The subject of today's lecture is stem field, splitting field, and algebraic closure. So consider a polynomial P in K[x], which which is irreducible and monic. So consider a polynomial P in K[x], which which is irreducible and monic. Let me give the definition: a stem field for P is an extension E, such that E contains a root of P. and is generated by this root. Well, such a thing exists: we can take just K[x] modulo P. This is a field, since P is irreducible. Well, on the other hand, any stem field E is isomorphic to such a thing. While it is easier to define the isomorphism in the other direction. Well, let's see, K[x]/P goes to E by the following formula: so, f goes f of alpha. Okay. To summarize, we have the following proposition: a stem field exist and if E and E prime a stem field exist and if E and E prime are two stem fields for P are two stem fields for P generated by roots alpha and alpha prime, where alpha and alpha prime are roots of P, then there exists a unique isomorphism from E to E prime taking alpha to alpha prime. Well, this is of course easy. So existence we have already seen. Uniqueness of the isomorphism is easy, since an isomorphism Uniqueness of the isomorphism is easy, since an isomorphism of K of alpha with something else With any field, well let's say with E prime is defined by With any field, well let's say with E prime is defined by the value it takes on alpha. Well, of course I have to be precise here. I said the exists an isomorphism but I must be precise and to say an isomorphism of what. Of course not only of abstract fields but of extensions of K, so of K-algebras. Then, it's just really defined by its value on alpha. So, what I have to prove still is the existence of such an isomorphism. But this is easy, because I have phi from K[x]/P to E. But this is easy, because I have phi from K[x]/P to E. I have psi from K[x]/P to E prime and I just take psi^(-1) times phi. This will be an isomorphism between E and E prime and of course, it will take alpha to alpha prime. So, here it takes x to alpha. So, here it takes x to alpha. Here it takes x to alpha prime. So my composition will take alpha to alpha prime. Okay, so in particular, a remark: In particular if a stem field contains two roots of P There exists a unique automorphism of it, taking one to the other. Well, first of all. Secondly, if E is a stem field of an irreducible polynomial, then, of course, the degree of E over K is the degree of P, and conversely. If the degree of E over K is the degree P and E contains a root of P then E is a stem field. then E is a stem field. Otherwise it's degree would be strictly greater then the degree of P. Well, and now I can give you some more irreducible criteria. Corollary One. So, if P is irreducible over K, well, sorry, rather P is irreducible over K if and only if it does not over K if and only if it does not have roots in extensions L of K have roots in extensions L of K of degree less or equal than n over 2 where n is the degree of P. How to prove such a thing? It's very easy. Proof: Otherwise if P is not irreducible, then it has a factor, a prime factor, irreducible factor Q of degree less or equal than n over 2 and one can take it's stem field as L and one can take it's stem field as L and one can take it's stem field as L. Well, in the other direction. Conversely, if P has a root alpha in L, then of course the minimal polynomial of alpha over K divides P. So P cannot be irreducible. Okay, one more irreducibility criterion. Let P be a polynomial over K, which is irreducible of degree n, and let L be an extension of degree m. If n and m are relatively prime then P is irreducible over L. Okay, you see, an irreducible polynomial might become reducible over an extension, but it doesn't happen if the degrees are relatively prime. Well, proof: if this is not the case Q divides P in L[x], let M be a stem field of Q over L. let M be a stem field of Q over L. let M be a stem field of Q over L. So, we have K in L in M. So, we have K in L in M. Well, M is L of alpha. So K of alpha, alpha is a root of Q so it's a root of P, is a stem field of P over K. is a stem field of P over K. So it's degree over K is equal to n, the degree of P. Well, on the other hand, if the degree of Q is d, then Well, on the other hand, if the degree of Q is d, then d is the degree of M over L because M is a stem field of Q over L. And so, the total degree, the degree of M over K, is m times d. Because it's the degree of L over K times the degree of M over L. But K of alpha is a subextension of M. So the degree n must divide md. And so, since n and m are relatively prime, then, n divides d, but d is less than n. So, n is equal to d and P is irreducible over L. So, n is equal to d and P is irreducible over L.