Fundamental Theorem of Algebra

Hello everyone. In this video, we're going to talk a little bit about the Fundamental Theorem of Algebra, but we're not going to see much of the Fundamental Theorem of Algebra, so let's just review the statement of it.

The Fundamental Theorem of Algebra says that every complex polynomial has a complex root.

That's the statement of the Fundamental Theorem of Algebra, and what this video will be about is to show that this isn't true over an arbitrary field. Well not over a general arbitrary field, but over other arbitrary fields, it might not necessarily be true.

So for example, we've already seen examples over $\mathbb{R}$ and you can imagine them over $\mathbb{Q}$, but here's an example. I want to give one for it all the $\mathbb{Z}_p$’s. So for all of our finite fields.

Question: Let $p$ be a prime. Prove that in $\mathbb{Z}_p[x]$, there exists a non-constant polynomial with no root in $\mathbb{Z}_p$.

So we're trying to find a polynomial over the integers modulo $p$ that's not a constant and that doesn't have a root in $\mathbb{Z}_p$.

So again, take a minute, try to come up with an example, see what you can do. This is not an easy problem by any stretch, but the solution is very short. It's tough because you don't have too much information to deal with.

Proof

Okay, so let's get started with the proof. So what we're going to do is we're actually going to enumerate all the elements of $\mathbb{Z}_p$, so let's call them $0$, $1$, $2$,...,$p-1$, okay, and these are all the elements of $\mathbb{Z}_p$.

Now we're going to consider the following polynomial.

$$p(x)=x(x-1)(x-2)...(x-(p-1))+1=\prod_{j=0}^{p-1}(x-j)+1$$

So when you rearrange this polynomial, you'll see that this is a polynomial in $\mathbb{Z}_p[x]$.

Further, this polynomial has no root in $\mathbb{Z}_p$ since…why does this have no root? Well if I plug in any $k\in\mathbb{Z}_p$, I'm going to get $1$, and that's the really, really clever idea to proving this question.

So plug in $0$, $1$, $2$, $3$,...,$p-1$, this polynomial, the first half of this, will be $0$. And $0+1=1$, so therefore if I plug in any $k\in\mathbb{Z}_p$, I get $p(k)=1$. Hence, this polynomial has no root in $\mathbb{Z}_p$. And that's it.

Okay, so again that the proof is really clever, but it's not obvious and it's not immediate how to come up with this idea. This idea is just kind of one that I've seen and I thought it's pretty cool so I thought I would show you, and I think it emphasizes FTA a lot, in the sense that FTA is very specific to the complex numbers.

There are other fields where FTA type theorems hold and they're very interesting to study and pure mathematicians do study these fields, but for us the only field where an FTA type theorem will hold is the complex numbers.

Summary

So again summary, we created a polynomial that when I plugged in all the elements of $\mathbb{Z}_p$, because there's only finitely many, I can do this create this finite polynomial that's $0$ whenever I plug in any of the elements of $\mathbb{Z}_p$, and just to get that there's no root, just add $1$ to the end of it, and then we always get that the value is non-zero.

Okay, so thank you very much for listening. Hopefully this gives you a little bit of insight as to what the FTA theorem says, and maybe understand a little bit better through a different field. Thank you very much.