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Hello everyone. Welcome 
to Week 11 part 2

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of Carmen's Core Concepts, 
my name is Carmen Bruni.

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In these video series, we go
through concepts in Math 135

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that we talked about through the week.

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This being part 2, it is the second part of

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my two-part video for Week 11.

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What I decided to do 
this week is I decided to

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break apart the theory 
from the applications.

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So in part 1, we talked a lot about 
theorems, and proofs, and statements,

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and here in part 2 we're going 
to talk about how to use the

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theorems to actually prove 
things in mathematics.

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So first I'd like to start off with 
the definition of irreducible

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This is something that has 
often been in assignments,

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and I think I'm going to 
move this into my lectures,

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and it talks about what it
means to be irreducible

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because often what we do in this course, or often 
what we do with polynomials, is we try to factor them.

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And the question becomes, 
“Well how far can I factor?”

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and the answer is, “Well you 
factor into irreducible polynomials.

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Well what's an irreducible polynomial?

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And here's the definition: so let F be a field.

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We say that a polynomial of positive degree

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in the polynomial ring over 
the field F is reducible

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in this ring when it can be written 
as the product of two polynomials,

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in F adjoin x, of positive degree.

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Otherwise we say that the 
polynomial is irreducible in

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the polynomial ring over the field F.

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For example, x squared plus 1 is irreducible
in the polynomial ring over the reals

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because it doesn't factor anymore.
There's no linear factors,

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but it's reducible over the 
complex numbers. There's

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clearly two linear factors, 
x minus i and x plus i

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So we can rewrite x squared plus 1 
as the product of those two factors,

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but we can't do that over R.

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There's no way to write it as the product 
of two polynomials of positive degree.

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So notice that if we're 
factoring a quadratic, it must -

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if we can factor a quadratic into a product 
of irreducibles it must factor as 2 linear

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components, and here we don't have 
that because again by the Factor Theorem,

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we know that this has - well we 
know this has no roots over R

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and so, by the Factor Theorem 
it can't have a linear factor.

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Okay great.

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So one thing I want to talk 
about now is that the field…

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the field that you're doing these
operations with matters, right?

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So in the previous example, 
we saw one example, but

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here's a little bit more complicated example. 

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So let's take this polynomial,
z to the 5, minus z to the 4,

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minus z to the 3, plus z 
squared, minus 2z, plus 2.

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And if we factor this over C, 
we get this factorization.

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So we can factor
as z minus i, z plus i,

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z minus the square root of 2, z plus 
the square root of 2, and z minus 1.

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But as we notice, not all of these 
elements live inside the real numbers.

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So over the real numbers, we can't actually 
factor the first two components here,

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z minus i and z plus i. So we 
have to bring those together,

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right, so that we get z squared plus 1,

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and I get z minus the square root of 2, and 
z plus the square root of 2, and z minus 1.

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So that's our factorization over R,

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and over Q, we don't 
have the square root of 2.

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Square root of 2 is irrational 
as we've seen in this course.

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So we have to merge the middle two 
products together to get z squared minus 2.

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So here we go, here's our 
irreducible factorization over Q,

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there's our irreducible
factorization over R, 

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and the top one’s our irreducible 
factorization over the complex numbers.

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So again, it does depend on what field you’re 
factoring over. This should make sense, right,

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not all of these fields have the same 
elements, so depending on where I

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factor, depends on how this 
factors as a polynomial.

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By the way, this answer would 
also be different over finite fields.

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I didn't talk about it, but if I 
factored this over, let's say,

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Z mod 5, I would get a different factorization.

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So for example, z squared 
plus 1 factors over Z mod 5.

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Let me think… 2 and 3 are 
both roots of z squared plus 1

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in Z mod 5, so there would be a z minus 
2, and a z minus 3 factor from this,

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and z squared minus 2, this is irreducible 
over Z mod 5. We'll talk about this in a minute

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Okay.

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This has no roots, I mean that's maybe 
the easiest way to say this for now,

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right, if I plug in the numbers from 
0 to 4, none of these give me a root.

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None of these are 0 and hence it has no roots 
since it has no linear factors by the Factor Theorem.

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Okay, so the field matters when we're 
factoring, it changes the answer a lot.