 In this video, I want to state and prove the so-called Eisenstein criterion, which is a criterion we can use to prove that certain polynomials are irreducible. So, suppose D is a unique factorization domain, and let P be a prime element in that unique factorization domain. Then the polynomial f of x given as a n times x to the n plus a n minus one times x to the n minus one, all the way down to a zero. If this is a polynomial in D a joint x, further suppose that P divides all of the coefficients of f except for the leading term, right? So, P will divide a zero, a one, a two all the way up to a n, but P doesn't divide a n. And also, so, you know, P doesn't divide a n right there, but also suppose that P squared doesn't divide a sub zero. So, again, let me check what's going on here. So, we have this prime, we have this prime number that doesn't divide the leading coefficient. It divides all of the other coefficients, but that prime squared doesn't divide the constant term. If those conditions are satisfied, then in fact, f must be a irreducible polynomial. And again, this is true for any unique factorization domain, such as the integers or Gaussian integers, something like that. So, we're going to prove this by contradiction. Suppose that f is in fact reducible, it has a proper factorization. So, it'll factor at the polynomial brx to the r all the way down to b zero and cs times x to the s all the way down to c zero. So, this is a proper factorization into polynomials. Neither of the two polynomials are constant polynomials. These are polynomials of smaller degree. Now, when you multiply these things out here, it's very easy to keep track of the constant term. The constant term a zero inside of f will only be produced by the product of the two constant terms of the factors, b zero and c zero. So, by assumption, P does divide a zero, but P squared doesn't divide a zero. So, what this tells us is because P is a prime element, then since P divides a zero, P divides one of these terms. P divides either P, b zero or c zero. But since P squared doesn't divide a zero, that means that P will divide only one of these two. Because if P divided b naught and c naught, then P squared would divide a naught. So, it's going to divide one and only one of these two coefficients. And so, without the loss of generality, we're going to assume that the prime P divides c naught, but P does not divide b naught in this setting. Alright, now let's look at the leading term. How is the leading term produced, a n here? Well, the only way we can get a n is to take the product of the two leading terms for these factors here. So, the coefficient a n is created by taking b r times c s. And remember, by assumption, P does not divide a n. So, since P is a prime element, well, we don't even need a prime element here. But since P doesn't divide a n, it means that P doesn't divide b r, nor can it divide c s. Because it divided one of those two, then it would divide a n. And that would be against our assumptions here. Now, since P does divide c naught, but P doesn't divide c s, that means there is a smallest with respect to the ordering here. So, going from the constant to x to x squared, the coefficients here. There's gonna be a smallest integer k such that P doesn't divide c k. And that could be the leading term, c s, maybe. But there's some smallest term where P doesn't divide c k. But it does divide c k minus one, c k minus two, c k minus three, all the way down to c zero. And so, I want us to look at the coefficient a k in that combination there. So, how is the coefficient a k of f produced? How is a k produced for f of x there? Well, we look at all the possible products of bi times c j, where i plus j adds up to be k, this convolution product here. So, we get things like b zero, c k, b one, c k minus one, b two, c k minus two, all the way down to b k, c zero. Now, some of these things we know to be divisible IP, don't we? Well, let's investigate here. Since we assume that c k is the smallest coefficient of this factor up here that's not divisible by P, this means that c is not as divisible by P, c one, c two, c three, c k minus two, c k minus one. These are all divisible by P. Therefore, this part of the sum is going to be divisible by P. So, if we move this to the other side of the equation, we're going to get a k minus b not c k is divisible by c. I mean, it's divisible by P, excuse me. But also, by assumption, a k is not divisible, I mean a k is divisible by P. And so, because of that, that implies for us here that since, you know, since we know that P divides a k, then we'd have to also have that P divides b not c k. But we get a problem here. We know because P is a prime, P divides b not or P divides c k, but we already know that neither of those two things happens. So, we get a contradiction. The contradiction came to the fact that f of x here has a factorization as a polynomial and therefore it must have been an irreducible polynomial. So, I want to show you just a quick example of that. Consider the following polynomial and view it as a rational polynomial. Really, though, we're viewing it as an integer polynomial. I really should say that because prime elements don't exist over a field. But z, which is a unique factorization domain that's not a field, it does have prime elements. But be aware that factorizations over zx and factorizations over qx are one of the same things. So, it doesn't make a bit of a difference here. So, if we look at the rational prime three, notice what happens here. The coefficient of x to the fifth, 16, is not divisible by three. But the other coefficients, nine, well, negative nine, three, six, negative 21, they're all divisible by three. And then the last number, negative 21, that's not divisible by nine. So, by Eisenstein's criterion, we have that this polynomial is in fact irreducible. Pretty nice, pretty slick. It's a lot easier than what we might have to do otherwise, right? We'd have to check because if we didn't use Eisenstein's criteria, what else would we do? We'd have to check that does this polynomial have a root or not? Because if it does, that root would have to then produce a linear factor. But even by the rational roots test, there's going to be a lot of possibilities we'd have to check. But then we'd have to also consider the possibility that it factors as a quadratic times a cubic polynomial. That would be a real, real pain to do. So, Eisenstein's is nice and slick in that regard. Now, I don't want us to get too deceived by Eisenstein's criteria. It's a nice simple thing to check. But how often does it come up, right? If you just get a random polynomial, you probably wouldn't expect that it would satisfy this condition about primes dividing almost everything except for the leading term. But p-square doesn't divide the constant term. The real value of Eisenstein's criterion is not in checking the reducibility of specific polynomials. Instead, it's its ability to generate irreducible polynomials of arbitrarily large degree. For example, I could take the polynomial x to the n minus p. And I can view this as like an integer polynomial, right? Or here p is a prime. In which case, yeah, sure, p doesn't divide the coefficient of x to the n because it's 1. It divides all the middle coefficients because they're 0. It divides the constant term because it's p. But it doesn't divide, it doesn't divide, I should say p-square doesn't divide p there. And so this is, in fact, an irreducible polynomial. And so this is a very relevant statement because this then provides a proof here. If you're trying to solve this thing, this would tell you that the nth root of a prime is actually, the nth root of a prime is not a rational number. In fact, you could replace p with any square free number. This would also be the case. In fact, as long as you take a number for which it's divisible i prime but not prime squared, that's not a perfect square as an integer. And perhaps you could think that the proof that Euclid had was much more elementary than that. I'll give you that, but it's kind of nice. And in particular, we can get arbitrarily large polynomial degrees which are going to be irreducible by Eisenstein's criterion. So it is a really cool criterion. I hope you liked it. If you learned something about Eisenstein's criterion or just about irreducible polynomials over a field in general, please like these videos. Subscribe to the channel to see more videos like this in the future. If you have any questions, please post them in the comments and I'll be glad to answer them when I can.