 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 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. So it has no zero divisors, and 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, 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 and 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 integers 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 still, 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. And 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 a 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 the 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. So 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. This is the class. So by this I mean the class of A in R mod P. Okay, this is by definition. And 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, if and only if A is not in P which means that A is not zero. Zero divisor is a non-zero element, a 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 they're productized in P. So that means if and only if P is not a prime idea of this. And so this equivalently to saying that it has no zero divisors if and only if P is a prime idea. And so this proves part one. Now, second one, slightly more interesting. So first we will assume that M is an idea such that R mod M is a field. Okay, 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 projection between the ideals containing. So if we take a ring and divide it by some ideal M, then there's a projection 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. And 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 equivalent, 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 that 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, this is, 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 in B in R, such that AB 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 AB 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 these, 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. A field is in particular an integral domain. So it's, so the ideal was also prime ideal by this, by the first part of the statement. So, okay, 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 and 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 put in particular case a field. I haven't thought about it. Yeah. But this query as a statement can just add something in a maximal ideal in a ring with your piece is a final ideal for it. Yeah, well, I mean, I have not thought about, you know, the, you know, I expect, 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, yeah. Anyway, so I, are you, so, so, but I mean, okay, if you want, so in, but you know, I have made this general assumption. So this statement implies that, but in commutative ring. So ring with one, okay? So if that makes you feel more at ease, I can do that. Okay, 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 divisibility of polynomials. This will 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 divisibility 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, you know, there exists an element C in R such that A times C is equal to B. Okay, that's not very surprising as 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 do otherwise and we write it as A does not divide B. Okay, 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 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 and follows that A divides B plus C. Okay, 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. Seven is equal to one times five plus two. And you can also do it for polynomials. You know, I think most like, I mean, 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 we have the, 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. However, you know, formulate it 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 zero. So not the polynomial, you know, which just is the constant polynomial zero. 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, equal to Q times G plus R, you know. And the R has smaller degree than G. And the degree of R is smaller than the degree of G. Well, all R is equal to zero. By definition, if you have the zero polynomial, it doesn't really have a degree. I mean, I could say the zero polynomial has degree minus one, then this would always hold, okay? Okay, 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 if, so we make some kind of induction. So if our polynomial F is equal to zero 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 zero 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. We subtract the leading coefficient, 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 this has, we multiplied by X to the M minus the degree of G, so the degree becomes the 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 and if bar is whether it's zero 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 zero. So thus the degree of F bar is strictly smaller than M or maybe F bar is also zero. So therefore we can, by induction, we can deduce that F bar has such a, 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, what is it, what I wanted? I don't know, with such that either R prime is equal to zero 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 G, the Q equal to Q prime plus, where did I write it, 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 Veta. Okay, 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 from the beginning of the lecture, I've said there exists a unique such thing. So what I've proven here is from the beginning so what I've proven here is first the existence. Okay, 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 it's 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 and the degree of R prime. So equal to zero or not one. So for each R and R prime, it holds that either they are zero 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 zero 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 zero. And so the degree, if you take the product of two polynomials the degree is much negative. So it follows the degree of R minus R prime is bigger equal to zero times the degree of G but this is impossible because both R and R prime have degrees smaller than 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. Okay 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 and we just see induction step is just one step and the algorithm for division with rest. So we can just do it if we have something like I don't know x to the three plus four x plus four x square plus x plus one and we want to divide it by x plus two. So we want to compute the division with rest of this. So we have to find, so x squared times the quotient of these two leading coefficients. So this is, we get, this is x squared. So this is, so we subtract this thing x to the three plus two x squared. So difference will be two x squared plus x plus one. Now we have here two x squared and we have x so we have to take two x and so we subtract now two x times this. So this is two x squared plus four x, okay? And so if we subtract it, what do we get? We get minus three x plus one and so in order to get we have here minus three and so this gives minus three x minus six and the difference is seven so the rest is seven. 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. Okay, 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. E one to E n, 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, right, so this was A n divides E i for all i equals one 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 if it is of A one to N, if it is a common divisor. Every common divisor of A one to N will divide R, then S divides R. So for instance, the greatest common divisor in O four and six will be two in the integers. Okay, 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. So Mark, greatest common divisors up to multiplying by a unit. No, if 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 K X we use this division with rest to compute the greatest, so a greatest common divisor. So in K X, 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 nonzero, whatever. Anyway, two nonzero polynomials. Now I write also R zero for F, R one 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 R zero, which is also R zero, is equal to Q one times G plus R two. So R two is the rest dividing F by G, or to put it in this form we have R zero is equal to Q one times R one plus R two. And now we reiterate this with R one and R two. So we put, so with them, so where, you know, as before we have that the degree of R two is smaller than the degree of G, which is R one or R two is equal to zero. And now we can reiterate this. So inductively we define, so I maybe can, so I do it once more, so I can say that R one is equal to Q two times R two plus R three with the degree of R three is smaller than the degree of R two or R three is equal to zero. This actually only obviously works if R two is not equal to zero, otherwise it stops here. And so inductively we have R, say I minus one is equal to Q I times R I plus R I plus one by division with rest, and the degree of R I plus one is smaller than the degree of R I or R I plus one is equal to zero. 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 zero. So this defines us these polynomials are zero, R one, R two, and so on until R n, 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 you, 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 zero. So this procedure stops if, so at some point when say R n minus one for some n, n minus one is equal to Q n times R n and the rest is zero, R n, 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, so this last one we get here is a common divisor. So claim is a greatest common divisor, R n 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 this sequence of equations. Here R i minus one is equal to Q i R i plus R i plus one. So if we start here, first start with this one. This one says that R n divides R n minus one. So we want to first show R n is a common divisor of f and g. So we see by definition we have that R n divides R n minus one by this last equation. And then we can look at the next one. We have if we look R n minus two is equal to Q i minus one, Q n minus one, R n minus one plus R n. So here you see that R n divides these two and so it divides R n minus two. So R n divides R n minus two. And so if we go to the next step, at each step we will have that if we go one further, we have that R n divides this one and this one therefore divides this one. So inductively we have R n divides R i plus one and R n divides R i plus it follows R i divides R i minus one. So it follows that R n divides all the R i and so in particular it divides R zero which is f and R one which is g. R n divides R zero, divides all R i and therefore R n divides R zero which is f and R n divides R one which is g. So it is a common divisor. And now we want to see it is a greatest common divisor. So here we have kind of to prove it's a common divisor we started kind of from the bottom from the R n 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 the greatest common divisor we have to see that S divides R n. So S divides f and g. So that means S divides f which is R zero and S divides g which is R one. And now we have defined we had that R zero is equal to say q one times R one plus R two. So S divides this one and S divides this one. So it divides also difference which is R two. So it follows that S divides R two. And now inductively we have this that we have R i minus one is equal to q i R i plus what's wrong. And we know and inductively we know that S divides R i minus one and S divides R i. So it divides this term and this term divides the difference which is R i plus one. So again we find that S divides all the i i. S divides R i in particular S divides R n and therefore R n 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, you know, 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. Okay, 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. It's 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 small 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. Okay, and k is a subring of k. Okay, so we make this assumption and as a side remark we note and 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 and 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. No, no, okay. 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 and there might be more greatest common divisors. So if the leading coefficient of H is an element in the smaller 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 their greatest common divisor they don't need to be equal. But here we have a criterion in this particular case and you know 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 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 by 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 now but if you look at it you just 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 a 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 zero well we can look at the leading coefficients so we have that 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 okay now we want to so much for this greatest common divisors now we want to you know 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 zero so let me do this so we have we can evaluate so definition so we have so let f is equal to sum i equals zero to n ai x to the i be a polynomial coefficients in k and say let a be a ring that contains k as sub ring so ring here again means commutative with one 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 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 i 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 itself so sum i equals zero 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 of any point in k and so obviously you know as this is all the structure here the equation is 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 everywhere at s 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 evaluation at s from kx to r as a polynomial f 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 actually an element of our k but later we will do it in other in particular in bigger fields and just introduce it generally here so now if you have that so an element say s in r will be called a zero of our polynomial if f of s is equal to zero so if this is a zero element in r and now now we so this was general we have this r 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 zero 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 zeros it has in k can be at most the degree of the polynomial okay 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 why I don't know I think it even holds like that so let's be polynomial in kx and a in k then then a is a zero of f if and only if x minus a divides f in kx so that means that we can write f equal to fx minus a times g where g is an element is another polynomial in kx okay so let's see how we can prove this difficult result well you know 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 so if x minus a divides f so then we have f is equal to x minus a 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 zero so this is zero so 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 zero 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 what some g plus some rest and the rest is either zero or its degree is smaller than the degree of x minus a which is one so that means f of a is equal to zero or r is a constant polynomial of a non-zero constant polynomial it means that r is an element in k okay now we again but we know now we can compute f of a which we happen to know it's zero again we already know how 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 zero so it means that r is equal to zero and therefore we find that indeed x minus e a divides g okay and now so as a corollary we find this in fact that the polynomial of degree and we can have as most in zeros or let think f be a polynomial coefficients in x which is not the zero polynomial then f has at most degree of f zero in k well obviously from what we have here that's trivial induction essentially so if whenever we have a zero 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 zero 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 okay so the degree is zero the number of zeros is also zero okay now we make the induction step so let f, f degree n plus one for some number 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 then we are done so we can assume that f has a zero so let's assume a in k is a zero of f well then obviously what we want to do is divide by x minus a because we know f is divisive 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 zero of f then we have f of b is equal to zero which is d minus a times g of b now if a is different from b so is a zero of f different from a then this is non-zero so it follows that g of b is equal to zero so thus f has at most n plus one zero namely all the zeros of g and in addition a ok so this is 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 what does it mean two minutes but 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