 So, good afternoon. You had some exam today? What? Awful? Well, I hope for the best. What? It was not so good. Well, anyway, we'll see. Okay, maybe... So, you're already unhappy? Yeah. Okay, I will have a look. Okay, so maybe now I start all the same. So, we wanted to talk about... I said you wanted to talk now about finite fields. This was just some kind of interlude, so that we see that finite fields, at least we can understand completely. And so, the idea is we want to classify the finite fields up to isomorphism. So, we know... So, if p is a prime number, then if I take zp, which is a field I call fp, with a usual multiplication, so the number's modulo p, with an additional multiplication, is a field, and it has p elements, which I can identify with the numbers from 0 to p minus 1. And so, now we want to show, up to isomorphism, so for every prime pump power, p to the n, there is a unique field with p to the n elements. So, when there are q, there is a unique field with p to the n elements. And there are no other finite fields. So, if a field is finite, it has to have p to the n elements for some prime p. Okay. So, first, we know that the characteristic of a finite field must be a prime number p, because the characteristic, if the characteristic is 0, then the field must be infinite. And so, let's say f, so maybe this is called a lemma. So, anyway, for the rest of this section, so let p be a prime number, and n a positive integer, and q equal to p to the n prime power. And so, let we take a field of characteristic p. Then, if we look at this, the following polynomial, the polynomial x to the q minus 1, so q is p to the n as here, has precisely q simple roots. So, simple zeros in its splitting field. So, over fq, over f. So, if we look at the zeros in any extension of f of this thing, there will be q simple roots. So, as many roots as a degree and all are different. This is very simple. I just have to see that there are no multiple roots. So, if a is a multiple root, then we have seen last time that both the polynomial and the derivative of the polynomial have to be vanish at a. So, then x to the p minus 1. So, if I take x to the p minus 1 applied to a is equal to 0 and if I take the derivative of x to the p minus 1 applied to a, this is equal to 0. But this polynomial is the constant polynomial minus 1. This is equal, identically equal to minus 1. So, therefore, this will never happen. So thus, we have no multiple roots. So, we can see that, we see that this polynomial has always q simple roots. And we will see that actually we find all the finite fields as the splitting fields of this polynomial over f p for different values of n. So, let's start with that. First, we want to show that there exists a field with q elements. So, let's say q which was p to the n elements for every prime power p to the n. Well, so we just take the following. We look again at this thing. So, let k be the splitting field of this polynomial x to the q minus 1 over f p finite field f p with p elements. And we put f to be inside the splitting field just a set of all the roots of all the zeros of this polynomial. Say a in, yeah, except I made a mistake and you should have noticed. I mean, there's a misprint in not in the notes but in what I've written here. Obviously, it's x to the p. I have x to the p minus x because you can see if I take the derivative of this one, this will be p which is 0. But we wanted to be, I wanted it to be minus 1 and this is if I have x here. If I take the derivative of x to the p minus x, this will be minus 1. So, this is the polynomial I want. But so, this maybe shows I wasn't too fast because you didn't notice. Anyway, so we take this polynomial here too. We take this splitting field of this. So, this is the set of all in k, a in k such that a to the q minus x to minus a is equal to 0. So, now we want to see that this is a field. So, claim f is a field. We know that it has q elements because that is just what we said here. There are q different roots. Well, you just have to see the field axioms are satisfied. So, obviously we have 1 to the q minus 1 is equal to 0. And we have, if we have a to the, so if a is an element of f and b is an element of f, then we have a to the q minus is equal to a and b to the q is equal to b. And so, if we take a b to the q, this is a times b by just applying this. So, it follows that a times b is an f. And if we take minus a to the q, we need to actually know, if we take a to the minus 1 to the q, this obviously is a to the q to the minus 1 is equal to a to the minus 1. And so, also a to the minus 1 is in f. And what is more interesting or what is, comes from the fact that we are over a finite field is that also the sum is there. So, if I take a plus b to the q. Now, you can compute this by the binomial coefficient obviously. So, this is a power r i sum i from 0 to q, q through i a to the i b to the q minus i. But it's a very easy fact to see that if q is a prime number that all these binomial coefficients are divisible by q or at least by p, they are I think also divisible by q if i is not equal to 0 or q. So, this is equal to 0. So, here we are computing in this is an element in this field. So, we have a field of characteristic p. So, p is equal to 0. Any multiple of p is equal to 0. So, this binomial coefficient will be equal to 0 unless i is equal to 0 or q in which case it's 1. So, this thing is just a equal to the q plus b to the q. And this we know is equal to a plus b. So, also a plus b is in f. And let me see what I have. And it's the same if we have instead of plus if you put the minus we get the same thing. So, here of i is equal to q. So, if q is equal to a power of 2 then the characteristic 2. So, a plus b is equal to a minus b. And if q is not that, is a power of an odd prime then q is odd. And then if I put the minus here I get the minus here. So, in any case I can put the minus here. The minus here. Okay. And so putting all this together we get thus f is a field with q elements. So, now what we want to see that f actually is equal to k that f is the splitting field of this polynomial. And as such it will be uniquely determined. So, at least we have now a field with q elements. And now we want to see. So, anyway it is clear just to remark this that if I already having said it f is a finite field. So, then it must have finite characteristic so assume of characteristic p. So, then I can look at the subfield generated by 1. So, then f contains the set of all n times 1 where n is in z. So, just this means you add 1 n times to itself as before as a power of the subset but it is a subfield because if you add or subtract such things or multiply them you stay in this realm as a subfield. And this set maybe I call this f 0 and f 0 is isomorphic or essentially equal to f p. It is a field with p elements. We have seen that the field has characteristic p. So, p times 1 will be equal to 0. And that is the smallest number for which is equal to 0. So, this thing here is just equal isomorphic to f p. And f then must be a field, you know f contains this as a subfield so f is an extension of this, of f p. And as it's a finite field it's a finite extension of f p. So, thus f is a finite extension of f p. So, in particular if it's a finite extension of f p it's a finite dimensional p vector space, f p vector space. So, in particular f is a finite dimensional f p vector space because we have always seen that the field extension and the bigger field is a vector space over the smaller one. And so, thus in particular the number of elements so say I write like this the number of elements of f is equal to the number of elements of f p which is p to the power the dimension of this to the n power n where n is the dimension of f as f p vector space which is just equal as we know to the degree of the field extension. Okay, so in particular we know that a finite field must have as an order the number p to the n for some n and the prime p and that it is a finite extension of it. And now we want to show that it actually must be as said the splitting field of this polynomial I've written their proposition. So, let f be a finite field with p to the n elements. p to n was q elements then we say it is the splitting field of this polynomial over f p x to the q minus x over f p. So, which then in particular means that funnily enough this will show that our finite field with f to the p with q elements consist precisely of all the zeros of this polynomial. So, we have to see basically we want to see that every element in f must satisfy this equation a in f satisfies a to the q minus a is equal to zero. So, obviously so we start very slowly. So, if a is equal to zero then this is true. So, we have already checked it for one element. So, now we have only q minus 1 to go. So, but now we look at f without zero so with the multiplication so the multiplicative group of f this is a group. So, is a finite group with a you know one element less obviously q minus 1 elements. Now, you know generally if you know we already have seen this near the beginning of the course if g is a finite group and we take any element g in g. So, a finite group say of order k and if g is an element in g well then I claim if I take g to the k this is equal to the identity element in g to the unit element. So, the neutral element in g and why is this now I can look at the subgroup generated by g is a subgroup and so as it is a subgroup its order is a divisor of the number of elements of g. So, divides this number k on the other hand we also know that the subgroup is just given by taking all the powers of g. So, we have that the order of g which by definition is the smallest in z bigger than 0 such that g to the n is equal to 1 is equal to the order of this group because we had seen that the elements are just 1 gg squared until g to this order minus 1. So, we have that so thus we have that the order of g divides our number k and now if we take g to the k so I can write k equal to L times the order of g so this will be equal to g to the order of g to the power L this 1 to the L is equal to 1 so therefore we see that g to the k is equal to 1 so anyway this I don't remember whether we covered this before but it's always true if you find a group and any element then if you take the power of the group you get 1. So thus now we go back to our situation we take an element a in f without 0 if you have an element a in f without 0 then if you take a to the q minus 1 then by what we have just seen this is equal to 1 so if you multiply this by a you have a to the q is equal to a so thus we have shown that for all elements a and f identity holds so now obviously as this polynomial x to the q minus x has degree q it can have at most q roots in its splitting field so it means the polynomial splits into linear factors over f x to the q minus x splits over f into linear factors now because we have all the 0's and it's the product of the factors of all the linear factors for all the 0's and so and on the other hand we have but then f is the splitting field of this polynomial because it consists only of 0's of it if you remember we had the statement that if we have a field over which the polynomial splits so we start with the small field k and we have a big field l so in l we have a polynomial in kx which splits over l then if we join all the 0's of this polynomial to the smaller field we get precisely the splitting field and this we have done here we have joined all the 0's of this polynomial to fq it happens to be that these are also all the elements of the field but that doesn't make a difference this thing does not contain contains only roots of this thing it splits and it doesn't contain anything which you cannot express in terms of roots because in fact it only consists of roots okay so we have found so where are we? so thus it is the splitting field so we can put it together so we know that the splitting field of a polynomial for a field is uniquely determined up to isomorphism so therefore this the finite field of p to the n elements is uniquely determined up to isomorphism so we can just put what we have learned in a theorem together so finite fields with say p to the n elements exist only finite fields with q elements exist only if q is a prime power p to the n and for each prime power p to the n there is up to isomorphism a unique field with p to the n elements namely the splitting field of this stupid polynomial so this is now x to the p to the n minus x over fq over fp okay so this is what we proved and so one can say that we have understood everything that there is to say about the classification of finite fields so so there is not so much we have this kind of list for every prime power there is one up to isomorphism and anyway so it's not such a complicated story so in some sense from now on we can mostly restrict ourselves to fields which are not finite okay so that's what I wanted to say about this is more like an aside show that one can at least completely understand the finite fields now we want to go on with the general theory and we want to introduce the Galois groups so later we will want to understand field extensions in terms of their Galois groups and the Galois group of a field extension L over K is the group of K-automorphisms of L so let me do this now so Galois groups we want to understand a little bit about them and we want to introduce them so as I said so if L over K is a field extension then the Galois group so the Galois group after Galois the French mathematician is of L over K is I write it Gal L over K is the group of K-automorphisms of L so it's clear that these are automorphisms of L so they form a group by a composition as automorphisms always do and you recall that a K-automorphism is an automorphism of L which is the identity on K which sends every element of K to itself so these form a group and we want to use this group to understand this field extension now in the moment that doesn't look particularly good because this seems a rather unconcrete thing you have this field extension and you have this group of automorphisms smaller that seems to be something very complicated which is impossible to understand so we want to see that you can say something about it so in particular if L is a splitting field of some polynomial with coefficients in K then this Galois group will turn out to be a subgroup of the group of permutations of the roots of this polynomial so it's certainly in particular will be a finite field but you know it's concretely described as a permutation group but this we have to prove and there's also another description so we will find two different descriptions of this Galois group as a permutation group which depending on the context are useful so let's first say I recall something we said in the group theory part so if you have a finite so for a finite set we had introduced what I call S of M the group of permutations of M which was just a set of all bijections of M to itself and obviously if M has N elements then this is isomorphic to the symmetric group on N letters we also write it so if number of elements of M is equal to N then SM is isomorphic so SM so and recall also that we had talked about actions of groups on a set and so just one word which we had used an action of G on a set M was called simply transitive we only need that now if for any two elements here there is an element in G which sends one to the other so if for all M1 and M2 in M there is there exists a unique element G in G with if I apply the action of G to M1 I get M2 so the action is you recall that the action was called transitive if there always exists a G which sends any M1 to any M2 and if it's unique it's called simply transitive and it's easy to see and we had seen it that it then follows that the number of elements of G is equal to the number of elements of M just fixing one element of M we can identify G we find the bijection of G and M this was all things that we had before and so now we have the first statement about the Galois group for a simple algebra extension theorem so we take a simple algebra extension and we want to identify say something about the Galois group so we say of degree N so that you remember that and so we have our element A so we take the minimal polynomial so then we find that the Galois group will be subgroup of the group of permutations of the roots of the minimal polynomial and actually it will be subgroup which acts simply transitively so it will have precisely as many elements as this thing has zeros so let's see it so so let P the set of zeros of this polynomial so the set of all B in KA such that F of B is equal to zero so this is the set of roots of F in KA so then we find that the Galois group of KA over K acts simply transitively on R in particular it is isomorphic to a subgroup of the symmetric group of R and the number of elements of the Galois group is equal to the number of elements of R and F was a minimal polynomial of A the field extension has degree N so this polynomial has at most N zeros so it means that this number R is smaller equal to N so we know that this Galois group of KA over K has at most N elements and it can be identified with the subgroup of the permutations of the roots of this polynomial okay so this we want to see so we first have to see that the Galois group will kind of permute these roots so that an element of the Galois group will always send a root to another root I mean to another or the same root but it doesn't send a root to something which is not a root so proof so we take an element in the Galois group and so if we write F equal to sum AI so F was a polynomial polynomial X to the I then if we if B is an element is an element in R so is a root of this polynomial then if we take Phi of F of B F of B is equal to 0 so this is Phi of 0 and thus 0 we can write this 0 in a more complicated way namely we apply Phi to the polynomial so this is Phi of sum I equal to 0 to N AI A B to the I and now Phi is a ring homophism so it pulls into the sum the AI are in K so it is the identity on them so this is equal to Phi of sum I equal to 0 no, like this AI Phi to the B I so this is nothing else as F of Phi of B so we see that if B is in R then also Phi of B is in R for any element in the Galois group so thus Phi of B is in R so in particular we see thus so if I take Phi and restrict it to R maps R to itself and obviously Phi is an automorphism of fields so it is a bijective map particularly it is an injective so also the restriction will be injective because Phi is and as R is a finite set and they have a finite injective map of a set to itself it is bijective so in other words so Phi restricts it to R is bijective so it means Phi restricts it to R is an element of the symmetric group on R now I want to see that the map which sends Phi to this restriction is actually and that this element is simply transitive so so we can say thus we have a map we have a group homomorphism because in both cases the group structure is by composition so it is the same which I couldn't call the restriction to R from the Galois group of L of K A over K to S of R so as I said we want to show that this is you know so as I said this is a group homomorphism because here the you know the composition of automorphism is just the automorphisms of maps and here is also the composition so it is just the same thing the group structure so Phi composed with P C restricted to R is obviously equal to Phi restricted to R composed with P C so we have a result this was the unique extension of automorphisms to simple algebraic equations which we proved at the beginning of our field theory adventure that so that there is a that for every root if whenever I have a root the two roots then there is always a unique extension of the identity so a unique K isomorphism of K A which sends one root to the other so we have seen for unique extension field isomorphisms to simple extensions that for for any two roots that for P 1 P 2 in R there exists the unique so this means unique Phi in the Galois group of K A over K with Phi of P 1 is equal to P 2 this was the theorem that we had so this precisely says that the this group the Galois group acts simply transitively on R because that says it for any so the action is by Phi applied to any D is Phi of P and the action is simply transitive so thus Phi there is the Galois group of K A over K acts simply transitively on R but if this is the case then it means also that this map here so that this restriction map must be injective because the the kernel of the restriction map consists of all elements in the Galois group which act as the identity here and according to what we have said here in particular this sent one P 1 to itself and the thing that the only thing that does this is the identity according to the fact that this acts simply transitive so the kernel consists only of so the identity on K 8 so the map that sends every element in K 8 to itself so well and so we have found that the Galois group of K A over K isomorphic to subgroup of S of R which acts simply transitively on R as it acts simply transitively on R it follows that the number of elements in this group is equal to the number of elements in R and as R is a set of roots of this polynomial F the polynomial F has degree N so the number of roots that it can have is at most N now you have to remember that the degree of the field so we had K A over K was field extension of degree N and F was a minimal polynomial so the degree of the minimal polynomial is N so the minimal polynomial has at most N roots in K A and this number R is smaller equal to N so we have precisely proven what we claimed so I want to so I want to give some examples very simple so moment we only know that the number of elements in the Galois group is at most N the field extension so let's look at two cases to see that actually different things can happen so if we take the Galois group of C the second root of 5 of Q or whatever it can also take second root of 2 over Q then you can see the polynomial of this is X squared plus X squared minus 2 and this splits into X plus second root of 2 and X minus second root of 2 and it is clear we have inside the we I claim that the Galois group this will be equal to no first we can have the identity and then we have the map which exchanges these two roots so that means if I have an element A plus B squared of 2 this can be sent to A minus B squared of 2 so it's easy to see that this element this map is field isomorphism and it's the identity on Q obviously and so this certainly these elements are in the Galois group and by what we just saw we see that the Galois group has at most two elements so this is the Galois group okay it's a very simple okay so if it was always as simple as that wouldn't have to worry now we can look at a more different example which instead take the third root of 2 then this Galois group is very small namely it contains just one element the identity so we see that the minimal polynomial of this thing third root of 2 over Q is you know obviously x to the third minus 2 but you find that in Q third root of 2 you know this polynomial has no other you know has no other roots in fact you know we had already an example where we had seen that you can if you want to find the splitting field if you find another root you have to actually add a complex number to it this e to the 2 pi i divided by 3 I think so here there's no other 0 of this polynomial so now the theorem says that the Galois group is a subgroup of the of the symmetric group on the roots there's only one root the only sub you know the symmetric group on one element is one element so it follows that the Galois group can only be the identity so but in some sense we will soon we now want to restrict our attention to so called Galois extension which are separated and separable and normal extensions and so in that case we find that the Galois group will always have order equal to the extension to the degree of the field extension and it will be easier to study so let me see so okay so now we want to introduce Galois extensions so a field extension L over K so this is a definition after all is called normal is a Galois extension if it is what we need we need finite or something well maybe I don't know whether that's actually part of the definition but we only deal with finite extension so finite field extension is called the Galois extension if it is separable and normal and you have to remember that in characteristic p every field extension is separable so then the condition would only be normal so we get as a corollary so let L over K be a Galois extension of degree N so then the Galois group is isomorphic to a subgroup of Sn so which acts simply transitively on the you know numbers 1 to N so the difference of the previous cases that now we really have and so in particular the number of elements in the Galois group is N so I could also say it like this the number of elements in the Galois group of L over K is equal to the degree of L over K which after all is N so before we had seen this thing where we have it simply that it is asomorphic to a subgroup of a symmetric group of on the roots and now basically we only have to see that there always will be precisely N roots of this minimal polynomial well let's see minimal polynomial of what so let's have a look so proof so we have we can our Galois extension you know by the theorem of the primitive element we can always take it as a simple algebraic extension for every separable extension there is a primitive element for every so by the theorem of the primitive element there is an element A in L such that you can write L equal to K of A so we find so our extension is a simple algebraic extension so we are in the situation of the previous theorem so we know that as so let's say let F be the minimal polynomial polynomial of A over K so this field extension L over K is normal and we have and F is an irreducible polynomial which has a 0 in L it follows that it splits splits over L into linear factors and as the field extension is separable and F is again an irreducible polynomial which has a 0 here you know all its roots are distinct of F in L are distinct so the set R the set of roots has precisely N elements all that we need to know so thus R which was set of all B in L such that F of B is equal to 0 has R elements R equal to N and so therefore the Galois group of L over K is isomorphic to subgroup of the symmetric group of R which is the isomorphic to a symmetric group of on N letters which acts simply transitive that was what the previous theorem said okay so this is this so now we come now we have another proposition we want to see so this would be mostly lemma we want to use that so until now so we know that this Galois group acts on this root of this polynomial but we want to also have some positive results about existence of elements in the Galois group and so we have the following statement so again we have a Galois extension and we take two elements A and B and L so then the statement is that there is an element in the Galois group which sends A to B if and only if A and B have the same minimal polynomial so we know precisely when the Galois group sends an element to another one this is if and only if A and B have the same minimal polynomial K I mean obviously it's normally very easy to find out what the minimal polynomial of an element is so it actually will be I think mostly use the other round so if we somehow can prove that two things so yeah we well anyway maybe I will say that later so now let's see how to see that it's not particularly difficult it's a variation of things we had before you know again about extension of field isomorphisms and so on so if A and so we have obviously two directions so first we assume that A and B have the same minimal polynomial we have to find an isomorphism we have to find an element in the Galois group now certainly we have that we can we have this statement of the extension of field isomorphisms to the for simple algebraic extensions so by the extension actually the unique extension of field isomorphisms for simple algebraic extension statement was that if two elements have the same minimal polynomial then there will be an isomorphism from K of A to K of B which sends a unique isomorphism from K of A to K of B when it sends A to B but we just need that there exists one we don't need the uniqueness so there is a is a K isomorphism C from K of A to K of B with is equal C of A is equal to B so now we have that L over K is a Galois extension so L is the splitting field of some polynomial with coefficients in Kx because it's a normal extension we had seen that if you have a normal extension an extension is normal if and only if it is the splitting field of some polynomial with coefficients in this model so as L over K is normal we have that L is the splitting field of some polynomial and we call it F in Kx over K now we know that if if something is a splitting field of a polynomial over some field then it will also be the splitting field of the polynomial over any intermediate field of F over K and the splitting field of F over KB because these are intermediate fields and so it is also splitting field over these but then we had this other extension theorem if you have an isomorphism between two fields then it can be extend so and you take a polynomial here and you take the image under this morphism of the polynomial here then it extends to an isomorphism of the splitting of the polynomial here the splitting field of that so in this particular case the polynomial is defined over K so if I apply Psi to this polynomial to the coefficients of the polynomial I get the same polynomial so therefore the extension the statement about the extension of isomorphisms to the splitting field says there is an extension of Psi to L which is the given map here by the extension of isomorphisms to the splitting fields there exists an isomorphism Phi from L to itself with Phi restricted to KA is equal to Psi so I hope you remember the statement here we were it was a somewhat complicated statement so we have we have a field we have two fields so I mean just just to remind you we have so F is in KX we have an isomorphism Phi from K to K prime and we apply so we take Phi star of F is the polynomial where you apply to the coefficients of the polynomial this isomorphism and then the statements in then we take L the splitting field of F over K and L prime the splitting field of F prime over K prime so I call this F prime I call it Phi star F over K prime then we had seen there exists an extension so there exists Phi from L to L prime with Phi restricted to K is equal to small Phi this was a general statement and we have applied it here what is called K there is K of A and what is called K prime there is K of B so what we have found so anyway we have this that Phi restricted to K is equal to Phi but this Psi was a K isomorphism so it means if I restrict Psi to K if I restrict it further to K it is the identity on K so it means that Psi is an element in the Galois group so as Psi restricted to K A is equal to Psi so Phi Psi restricted to K is equal to identity on K it follows that Phi is an element in the Galois group over K and by definition sending A to B so one direction we showed have the same minimal polynomial then there is an element in the Galois group which sends one to the other so now we want to show the other direction which is a bit simpler maybe I hope you maybe I will wipe it out here so that you can remember the statement that's okay so we want to show the converse so we take an element in the Galois group which sends A to B so now we have to find that A and B have the same minimal polynomial well that's not so difficult we take F and G minimal polynomial of A over K and G minimal polynomial of B over K then I can use the same simple observation as before which I maybe write down once more so we have 0 is obviously equal to F of A therefore it's also equal to phi of F of A and by the same computation as before this is the same as F of phi of A which is F of B so we see that B is also 0 of F but G is the minimal polynomial of B so it follows that G divides F G divides F but obviously I could have done the other round I could exchange the role of A and B and say the same for G of B is a 0 so phi of G of B is a 0 of G and by taking phi to the minus 1 so in the same way exchanging the role of A and B we have that F divides G so as F and B F and G are both monochromes it means they are equal so this proves this proposition and so much time up, basically no time so let me see okay so I will maybe at least I can even prove it so the next result so maybe I first make a definition so with the the next result which is quite easy to prove will be one of the will be half of the main theorem of Galois theory and basically we have worked until now to prove it quite simple but anyway maybe I first make a definition so let I really want to make a definition the Galois group the Galois extension and let G be a subgroup of the Galois group then we can consider a fix set of this thing so then that fix of G be the set of all elements A in L such that V of A is equal to A for all so we have we'll see later that this is actually always a field but it doesn't matter now so we have the we can look at the set of all elements in L which are fixed under all elements in G where G is a subgroup so the moment we are only interested in this in the case G is a subgroup of the Galois group okay so maybe I call it maybe I'll also call it Phi because the Galois group is always called Phi okay so the theorem that we want to show we'll later see that it's an important step is so if you have a Galois extension we can ask ourselves what is this fixed field if we take the whole Galois group so then the fixed of the Galois group of L over K this is just K so an element in the Galois group will be an element in the field a big field will be fixed under the whole Galois group if and only if it is in the smaller field of the field extension let me see I think that's fairly we can still prove it it's fairly simple so we have two inclusions so this inclusion is clear because obviously every element in K is fixed under the Galois group by definition the elements in the Galois group are the identity on K and now conversely so we assume that Phi of A so A is an element in L that Phi of A is equal to A for all Phi in the Galois group K so we take the minimal polynomial of A over K that even makes sense if A lies in K because then it's just a linear polynomial so as L over K is a normal extension we know that the Galois group splits that F splits over in L so any other root of F in L then we have seen there exists an element in the Galois group so it sends A to B in the Galois group with Phi of A is equal to B it's actually here so we have made the assumption that our element that Phi of A is equal to A for all elements in the Galois group thus A is equal to B by our assumption that A is fixed under all elements in the Galois group so it means that all the roots of F are equal on the other hand L over K it was a Galois extension, it's also a separable extension and F was a minimal polynomial so F is irreducible in K and has obviously 0 so in L so it follows that it has no multiple roots in L so for one thing all the roots are equal on the other hand it has no multiple roots that means there's only one root this polynomial is a polynomial of degree 1 a linear polynomial and therefore A is in K and this proves this result okay so this maybe is enough as you see this was not so particularly difficult but one always you know one always kind of uses the same thing but always in a each time in a slightly different way so anyway so this is this and next time we will at least start to finish the proof of the principle theorem of Galois theory okay so any questions okay thanks