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Hello everyone. In this video, 
we're going to talk a little bit about

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the Fundamental Theorem of Algebra, but

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we're not going to see much about the Fundamental Theorem 
of Algebra, so let's just review the statement of it.

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The Fundamental Theorem 
of Algebra says that every

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complex polynomial has a complex root.

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That's the statement of the Fundamental Theorem 
of Algebra, and what this video will be about

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is to show that this isn't true

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over an arbitrary field.

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So we've already - well

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not over a general arbitrary field, 
but over other arbitrary fields,

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it might not necessarily be true.

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So for example, we've already seen examples 
over R and you can imagine them over Q,

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but here's an example. I want 
to give one for it all the Z p’s.

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So for all of our finite fields.

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So let p be a prime, and prove that in Z p

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adjoin x, there exists a non-constant 
polynomial with no root in Z p.

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So we're trying to find a polynomial 
over the integers modulo p

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that's not a constant and that 
doesn't have a root in Z mod p.

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So again, pause the video, take a minute, try to 
come up with an example, see what you can do.

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This is not an easy problem by any 
stretch, but the solution is very short.

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It's tough because you don't have 
too much information to deal with.

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Okay, so let's get started with the proof.

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So what we're going to do is 
we're actually going to enumerate

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all the elements of Z p, so let's call them 
0, 1, 2, all the way up to p minus 1,

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okay, and these are all the
elements of Z mod p.

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Now we're going to consider 
the following polynomial.

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So we're going to define p of x 
to be the polynomial given by

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x, x minus 1, x minus 2, all the way 
to x minus p minus 1 in brackets,

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and we're going to add 1 at the end.

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So if you'd like to write it in product 
notation, you can do so by doing this.

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So that's the product of

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x minus j for all j between 
0 and p minus 1 inclusive, 

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and we're going to add
1 to the end, okay?

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So when you rearrange this polynomial,

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you'll see that this is a polynomial in Z p of x.

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So notice that

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p of x is inside

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Z mod p adjoin x.

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So that's a true sentence 
that p of x lives inside

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this set of polynomials.

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Further, this polynomial

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has no root

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in Z mod p since…

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why does this have no root? Well

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if I plug in

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any k

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inside Z p,

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I'm going to get 1,

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and that's the really, really clever 
idea to proving this question. 

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So if I take

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any k, so between -

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so plug in 0, 1, 2, 3, all 
the way up to p minus 1,

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this polynomial, the first half of this, will be 0.

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And 0 plus 1 is always going to be 1

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so therefore if I plug in any k 
inside Z p, I get p of k equals 1.

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Hence by the Factor - well do I need 
to say by the Factor Theorem?

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I don't really. Hence, this polynomial

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has no root

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in

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Z mod p.

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And that's it.

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Okay, so again the proof is really clever,

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well this proof is really clever, but it's not…

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it's not obvious and it's not

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immediate how to come up 
with this idea. This idea is just

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kind of one that I've seen and I thought it's 
pretty cool so I thought I would show you,

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and I think it emphasizes FTA a lot,

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in the sense that if you do have 
another field like something - 

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FTA is very specific to the complex numbers.

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There are other fields where 
FTA type theorems hold

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and they're very interesting to study and
pure mathematicians do study these fields,

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but for us the only field where an FTA type 
theorem will hold is the complex numbers.

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So again summary, we created 
a polynomial that when I

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plugged in all the elements of Z p, 
because there's only finitely many,

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I can do this create
this finite polynomial

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that's 0 whenever I plug in 

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any of the elements of Z p,

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and just to get that there's no
root, just add 1 to the end of it,

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and then we always get 
that the value is non-zero.

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Okay, so thank you very much for listening.

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Hopefully this gives you a little bit of insight as 
to what the FTA theorem says, and maybe

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understand a little bit better through 
a different field. Thank you very much.