Week 10-Lecture 52

So, in the last lecture we saw that every field has an algebraic closure Now the question is how many So now we prove sort of uniqueness, it is not unique but it is unique up to nK isomorphism So let me state it precisely, then really prove it So theorem, this will also Steinitz, let K be a field, then, 1 let E over K be a field extension with E algebraically closed In the last lecture we indicated that how to construct such E. Okay Then, let L over K be an algebraic extension, may not be finite, then there exists an K embedding from L to E Sigma K embedding means, see, this is an algebraic extension, this is contained here, so we can extend this, this Sigma is K Linear and embedding means it is a field homeomorphism So such an embedding exists for every algebraic extension, that is Part-1 And part 2 we say that, if E and E Prime are two algebra closures of K, then there exists a K isomorphism Sigma from E to E Prime, that means this computes with its K Linear and it is a field homeomorphism So that means this Sigma is indeed an element in Hom K alg E, E Prime This is K algebra homeomorphism from E to E Prime, which is also, which is K Linear, that is K algebra homeomorphism So that means that E and E Prime are unique up to this K algebra your isomorphism So let us prove this So proof, again proof will involve Zorn’s lemma So, proof All the theorems, one cannot prove without using the Zorn’s lemma So, consider the set M, this is the set M of all pairs of M, Sigma such that M is an intermediary field of L and K and the Sigma is the K embedding from M to E, K embedding

That is it is an element in Hom K alg, M to E, this This is our set M. Remember, what we want to do is a following Let me draw a diagram L over K, this is the given algebraic extension and we have given E, E is given here, E is contain this K is a field extension of case and this E is algebraically closed, that is what we have given and we want to extend it here, this Sigma we are looking for, K algebra homeomorphism or K embedding This is what we are looking for So now what we have done is, we have taken all the intermediary fields M which have the extension and we are going to maximise it and we will prove that this process will reach M. And precisely that means choosing the maximal element And choosing a maximal element one needs to apply Zorn’s lemma and one has to have an order And we have to check that order set is inductively ordered That means every chain has an upper bound there And one notes that there are maximal elements and those maximal elements will give as the required thing This also shows that this process is not very natural and this may not, the embedding may not be unique, there may be many embeddings, that is it Let us now justify that we can apply Zorn’s lemma to this set M, that we have defined So we need an order 1st So, define, so they set M is ordered by the following order So take 2 elements in M, one is M, Sigma, the other is M Prime Sigma and then we say that this is bigger equal to this, if and only if, this is the definition, this is if and only if, M is contained in M Prime and Sigma Prime is restricted to M equal to Sigma So, with this we need to check now that this M is inductively ordered So 1st of all this M is nonempty because this K, there is natural inclusion iota from K to L because this L is given to be extension of K, so there is a natural inclusion So this pair belongs to this M. Therefore M is nonempty and inductively ordered means that is every chain, what is the chain, chain is a totally ordered subset, has an upper bound in M, this is what we want to check Alright, so let us that, let us call that chain to be C, C is the pairs Mi, Sigma i, i in I, so this I is totally ordered and we want to check that it has an upper bound And obviously what we will claim is you take the union of these Mi’s, i in I, let us call this as L0 So 1st of all note that this is a field, this is a subfield of, so note that L0 is a field and L0 is in between K contained in L0 contained in L. So, field is clear because if you take

to elements here, both of them will lie in some Mi, one will lie in Mi, the other will lie in Mg but I is totally ordered, therefore I can assume both lie in the same but all these Mi’s there who is a field, therefore this is a field, this is clear Another thing is we also can extend that, we can extend that Sigma is, we will define Sigma 0 from L to E, L0 to E. How do we define this? Take any X in L0 and choose i so that X belongs to Mi, i in I and now define, we will be given this Sigma i So define Sigma 0 of X to be equal to Sigma i of X But this will not depend on the choice of this y because this is the chain That means Mi, Sigma i and if you were to arrange Mj, j in J and as you assume this i is totally ordered, so J is bigger, then Mj, Sigma J, this is contained here and then therefore this is also equal to Sigma J So it is well-defined, it is no problem And remember everywhere we are choosing that this is a chain So therefore anyway, we have therefore got hold of this element L0, Sigma 0, this is again in M and which is clearly an upper bound, upper bound for the chain Mi, Sigma i And now we will, obviously we will claim that this L0 equal to L, we want to claim this equality here, this is what we will prove Alright, so suppose on the contrary L0 is not full I, full L. Suppose L0 is properly contained in L. Then what can we do, we can choose an element which is not in L0 and which is in L. So choose x in L which is not in L0 and remember this L is L over K is algebraic That is given to us and this x is given there in L. So minimal polynomial of x over K makes sense, this is minimal polynomial of x over K, this is not over K, over L0 This is a polynomial in L0 X, x over K is algebraic, therefore L over L0 is also algebraic because L0 contains K and therefore minimal polynomial makes sense and it is a polynomial in L0, it is a polynomial in L0 X So look here, this is L, this is L0X, this simple extension of L0 which is contained in L, this is L0 here, and this is K here and we have an embedding here to E. And we are looking for a contradiction now, this is proper Anyways, so therefore this polynomial is in L0 over K and we have L0 to E there is an extension, this is Sigma 0 So therefore this L0, this is, this is, E is algebraically closed and this L0 is contained in this, so therefore I can always find, if I take the image of this, if I look at mu x, L0 and take, + Sigma to this, Sigma 0 to this, so you get a polynomial in E, this is a polynomial in E [X] And E is ultimately closed, therefore this polynomial will have 0, so therefore I can always extend So this polynomial has a zero, so y is a zero of Sigma 0 of mu x, L0

So I have these, the image of this So, let me identify, you do not have to keep saying this one, or you have this Sigma 0 of L, this is a subfield here which is the image of this and to this I adjoint y, so that these 2 films are isomorphic, that is just given by this Sigma 0, this is a subfield here So therefore if I take this, this composition, this, that will give me an extension, that will give me an embedding, let us call this embedding as, this is an embedding, so let us call this as rho So therefore I got an embedding rho from L0 x, this field to E. And obviously rho restricted to L0 is given Sigma 0 Therefore, what we check, what we check that that this pair L0, Sigma 0, this pair and L, x, L0 x, this rho, this pair obviously, this is strictly bigger, this is strictly bigger, that this is also in M but this was the maximal element in M. So a contradiction to the maximality L0, Sigma 0 So this proves that that L0 has to be L, this is K, this, and then that Sigma 0 is an embedding in E So therefore we have an embedding of, this is K embedding, so there is a K embedding, so that this diagram is commutative So, that proves the 1st part, so this proves 1 Call it, so now we have to prove 2, I just wanted to show you 2 is, if E and E Prime are two algebraic closures of K, then there exists a K isomorphism such that E to E Prime, this is the K algebra, isomorphism from E to E Prime All right So we have given 2 algebra closures E over K and E Prime over K, they are algebraic closures of K So that means 2 things, that is E over K is algebraic and E is algebraically closed, this means algebraic closure So similarly for E Prime So, the Part-1 shows that I am going to apply to E. So applying 1 to algebraic extension E over K Now I am looking at E over K has algebraic extension and I am looking at E Prime as algebraically closed field and I want to apply 1 So, that will tell us E here, this is algebraic extension of K and this is algebraic closure, so this is algebraically closed field, algebraically closed, so I can extend this to embedding Sigma, K embedding This is by Part-1, now I want to show that this Sigma is indeed an isomorphism This is an embedding and now therefore E, if I take the algebraic closure of, so take, so this, take let, so the algebraic closure of K, this way of E, of K in E Prime, I have

this So let me write on the next page So we have this diagram, so E, E to Sigma, E to E Prime, we have the Sigma, I want to show the Sigma is an isomorphism So, look at Sigma of E. See, it is a field, therefore I only have to check it is surjective So this is the image of Sigma and this one, now Sigma E is algebraically closed, E is algebraically closed, therefore Sigma E is also algebraically closed This is also algebraically closed and all this contains K, this is algebraic extension So if I take this field and take its algebraic closure in E Prime, so algebraic closure of Sigma E in E Prime Algebraic closure of Sigma E is also algebraically closed and this is contained in a prime, it contains K, so but this is algebraic closure, so that will show that this is equality here So therefore we note that Sigma E is E Prime That means this Sigma is surjective and that means it is a K isomorphism, that is what we wanted to prove Alright, now I want to make 2 remarks, 2 very important remarks, one this K isomorphism, K isomorphism Sigma in 1, remember that is L over K algebraic, is algebraic and this is algebraically closed field, E is algebraically closed Then we have extended this here, this Sigma which is a K embedding, this is not unique in general, as I remarked earlier also So therefore in 2 also, the K isomorphism in 2, that is also not unique Because all this we got by using Zorn’s lemma and by choosing a maximal element and partially ordered set may have more than one maximal element So therefore, it is not unique in general, unless, unless E equal to K. Anyway, but if you do not have this, that means if you feel is not algebraically closed, then the algebraic closures, algebraic closures of K are not unique in general, unless E, unless K is algebraically closed So, therefore we cannot say the, we cannot set the, we are not allowed to say the So we will keep saying an algebraic closure, that is one remark This is very very important because one may get a feeling that these are unique Okay, the next remark, remark 2, if I want to study algebraic, so study of algebraic extensions of a field we study, this study, we will reduce

it to, if I choose, E over K algebraic closure of K, an algebraic closure, so choose an algebraic closure of K, that we know it exists now and then this, therefore F of intermediary field of E over K makes sense So these are, this is a set of intermediary fields, subfields in between E and K So if you want to study algebraic extension of this, it is enough to study this Study of algebraic extension of a field K can be reduced to the study of the set E over K intermediary subfields of E over K where we have chosen algebraic closure of K So, this I have not used earlier but this is very important in the study of this Okay, now the next one is, so I would rather stop here and prepare for proving that the field of complex numbers is algebraically closed And for that proof I am going to use the fundamental theorem on symmetric polynomials, this I will do it in the next lecture Thank you