the ring divided by a prime ideal, you get an integral domain and so it has no zero divisors. If you take the quotient of a ring by a maximal ideal, it will be a field. The quotient will be a field. So the definition obviously however is in terms of the ideal itself. So let me write it. So we take a ring, say commutative ring with one. So an ideal say P in R will be called a prime ideal. If it has the property that whenever the product of two elements of R lies in the ideal, then one of the two elements already lies there. So if whenever we have A times P is an element of P, for A in P elements of R, we have A in P or B in P. Note that by the definition of an ideal, if we have a product of two elements in the ring, one of which lies in the ideal, then the product lies in the ideal. So but now for prime ideal also the converse holds. So this is the definition. So for instance as an example, so in the integer Z, if P is a prime number, then the ideal P which is the same as PZ is a prime ideal. Basically, the elements here are all numbers divisible by P, and a product of two integers is divisible by a prime number if and only if at least one of them is. Now let's come to the definition of a maximal ideal. So a maximal ideal is as a name says, an ideal which is maximal, so which is not contained in a bigger ideal except for the whole ring. So an ideal M in a ring R is called a maximal ideal. Let me just write maximal if there is no ideal which is in R, which lies strictly between M and R. So with M is contained in I and really smaller and this. So the only ideal of R in which the maximal ideal is contained is R itself. Now I want to state the thing which I said in the beginning, the characterization in terms of the quotient of maximal and prime ideals. It's a rather simple fact. Maybe I'll just call it proposition. So we have R ring, so a commutative ring with one. So first an ideal P in R is a prime ideal, if and only if the quotient R mod P is an integral domain. So that means it has no zero divisors. That's essentially a directory formulation of the definition as we will see. An ideal M in R is a maximal ideal, if and only if the quotient is a field. So that also we will use this quite often later when we want to make field extension. So we want to find for given field, we want to find a bigger field in which the field is contained. So we will take some ring which is a polynomial ring, and we quotient by certain maximal ideal and the quotient will be another field. So and we will see that later when we talk about fields. So let's prove it. So first one is essentially completely trivial, it's just a reformulation. So we have assumed that R is a commutative ring with one. So that's part of the definition of an integral domain, the only thing that we need it has no zero divisors. We have to see and let me just see this commutative ring with one. So that then does it follows also R mod P. So I do that one is commutative with one. This mean the product structure, the one here is the class of the one before. Now let's take, so if we're given A and B in R, we know that the class of A will be equal to zero, if and only if A is an element of P. Now this is the class, so by this I mean the class of A in R mod P. This is by definition. So if I take the product of two elements in the quotient, this by definition is this, and so this will be equal to zero, this is equal to zero if and only if A times P is in P. So I mean these are all of these things. So thus we find that A is a zero divisor, if and only if A is not in P, which means that A is not zero, is your device is a non-zero element in non-zero multiple of which. So it's a non-zero element which we multiply with another non-zero element to get zero, and there exists another B in R without P such that AB is in P. So that means R has zero divisors if and only if. There exists elements A and P in R without P such that the product is in P, so that means if and only if P is not a prime ideal is. So this is equivalent to saying that it has no zero divisors if and only if P is a prime ideal, and so this proves part one. Now, second one, slightly more interesting. So first we will assume that M is an ideal such that R mod M is a field. Then it follows as R mod M is a field that the only ideals in R mod M are the zero ideal and the whole of R mod M. So we had this statement that therefore if we take a quotient, there was a bijection between the ideals containing, so if we take a ring and divide it by some ideal M, then there's a bijection between the ideals in R which contain M and the ideals in R mod M. So the statement then says that this is equivalent. So the ideals in R which contain M are precisely the inverse images of these two. So it means this statement is equivalent to the only ideals in R which contain M, R, M, and R. By definition, this is precisely the definition of a maximal ideal. So in particular, if we follow this, we have seen that if R mod M is a field, then M is a maximal ideal and we find it is equivalent if we can reverse this step. Because these two are equivalents, but here we have to see that this step is also an equivalent. So therefore, so thus, it's enough to prove the following lemma, which is actually also quite simple. So let R be commutative ring with one. So let R be ring whose only ideals are the zero ideal and R. So with zero and R is only ideals. Well, then we have to see that it is a field. Well, so what does it mean? If you have a commutative ring with one to be a field, it means that every non-zero element must be a unit. So it must have a multiplicative inverse. So we have to see that. So in order to prove this statement, we have to take an element A in R, a non-zero element in R, and we have to see that it has an inverse. So that A, let's say there exists a B in R, such that A B is equal to one. Well, now we have to somehow connect this to these ideals, because here we only have a statement about ideals. So the only ideal we can make out of our element A is obviously the ideal generated by A. So this is an ideal. And obviously, A is contained in this ideal. So it's not equal to the zero ideal. But the only ideals in R are zero and the whole of R. So thus, the ideal generated by A is equal to R. So in particular, we have that one is an element of A. But you have to remember what is A? The ideal generated by an element is just the set of all A B, such that B is in R. So in other words, it means that there exists a B in R, such that A B is equal to one. And so we have found our inverse. As you see, it's quite simple. So in some sense, it's almost a reformulation of the definition of maximal and prime ideal that the quotient in one case is a field, in the other case an integral domain. But as we shall see, the statement with this, as already mentioned, this kind of statement that if you have a maximal ideal, then the quotient by it is a field will be a method to construct new fields. Where am I? So as a corollary, we find that every, I mean, it would be easy anyway, but every prime ideal, every maximal ideal is a prime ideal. Now we know that if I take the quotient by a maximal ideal, it is a field. The field is in particular an integral domain. So the ideal was also prime ideal by the first part of the statement. So OK, now we want to look at very particular rings, which are just polynomial rings over a field, which will later interest us when we also do field extension in Galois theory. So maybe I can start here. So let small k be a field. And we want to study the polynomial ring kx. So kx is the polynomial ring which we had introduced before. We had introduced the polynomial ring with coefficients in any ring. And now we have to do it in particular case of field. What? I haven't thought about it. This glory as a statement can just add something every maximal ideal in a real ring beauty is a prime ideal for it. Yeah, well, I mean, I have not thought about it. You know, I don't expect, you know, I haven't thought about it, but I don't expect it to be true if the ring is not commutative with one. But you know, I think that. Yeah, anyway, so I. As a statement. So but I mean, OK, if you want, so in. But you know, I have made this general assumption. So this statement implies that. But in commutative ring. So what? OK, so if that makes you feel more at ease, I can do that. OK, so we look at this polynomial ring. And we will actually be, we want to study this polynomial ring Kx of polynomials with coefficients in K more carefully. In particular, we will be interested in questions of divisibility, so when one polynomial is divisible by another. And so before doing this, I want to introduce generally, so study the visibility of polynomials. So we'll be. Actually, most likely this will be a review of something that you have learned in high school, but at any rate, it's, you know, we want to have it in our language. So first, then I introduce the visibility in general for any integral domain. So definition, so let R be an integral domain. So then we take two elements, AB in R. We say that A, you know, it's the obvious definition, A divides B in R, which is denoted A divides B. If the obvious thing holds, there exists an element C in R such that A times C is equal to B. OK. That's not very surprising. There's a definition. And otherwise, we see, so if there's no such C, then we say that A does not divide B, obviously. And we write it as A does not divide B. OK. And there are some obvious properties. So for instance, we have that. So A, so obvious properties. So by definition, A divides B, if and only if B is an element of the ideal generated by A. And well, and so we also have that and it follows from this that if A divides B and B divides C, then it follows that A divides C. That's also trivial. From the definition, and if A divides B and A divides C, then follows that A divides B plus C. OK. So obviously, there's nothing to say. This is completely obvious from the definition. Now, for the divisibility of polynomial, I have to kind of remind you of R1. So you have something that, as I said, you learned at school. You first learned it for the integers. If you have two integers, you can divide one integer by another with the rest, whatever. 7 is equal to 1 times 5 plus 2. And you can also do it for polynomials. I think I'm pretty sure that you have made such computations of dividing one polynomial by another and then sometimes there was a rest by the typical algorithm, which is sometimes called the Euclidean algorithm, but I will reserve that name for something else. So the divisibility of polynomials is governed. Or you can check it by the algorithm of division with rest. That I think you must have all seen. I, however, formulated precisely and prove that it works. So theorem, division with rest. So let's take two polynomials with coefficients in a k. So k is still our field. And we assume that the second one is not 0. So not the polynomial, which just is the constant polynomial 0. So then there exists unique polynomials q in kx. This stands somehow for the quotient. And r in kx, which would be the rest, such that we can write f, f equal to q times g plus r. And the r has a smaller degree than g. And the degree of r is smaller than the degree of g. Well, r is equal to 0. By definition, if you have the 0 polynomial, it doesn't really have a degree. I mean, I could say the 0 polynomial has degree minus 1. Then this would always hold. So this is the statement. And you prove this by dividing with rest by the usual algorithm. But I want to prove this by an inductive argument, which also tells you again what the algorithm is. So we make some kind of induction. So if our polynomial f is equal to 0, or the degree of f is smaller than the degree of g, well, then we are already done. We can just put q equal to 0 and r is equal to f. And trivially, it's fine. So this is somehow beginning of the induction. So if the degree of f is bigger than the degree of g, then we proceed by induction on the degree of f, which I want to call m. And so we have to somehow find a way to replace f by something of lower degree and use the statement for that. And we basically just do the obvious thing. Subtract the leading coefficient with the leading term. So let's see. So let a be the leading coefficient of f. So just remind you, so if f, say, has degree m, then this means it's the coefficient of x to the m. And b, the leading coefficient of g. And we write down another polynomial, which I call maybe f bar, which is just we subtract a suitable multiple of g from f in order to lower the degree. So let's see. We take f minus a divided by b times x to the m minus the degree of f of g times g. Now, the degree of g we know is smaller equal to m, because that was our assumption here. So this is a positive number. So this thing is actually polynomial with coefficients in k. So this is an element in kx. And this is a polynomial of degree m. This is a polynomial of degree m, because this had the degree m minus the degree of g. And then we, this had the degree of g. And it says, if we multiply it by x to the m minus degree of g, so the degree becomes degree of g. So you have two polynomials of degree m. So it follows that f bar has at least at most degree m. So we have the degree of f bar. So it's smaller equal to m. And in order to check whether it's m or smaller, we have to see what the coefficient of x to the m in f bar is, whether it's 0 or not. So let's compute it. So the coefficient of x to the m in f bar is what? Well, so we just compute it here. Here it is the coefficient of x to the m in f, which is a minus a divided by b, the coefficient of x to the m of this thing, which is the same as the coefficient of, as the leading coefficient here, times this. So this is a divided by b times b. And so this is 0. So thus, the degree of f bar is strictly smaller than m, or maybe f bar is also 0. So therefore, by induction, we can deduce that f bar can be written in this way. We have that f bar is equal, say, to q prime times g plus r prime with the degree of q prime is, say, whatever I want it. No, I don't know. With such that either r prime is equal to 0 or the degree of r prime is smaller than the degree of g. Well, now I just put the r that I wanted to find here to be r prime. And the q equal to q prime plus this thing, a divided by b, x to the m minus c degree of g. Then by definition, I mean by what we have just defined here, we find that, obviously, r prime still fulfills this condition. And we have that f is equal to q g plus r. And we had that. OK. So this is the division with rest. And so I should maybe say, I've said there exists a unique such thing. So what I've proven here is, first, the existence. And so I still have to prove the uniqueness, but that's kind of trivial. So assume we can write this thing in two different ways. So assume f is equal to q g plus r and is also equal to q prime g plus r prime. Now these are different ones. With the same assumption as before, that these are polynomials and the degree of r prime. So it's equal to 0 or not 1. So for each r and r prime, it holds that either they are 0 or the degree is smaller than the degree of g as before. So then we have that. So these things are equal. So I can take the difference. So 0 is equal to q minus q prime times g. I take the difference of these plus r minus r prime. Or in other words, r minus r prime is equal to q minus q prime times g. So if q is different from q prime, then it follows that the degree of q minus q prime is bigger equal to 0. And so the degree, if you take the product of two polynomials, the degree is much bigger. So it follows the degree of r minus r prime is bigger equal to 0 times the degree of g. But this is impossible because both r and r prime have degrees smaller than the degree of g. So this is a contradiction. So thus q is equal to q prime. But now if q is equal to q prime, then r is just f minus this and the same for r prime. So also r is equal to r prime. OK, so this is simple. And as I said, if you look at how we prove this thing, we actually see that we have given an algorithm for making the division with rest. And the algorithm is precisely the one you learned at school. So if we take, so the proof gives an algorithm. We just see induction step is just one step in the algorithm for division with rest. So we can just do it if we have something like x to the 3 plus 4x plus 4x squared plus x plus 1. And we want to divide it by x plus 2. So we want to compute the division with rest of this. So we have to find x squared times the quotient of these two leading coefficients. So we get this is x squared. So we subtract this thing, x to the 3 plus 2x squared. So the difference will be 2x squared plus x plus 1. Now we have here 2x squared and we have x. So we have to take 2x. And so we subtract now 2x times this. So this 2x squared plus 4x. OK? And so if we subtract it, what we get? We get minus 3x plus 1. And so in order to get, we have here minus 3. And so this gives minus 3x minus 6. And the difference is 7. So the rest is 7. So this is how one does this division with rest. I mean, I think you have done this many times. It's also the same way how you do it with numbers. OK. Now when we have been talking about divisors, so we can also talk about common divisors and greatest common divisors. So let me define what greatest common divisors is and then use the Euclidean, this division with rest to compute the greatest common divisor of two polynomials. So definition. So r is, say, maybe still an integral domain. And we assume we have some elements, e1 to en, some elements in r. So an element r in r will be called a common divisor if it divides them all. Well, if a divides, if r, so this was en divides ei for all i equals 1 to n. And the greatest common divisor is one which is in a suitable sense, the largest of all common divisors. This actually means that every common divisor has to divide it. So a common divisor which is divided by all other common divisors. So r in r is called a greatest common divisor of e1 to en if it is a common divisor. And every common divisor of e1 to en will divide r. Then s divides r. So for instance, the greatest common divisor of 4 and 6 will be 2 in the integers. OK. So where are we? So it's clear by definition that the greatest common divisor is almost unique. It's not completely unique, but it's unique up to multiplying by a unit. Greatest common divisors up to multiplying by a unit. You have two greatest common divisors, s and r. Then s divides r and r divides s, which implies that s must be r multiplied by a unit. So now in our polynomial ring kx, we use this division with rest to compute a greatest common divisor. So in kx, use division with rest. This actually would also work in the integers with rest to compute a greatest common divisor. How does that go? So again, we take, so this is a polynomial mark. So let f and g be two polynomials. And we assume that g is, maybe they are both non-zero, whatever. Anyway, two non-zero polynomials. Now I write also r0 for f, r1 for g. And we make division with rest. So we divide f by g with rest. So we write f, so by division with rest, we have f, which is also r0, is equal to q1 times g plus r2. So r2 is the rest dividing f by g. Or to put it in this form, we have r0 is equal to q1 times r1 plus r2. And now, we reiterate this with r1 and r2. So we put, so as before, we have that the degree of r2 is smaller than the degree of g, which is r1. Or r2 is equal to 0. And now, we can reiterate this. So inductively, we define, so I do it once more. So I can say that r1 is equal to q2 times r2 plus r3 with g of r3 is smaller than the degree of r2, or r3 is equal to 0. This actually only obviously works if r2 is not equal to 0. Otherwise, it stops here. And so inductively, we have r, say, i minus 1 is equal to qi times ri plus ri plus 1 by division with rest. And the degree of ri plus 1 is smaller than the degree of ri. Or ri plus 1 is equal to 0. So we can inductively define all these polynomials as long as the rest that we get here is non-zero. Because obviously, we cannot divide by 0. So this defines us these polynomials, 0, r1, r2, and so on, until rn, until at the last step, it divides it. You can see this cannot go on forever, because the degree gets almost always smaller. So at some point, at the worst case, so the degree would become negative if it would go forever, which is not possible. So at some point, it must be that the rest is 0. So this procedure stops if, so at some point, when, say, rn minus 1, for some n, rn minus 1 is equal to qn times rn. And the rest is 0, rn, I don't know. So at some point, we will have this. And now the claim is that this last, so the rest of the previous division, this last one we get here, is a common divisor. So claim is a greatest common divisor, rn is a greatest common divisor of f and g in kx. So now we just have to prove that it's first a common divisor, and then that every other common divisor divides it. And we do this somehow out of this procedure. So we have always this equation, so we have always have this sequence of equations. Here, ri minus 1 is equal to qi ri plus ri plus 1. So if we start here, we first start with this one. This one says that rn divides rn minus 1. So we want to first show rn is a common divisor of f and g. So we see by definition, we have that rn divides rn minus 1 by this last equation. No, yes. And then we can look at the next one. We have, if we look, rn minus 2 is equal to qi minus 1, qn minus 1, rn minus 1 plus rn. So here you see that rn divides these two, and so it divides rn minus 2. So rn divides rn minus 2. And so if we go to the next step, at each step, we will have that if we go one further, we have that rn divides this one and this one, therefore divides this one. So inductively, we have rn divides ri plus 1, and rn divides ri, thus it follows ri divides ri minus 1. So it follows that rn divides all the ri. And so in particular, it divides r0, which is f, and r1, which is g, rn, of course. rn divides r0, divides all ri, and therefore rn divides r0, which is f, and rn divides r1, which is g. So it is a common divisor. And now we want to see it is a greatest common divisor. And here we have kind of to prove it's a common divisor, we started kind of from the bottom from the rn, and then we worked our way up. And to prove it's the greatest common divisor, we start from the top and work our way down through these equations. So to see, so let S be a common divisor of f and g. In order to show it's a greatest common divisor, we have to see that S divides rn. OK. So S divides f and g. So that means S divides f, which is r0, and S divides g, which is r1. And now we have defined, we had that r0 is equal to say q1 times r1 plus r2. So S divides this one, and S divides this one. So it divides also a difference, which is r2. So it follows that S divides r2. And now inductively we have this, that we have i minus 1 is equal to qi ri plus, what's wrong? And we know, and inductively we know, that S divides ri minus 1, and S divides ri. So it divides this term and this term, so it divides the difference, which is ri plus 1. So again, we find that S divides all the ii. S divides all ri, in particular, S divides rn. And therefore, rn is the greatest common divisor. So this method of finding the greatest common divisor is sometimes called the Euclidean algorithm. Sometimes also just the division with rest is called the Euclidean algorithm. Anyway, so supposedly it's, I think it must be written down in the elements of Euclide. So it's not the newest of all, maybe just for integers though. OK, so I want to use this to give a, I mean, as a consequence, I want to prove something which doesn't look very exciting at all. Which, however, we will use later to prove a very important theorem, is a theorem of the primitive element. So it's a rather trivial fact, which, however, we will want to use for something important. The statement is the following. Let k be a field, and let k be a subfield. So that just means that k is a field, and small k is a field, and small k is a subring of the large k. And k is a subring of k. So we make this assumption. And as a side remark, we note, I maybe should have written it before. Maybe I state the corollary afterwards. So note that in this case, if you take the polynomial ring with coefficients in k, in small k, this will be is a subring of the polynomial ring with coefficients in the large k. That's clear. If you take part of any two polynomials with coefficients in small k, it will lie here. But obviously, so this is obvious. So now the corollary under these assumptions. So let's take f and g to be two polynomials with coefficients in the larger field, kx. And we form. So the greatest common divisor of these two polynomials is not unique, but it's unique up to multiplying by a unit. So let H be a greatest common divisor of, I'm not wondering whether what I say is correct. Let me just check. So I think, no, no. OK. So actually, I just take the polynomial here. So now I take the greatest common divisor of these two polynomials as polynomials in the larger field. So I mean, in principle, in the larger field, in this, there are more common divisors. There might be more greatest common divisors. So if the leading coefficient of H is an element in the smaller k, small k, then H is a polynomial in small kx. Now, this is almost completely trivial, as we also will see from the proof. But at least in theory, if you just think of it, if you have a ring with two elements in a ring, you take their greatest common divisor in a ring, and you take a bigger ring and take their greatest common divisor, they don't need to be equal. But here, we have a criterion in this particular case. And if the ring is bigger, there might be just more elements. But here, it is like this. And so just to remind you, I had used this Euclidean algorithm, so division with rest, to compute a greatest common divisor of two polynomials. But it's not unique. It's one way to get the greatest common divisor. But the greatest common divisor is only well-defined up to multiplying by a unit. So there are. But then, we will now see that this is basically trivial by this Euclidean algorithm. So we do have one greatest common divisor given by the Euclidean algorithm. So let L be the greatest common divisor of F and G computed. So as it is computed by the Euclidean algorithm, so this division with rest, the Euclidean algorithm in large kx. So there's one way how I can compute the greatest common divisor of F and G. I mean, one instance of it by just applying this Euclidean algorithm. I get some polynomial. But if you look at it, how do you compute this greatest common divisor? You do it by successively doing division with rest. So if you start with two polynomials in small kx, you stay in small kx. It's always you take the rest, dividing one by the other. The fact that the ring in which you kind of think you are is the other kx doesn't affect the algorithm at all. So as L is computed by repeated division with rest from F and G, it follows that L is a polynomial with coefficients in small k. And now if I take any greatest common divisor of F and G, then it follows that kh must be equal to L up to multiplying by unit. But the units in a polynomial ring are the non-zero constants. So it follows that h is equal to A times L for A, an element in k without 0. Well, we can look at the leading coefficients. So we have let hn, the leading coefficient of h, ln, the leading coefficient of L. So then, no, I just multiply by a constant. I have hn is equal to A times ln. So if this leading coefficient is also in k, then by dividing by this, we find that A is in k. So if hn is in the small k, then it follows that A is in k. And so the polynomial, h is obtained from L by multiplying by an element in k. So it lies in kx. So I maybe should have called it k and L. But so there's a small k and the big k. And thus, we have that h, which is equal to A times L, is in kx, in small kx. So it's very simple. But as I said, we will see that it's actually useful. Now we want so much for this greatest common divisors. Now we want to, if you have a polynomial, you can evaluate it at an element in k or at an element in a bigger ring than k in a ring that contains k. And once you can evaluate it, you can also talk about zeros of the polynomial, namely, the elements in, for instance, in k where the polynomial is 0. So let me do this. So we can evaluate. So definition. So we have, so let f be equal to sum i equal 0 to n, ai x to the i be a polynomial with coefficients in k. And say let R be a ring that contains k as subring. So ring here, again, means commutative with 1. What? What? k is a field, yeah. So k is always a field in the whole story. So I started in the beginning that k was a field and kx was a polynomial ring. But so I mean, I don't really need that here. Let k be a field. But I had all the time assumed that k was a field. And R is a ring. So then, so for an element S in R, we can define f of S. Well, it just means you replace the variable x by the element S. And this x to the i means really multiplying S i times 3 itself. So sum i equal 0 to n, ai S to the i. So this is an element in R. So we can take the value of f at any element in this ring, R. In particular, at any point in k. And so obviously, as this is all the structure here with multiplications compatible with this, if we take the sum of two polynomials and we have evaluated at S, this will be equal to f of S plus g of S. Because the sum is just by adding the corresponding coefficients. And so it does the same. And the product of polynomials is defined in such a way that if I take the product of two polynomials and evaluate at S, this is f of S times g of S. This is why one had this crazy definition. So we see that evaluating at the point S is actually a ring homomorphism. So thus, I could call it the evaluation at S from kx to R, which sends a polynomial f to f of S is a ring homomorphism. So we have which one would call the evaluation at S. So for the moment, we will use this evaluation only for S action element of our k. But later, we will do it in other links, in particular in bigger fields. I just introduce it generally here. So now if we have that, so an element, say S in R, will be called a 0 of our polynomial if f of S is equal to 0. So if this is a 0 element in R. So this was general, we have this R. Now we go back to just having k. So take S in k. So then the first not very surprising result that I want to show you is that if an element A in k is a 0 of f, then f is divisible by x minus a. This is a very exciting new result. And then we will use this fact to prove that if you have a polynomial in kx, then the number of 0s it has in k can be at most the degree of the polynomial. And it's quite easy to see how one would prove that from this. So maybe I call proposition. So let f be a polynomial in kx. So maybe, I don't know. I think it even holds like that. So let's be a polynomial in kx and A in k. Then A is 0 of f if and only if x minus a divides f in kx. So that means that we can write f equal to x minus a times g where g is an element is another polynomial in kx. OK, so let's see how we can prove this difficult result. Well, the only thing we know is division with rest. So we do division with rest. So maybe first we do the trivial direction. And then we do the almost trivial one. So if x minus a divides f, so then we have f is equal to x minus a times g. And so if I take f of a, then we can just put a for x. This is a minus a times g of a. And obviously this is 0, so this is 0. So this direction is clear. So now let's look at the other one. For this we use division with rest. So we assume that f of a is equal to 0 for some a in k. So now we make division with rest by x minus a. So we have f is equal to x minus a times some g plus some rest. And the rest is either 0 or its degree is smaller than the degree of x minus a, which is 1. So that means r is equal to 0, or r is a constant polynomial, of a non-zero constant polynomial. That means that r is an element in k. OK, now we again, but we know now we can compute f of a, which we happen to know it's 0. Again, we already know how compute f of a compute such a product. This is a minus a times g of a plus r of a, but r is constant. So it's just r. So this part is 0. So it means that r is equal to 0. And therefore, we find that indeed x minus e a divides g. OK. And now, so as a corollary, we find this fact that the polynomial of degree n can have as most n zeros. Or let think f be a polynomial coefficient in x, which is not the 0 polynomial. Then f has at most degree of f zeros in k. Well, obviously from what we have here, that's trivial induction, essentially. So if whenever we have a 0, we divide a, we divide by x minus a, until we are left with nothing. So we make induction on the degree of f. So if the degree of f is equal to 0, then f is a constant polynomial. And it's a non-zero constant polynomial. f is a non-zero constant. Therefore, it has no zeros, because it's always the same. So that case is OK. So the degree is 0. The number of zeros is also 0. OK, now we make the induction step. So let f f degree n plus 1 for some n. So now, we want to show that f has at most as many zeros. So we start at this by saying, so if f has no zeros, then obviously our statement is true. And then we are done. So we can assume that f has a 0. So let's assume a in k is a 0 of f. Well, then obviously what we want to do is divide by x minus a, because we know f is divisible by it. f is equal to x minus a times g, where g is a polynomial in kx and the degree of g is one less than the degree of f. Thus we know that g has at most n zeros by induction. And so if, say, b is a 0 of f, then we have f of b is equal to 0, which is b minus a times g of b. Now, if a is different from b, so is a 0 of f different from a, then this is non-zero. So it follows that g of b is equal to 0. So thus, f has at most n plus 1, 0, namely all the zeros of g and addition a. OK, so this is in k. So this was a rather simple thing. Let me see how many minutes I have. So this was the end of this section. It's not quite clear. I think if it's fine with you, I would stop now because I wouldn't. What? What does it mean? Two minutes. OK, so another time I will maybe get it back with interest. And so I will stop now. Thank you. So I don't want to start a new chapter now.