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 about 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.
So we've already - 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 R and you can imagine them over Q,
but here's an example. I want to give one for it all the Z p’s.
So for all of our finite fields.
So let p be a prime, and prove that in Z p
adjoin x, there exists a non-constant polynomial with no root in 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 Z mod p.
So again, pause the video, 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.
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 Z p, so let's call them 0, 1, 2, all the way up to p minus 1,
okay, and these are all the elements of Z mod p.
Now we're going to consider the following polynomial.
So we're going to define p of x to be the polynomial given by
x, x minus 1, x minus 2, all the way to x minus p minus 1 in brackets,
and we're going to add 1 at the end.
So if you'd like to write it in product notation, you can do so by doing this.
So that's the product of
x minus j for all j between 0 and p minus 1 inclusive,
and we're going to add 1 to the end, okay?
So when you rearrange this polynomial,
you'll see that this is a polynomial in Z p of x.
So notice that
p of x is inside
Z mod p adjoin x.
So that's a true sentence that p of x lives inside
this set of polynomials.
Further, this polynomial
has no root
in Z mod p since…
why does this have no root? Well
if I plug in
any k
inside Z p,
I'm going to get 1,
and that's the really, really clever idea to proving this question.
So if I take
any k, so between -
so plug in 0, 1, 2, 3, all the way up to p minus 1,
this polynomial, the first half of this, will be 0.
And 0 plus 1 is always going to be 1
so therefore if I plug in any k inside Z p, I get p of k equals 1.
Hence by the Factor - well do I need to say by the Factor Theorem?
I don't really. Hence, this polynomial
has no root
in
Z mod p.
And that's it.
Okay, so again the proof is really clever,
well this proof is really clever, but it's not…
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 if you do have another field like something -
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.
So again summary, we created a polynomial that when I
plugged in all the elements of 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 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.