Hello everyone. Welcome to Week 11 part 2
of Carmen's Core Concepts, my name is Carmen Bruni.
In these video series, we go through concepts in Math 135
that we talked about through the week.
This being part 2, it is the second part of
my two-part video for Week 11.
What I decided to do this week is I decided to
break apart the theory from the applications.
So in part 1, we talked a lot about theorems, and proofs, and statements,
and here in part 2 we're going to talk about how to use the
theorems to actually prove things in mathematics.
So first I'd like to start off with the definition of irreducible
This is something that has often been in assignments,
and I think I'm going to move this into my lectures,
and it talks about what it means to be irreducible
because often what we do in this course, or often what we do with polynomials, is we try to factor them.
And the question becomes, “Well how far can I factor?”
and the answer is, “Well you factor into irreducible polynomials.
Well what's an irreducible polynomial?
And here's the definition: so let F be a field.
We say that a polynomial of positive degree
in the polynomial ring over the field F is reducible
in this ring when it can be written as the product of two polynomials,
in F adjoin x, of positive degree.
Otherwise we say that the polynomial is irreducible in
the polynomial ring over the field F.
For example, x squared plus 1 is irreducible in the polynomial ring over the reals
because it doesn't factor anymore. There's no linear factors,
but it's reducible over the complex numbers. There's
clearly two linear factors, x minus i and x plus i
So we can rewrite x squared plus 1 as the product of those two factors,
but we can't do that over R.
There's no way to write it as the product of two polynomials of positive degree.
So notice that if we're factoring a quadratic, it must -
if we can factor a quadratic into a product of irreducibles it must factor as 2 linear
components, and here we don't have that because again by the Factor Theorem,
we know that this has - well we know this has no roots over R
and so, by the Factor Theorem it can't have a linear factor.
Okay great.
So one thing I want to talk about now is that the field…
the field that you're doing these operations with matters, right?
So in the previous example, we saw one example, but
here's a little bit more complicated example.
So let's take this polynomial, z to the 5, minus z to the 4,
minus z to the 3, plus z squared, minus 2z, plus 2.
And if we factor this over C, we get this factorization.
So we can factor as z minus i, z plus i,
z minus the square root of 2, z plus the square root of 2, and z minus 1.
But as we notice, not all of these elements live inside the real numbers.
So over the real numbers, we can't actually factor the first two components here,
z minus i and z plus i. So we have to bring those together,
right, so that we get z squared plus 1,
and I get z minus the square root of 2, and z plus the square root of 2, and z minus 1.
So that's our factorization over R,
and over Q, we don't have the square root of 2.
Square root of 2 is irrational as we've seen in this course.
So we have to merge the middle two products together to get z squared minus 2.
So here we go, here's our irreducible factorization over Q,
there's our irreducible factorization over R,
and the top one’s our irreducible factorization over the complex numbers.
So again, it does depend on what field you’re factoring over. This should make sense, right,
not all of these fields have the same elements, so depending on where I
factor, depends on how this factors as a polynomial.
By the way, this answer would also be different over finite fields.
I didn't talk about it, but if I factored this over, let's say,
Z mod 5, I would get a different factorization.
So for example, z squared plus 1 factors over Z mod 5.
Let me think… 2 and 3 are both roots of z squared plus 1
in Z mod 5, so there would be a z minus 2, and a z minus 3 factor from this,
and z squared minus 2, this is irreducible over Z mod 5. We'll talk about this in a minute
Okay.
This has no roots, I mean that's maybe the easiest way to say this for now,
right, if I plug in the numbers from 0 to 4, none of these give me a root.
None of these are 0 and hence it has no roots since it has no linear factors by the Factor Theorem.
Okay, so the field matters when we're factoring, it changes the answer a lot.