 In this video we are going to state and prove a very important theorem in Gawa theory, commonly referred to as Kroniger's theorem. Kroniger's theorem tells us that if we have any field f and we have any irreducible polynomial p of x in the polynomial ring f of joint x, then there exists a field extension. In fact, it's a finite extension e over f such that the polynomial when viewed as a polynomial in e of joint x, because after all that e of joint x contains f of joint x, if you view p of x as a e polynomial, it has a root. And so the irreducible polynomial has a root in e. So in general terms here, Kroniger's theorem tells us that every irreducible polynomial has a root in some finite extension of the field. Alright, so let's begin here. Suppose that p of x is an irreducible polynomial, therefore it's not constant. Irreducible polynomials don't describe zero, it doesn't describe units, which of course zero in the units of the polynomial ring f of joint x are exactly the constant polynomials. That is those numbers from f since f is a field. And therefore, we can assume that the degree of the polynomial p here is positive, let's call it n for the sake of it. Now recall, from previous work, if you take the polynomial ring whose coefficients come from a field, we know that this is a Euclidean domain. In particular, every Euclidean domain is a principal ideal domain. So f of joint x is a principal ideal domain. In a principal ideal domain, we proved that a ideal, which are necessarily principal, every ideal in a principal ideal domain is maximal if and only if it is generated by a irreducible element. So since p of x is an irreducible polynomial, the principal ideal generated by this irreducible polynomial is going to be maximal inside of the principal ideal f of joint x. Now, we've also shown, I just love how the breadcrumbs, we can follow our way in this situation here, right? All this beautiful theory here, f of x is Euclidean domain. Every Euclidean domain is a principal ideal domain. Ideal generated by irreducible elements in a PID is maximal. And then, if you have a maximal ideal and you mod it out, you get a field. So f of joint x mod i gives you a field because a commutative ring modded out by a maximal ideal gives you a field. And so this is our candidate, E. E is the field that we're going to describe here. All right, so there's a few things we have to then say about E. Is it a field, but is it a field extension? Can I view F inside of E? And the answer to that is yes. In a very natural way, we can view uptoisomorphism that F is a subfield of E. So up to relabeling, F is a subset of E. And this is how we're going to do it. Imagine we'd have two elements in the field F, call them A and B. These can be viewed as constant polynomials in F of joint x. And suppose that A and B are different elements. I want you to consider the cosets A plus i and B plus i, which will be elements of the field E. Now, if they were true that A plus i is equal to the coset B plus i, the two cosets are the same, only if A minus B is an element of the coset. Okay? This is, of course, the equivalence that we have for our quotient ring right here. So A minus B is an element of i in that situation. This is the principal ideal generated by P. So A minus B, how can it be inside of that ideal? The degree of P, remember, was N. And we showed previously that in the Euclidean domain, remember F of joint x is Euclidean domain, which Euclidean domains are principal. How you proved that Euclidean domains were principal is you showed that every ideal is generated by an element of minimum norm that's non-zero. For polynomials, their norm is its degree. So the generator P of x must be a polynomial minimum degree in here. So how in the world can we produce A minus B here? Since A minus B, since these are different elements, their difference is not zero. So this is a polynomial of degree zero, right? But that's less than N, which is going to get us a contradiction. A minus B can't be in there. So in fact, it must have been that A plus i is not equal to B plus i. These are distinct cosets in the quotient ring E, which is F of joint x mod out the principal ideal generated by this irrational polynomial. So these two cosets are distinct. So if you take the identification where an element of the field F is associated to its coset A plus i, that is an injection. It's a one-to-one map. So without the loss of generality, we can say that F is a subset of E because there is an isomorphic copy of F inside of E. So E is in fact an extension field. We can say that E extends the field F. So that was the first thing we have to show. The second thing we have to show is that it has a root, like we said it was going to, okay? And to do that, we're going to take the element alpha to be the coset associated to the element x. So x plus i, that indeterminate, it's quotient inside of the ring E here. We're going to call that element alpha for a moment. And so if the polynomial P of x has the coefficients A0 plus A1x plus A2x squared plus A3x cubed all the way up to Anx to the end here. Notice that each of these coefficients are coefficients in F. What would evaluation at alpha mean? It would mean you substitute each and every one of those x's with an alpha here. So P of alpha just means you look at this value here. But what is alpha in inside of, we're not viewing this as a polynomial here, we're viewing this as the evaluation inside of E. Alpha is x plus i. So we end up with this thing right here inside the quotient ring. So we get A1 plus A1x plus i plus A2x plus i squared plus A3x plus i cubed all the way up to An times x plus i to the end of the power. Now because we are in the quotient ring effigient x mod out by i, this here is equivalent to A0 plus A1x plus A2x squared all the way up to Anx to the end plus i. So for which notice this right here is our polynomial P of x. It's the polynomial P of x. So since P of x is inside of i, this coset is just i itself. And so this shows us that P evaluated alpha is in fact equal to zero inside of the field extension E. The other thing I didn't mention here is that this was a finite extension. And why is this a finite extension? Well, this is something we'll go into a little bit later. We'll talk about this in the last video of this lecture. So I'll just, I'll leave it, I'll leave it that bit open for just a moment here. But the point of Kronecker's theorem is that every polynomial irreducible polynomial has a root in perhaps a field extension. Okay. An important definition connected to Kronecker's theorem is the following. Let E over F be a field extension and let S be some subset of E. Then we can define F adjoin S to be the smallest subfield of E containing F and S. Which such a thing exists because it's going to be the intersection of all subfields of E containing F and S there. Which since E is included in that list, this intersection is non-empty. It contains F, it contains S. There's some field that has that satisfaction. So these things do in fact exist. In the case that S has the form alpha one, alpha two, all the way up to alpha N. Then we would write this as F adjoin alpha one, alpha two, all the way up to alpha N. So this is the same thing. This is notation we've used previously. We never really officially defined it. It's now official. Now in the case that you have just one element, this would look like F adjoin alpha. This is referred to as a simple extension. All right. And Kronecker's theorem, when you have the notion of a simple extension, then rewrites this as every, we can rebrand it in the following way. Every irreducible polynomial over a field has a root in a field extension. And that field extension is necessarily a simple extension. And what is that simple extension? We take the field adjoined a root to that polynomial. But the thing is this idea of a simple extension has some idea of some larger extension. Kronecker's theorem actually does it in a constructive way that we don't need to know what the larger field is. We can build it from scratch by using the quotient ring. And thus we can create a unit to add there. But Kronecker's theorem gives us these simple extensions. Let me give you one simple example here. So take the polynomial x squared plus one and view this as a rational polynomial q adjoin x. So then by Kronecker's theorem, if you take the field, excuse me, if you take the polynomial ring q adjoin x and mod out by x squared plus one, this gives you an extension field of q that has a root for x squared plus one. I claim that this field is isomorphic to what we have been calling the field q adjoin i, where viewed as a complex number, i is a square root of negative one, which is then a root of the polynomial x squared plus one. If we view this as a complex polynomial, right? So q adjoin i is a simple extension, a simple extension here. And so we're going to get that this field we construct with Kronecker's theorem is equivalent to this. And this is going to show us that this field extension has in fact degree two. And so when you add the root of a polynomial, this example we did here generalizes into larger settings. And I claimed earlier that it was a finite extension. Why is Kronecker's theorem always given as a finite extension? Well, something we're going to see a little bit later is that the field formed in Kronecker's theorem will have an extension at most the degree of the polynomial, which in particular if it's an irreducible polynomial, the degree of the extension will equal the degree of that polynomial. But I'm getting a little ahead of myself. We'll define that at the end of this lecture.