 So we've previously defined the notion of an algebraic field extension. We're now ready to define what we mean by an algebraically closed field. So suppose E is a field. We say that E is algebraically closed if all non-constant polynomials in E have a root. Clearly, if you have a constant polynomial, they don't have any roots, so that's the obvious exception, but every other polynomial in E has a root. Clearly, linear polynomials always have roots, not a big deal, but we've seen many examples of fields and polynomials which are irreducible. If you're irreducible, of course, you don't have a root, okay? If your field is algebraically closed, then you have a root for every polynomial. A very important example of such a field will be the complex numbers. By the fundamental theorem of algebra, the complex number field is an algebraically closed field. Every non-constant polynomial has, every non-constant complex polynomial has a complex root. And so the fundamental theorem of algebra in our language would just be O, C is an algebraically closed field, right? That's not how it's usually defined in a college algebra class like math 1050, but in the language of abstract algebra, the fundamental theorem of algebra says that C is an algebraically closed field. Now, a very important property of algebraically closed field is the following. And this is often stated and or proven for the complex numbers, like I said, in a college algebra setting or a pre-calculus setting. A field F is algebraically closed if and only if every non-constant polynomial in F of joint X factors into linear factors. So to be algebraically closed means that every polynomial has a root. We're saying that also means that everything can be factored into linear factors. This terminology here that you factored into linear factors, we say that the polynomial splits. So a field is algebraically closed if and only if all polynomials split. Clearly non-constant polynomials can't have a linear factor. Those ones are obviously excluded, but for non-constant polynomials, they always split, okay? So let's prove this one. One direction is much easier than the other. We're gonna take the harder direction first. Assume that F is algebraically closed and take a non-constant polynomial F of X that belongs to the polynomial ring F of joint X, like so. Now, by assumption, because F is algebraically closed, the polynomial F of X has a root. Let's call that root R sub one. Then by the factor theorem, if R one is a root, that means F of X factors as X minus R one times some other polynomial F one, a joint X, like so. For which, notice that F one is a polynomial with F coefficients, and clearly the degree of F one is gonna be exactly one less than the degree of F of X right here. And so this is the moment in the proof where we scream induction, which if we're gonna do induction, we should have a base case. The base case is, of course, just linear polynomials. So a linear polynomial, something like, you know, MX plus B. This is our base case. Linear polynomials clearly factor into linear polynomials. Linear polynomials always split without any assumption about algebraic closure or not. But that's our base case. And so then we're like, okay, if every polynomial degree less than N splits, then take a polynomial degree N by assumption since it's algebraically closed. It has a linear factor, and then it has a polynomial with N minus one degree by the inductive hypothesis. Every polynomial degree N minus one splits into linear factors. So you get a bunch of linear factors coming from F one. You would join that to the linear factor X minus R one. And then by the big, by the boom, you then have a linear factorization of F, and that then proves the first direction. So it's an induction argument using the factor theorem, which gives you a linear factor because we have a root by the assumption of being algebraically closed. The other direction is pretty clear because if a polynomial factors into linears, so you get X minus R one times X minus R two all the way down to X minus R N, maybe there's some coefficient in front, A, we just factor out everything. Then clearly, there's lots of roots here. R ones are root, R twos are root, R Ns are root. Now, admittedly these could all be the same number, right, the root's multiplicity could be repeated. That's a possibility, but nonetheless, because we have a linear factor, we have a root. Again, the factor theorem in play there. And so every non-constant polynomial has a root. So this idea of polynomial splitting, actually on the surface appears to be a stronger condition than being algebraically closed where you have just one root guaranteed and therefore you have one linear factor. But by induction, the two notions are actually logically equivalent to each other. All right, another very important property about algebraically closed fields. And this is gonna be a big deal as we talk about the algebraic closure in the next and final video for this lecture here. An algebraically closed field has no proper algebraic extensions. Now, that doesn't mean that there's no proper field extensions. It's just if you have an algebraically closed field, then if you would join any new elements to the field, those new elements have to be transcendental. And this is actually where the name algebraically closed comes from, that it contains all of the algebraic elements of the field. And if you were to join something new, it has to be transcendental. You can't add a new algebraic element. Hence why it's algebraically closed. You can never be transcendentally closed because you can always just add a new element that is just some new thing that's not belong to the field. It doesn't have any algebraic relations. But there is a limit of how many algebraic elements you can include. And the field is algebraically closed when you've hit that limit, all right? So suppose F is an algebraically closed field. It suffices to show that there's no element alpha such that alpha, F join alpha over F is algebraic. Because if you had some field E that extends F and this is an algebraic extension, that means every element in E is algebraic over F. Take some element alpha that belongs to E that doesn't belong to F and then look at the field F join alpha, right? This would be an algebraic extension and this would exist because of this algebraic extension. So it does suffice to look at a simple algebraic extension in that situation. Now consider the minimal polynomial mu of alpha over F. By definition, the minimal polynomial is an irreducible polynomial. But our field is algebraically closed. What we just saw from the previous theorem is that every polynomial, non-constant, will factor into linears. So that tells you that if your degree is two or higher, you cannot be irreducible because you factor into linears. So the only way you can be irreducible over F over an algebraically closed field is that you yourself have to be a linear polynomial. So mu has the form mu equals X minus alpha. But if the polynomial is X minus alpha, so I should also mention that the minimal polynomial has alpha as a root. So alpha, mu has to be a linear polynomial with alpha as a root. The only, and it's also monic, right? The leading coefficients one. When you take all of those conditions together, the only possibility for the minimal polynomial is X minus alpha. But the coefficients of the minimal polynomial come from F. The only way that X minus alpha can be a polynomial over F of joint X is because alpha, which is one of the coefficients, well, negative alpha is, but that's good enough. Alpha has to be an element of F in that situation. So this algebraic element belongs to F. And so we get no proper algebraic extensions of a algebraically closed field, hence the name. And our next video, like I said, we're gonna improve the existence of a so-called algebraic closure that if you have a field, then you could have joined all of the algebraic elements to that field and form what we call the algebraic closure, which will be an algebraically closed field. And will be, in fact, the smallest algebraically closed field that contains that field.