# Homework Help: Abstract Algebra: Prove two kernels are the same

1. Jan 2, 2014

### philbabbage

1. The problem statement, all variables and given/known data

Prove that $(\mathbb{Z}/2\mathbb{Z})[x,y]/(x^2 + xy + x^2, x^2 + x + 1, y^2 + y + 1)$ is
isomorphic to $F_4[z]/(z^2 + z + 1)$ by showing that the kernel of

$\phi : (\mathbb{Z}/2\mathbb{Z})[x,y] \to (\mathbb{Z}/2\mathbb{Z})[x,y]/(x^2 + xy + x^2, x^2 + x + 1, y^2 + y + 1)$

is the kernel of

$\psi : (\mathbb{Z}/2\mathbb{Z})[x,y] \to F_4[z]/(z^2 + z + 1).$

2. Relevant equations

First Isomorphism Theorem

Err, this is somewhat long. There might be other problems here, but
the thing that I know I'm struggling with is showing that the kernel
of psi is in the kernel of phi.

3. The attempt at a solution

I want to show that each kernel contains the other.

I'm shooting in the dark here, but first I'll show that the kernel of
psi is contained in the kernel of phi.

This means setting up the homomorphism
$\phi : (\mathbb{Z}/2\mathbb{Z})[x,y] \to (\mathbb{Z}/2\mathbb{Z})[x,y]/(x^2 + xy + x^2, x^2 + x + 1, y^2 + y + 1)$

I'm going to define $I = (x^2 + xy + y^2, x^2 + x + 1, y^2 + y + 1)$ here.

So I took all polynomials in $(\mathbb{Z}/2\mathbb{Z})[x,y]$ and tried factoring them to put them in terms of I:

First the x's.

$(a_{n}x^{n} + a_{n-1}x^{n-1} + \ldots + a_{0}x^{0})y^{0}$
$\vdots$
$(a_{n}x^{n} + a_{n-1}x^{n-1} + \ldots + a_{0}x^{0})y^{n-1}$
$(a_{n}x^{n} + a_{n-1}x^{n-1} + \ldots + a_{0}x^{0})y^{n} + I$

$(a_{1}x + a_{0})y^{0}$
$\vdots$
$(a_{1}x + a_{0})y^{n-1}$
$(a_{1}x + a_{0})y^{n} + I$

Then the y's

$(a_{1}x + a_{0})y^{0}$
$(a_{1}x + a_{0})y^{1} + I$

Which I multiplied out

$a_{1}x + a_{1}xy + a_{0}y + a_{0} + I$

But now I'm not sure what to do with my result, or if this is particularly meaningful.

Showing that the kernel of phi is in the kernel of psi was much easier
for me. I defined psi:

$\psi(f) = f(z,z^{-1})$

Showed that it is a homomorphism:

$\psi(a)*\psi(b) = (a)(z, z^{-1}) * (b)(z, z^-{1}) = (a*b)(z,z^{-1})$
$\psi(a*b) = (a*b)(z, z^{-1})$

$\psi(a)+\psi(b) = (a)(z, z^{-1} + (b)(z, z^-{1}) = (a+b)(z,z^{-1})$
$\psi(a+b) = (a+b)(z, z^{-1})$

And applied psi to each of the three polynomials x^2 + xy + y^2, x^2 +
x + 1, x^2 + x + 1 in an effort to show that they were the kernel of
psi. I don't show this part, but each polynomial was equal to the
ideal in $F_4$.

Last edited: Jan 2, 2014
2. Jan 2, 2014

### Mandelbroth

Do you mean $(\mathbb{Z}/2\mathbb{Z})[x,y]/(x^2 + xy + x^2, x^2 + x + 1, y^2 + y + 1)$ here, in your problem statement? I'm guessing you do, from your approach, and the problem would be weird if not.

3. Jan 2, 2014

### philbabbage

Yes, I do. Let me change that. Actually, I think I do. Double check me, since I am quite new to both the subject and the notation. By $\mathbb{Z}_2$ I meant the integers mod 2 (although I realize I had written $Z_2$, which would definitely be weird), which I suspect is the same thing as $(\mathbb{Z}/2\mathbb{Z})$.

I'm constructing two homomorphisms from all polynomials with indeterminates x and y in the integers mod 2. One homomorphism goes to the galois field of 4 elements, the other goes to the polynomials in the integers mod 2 with ideal (x^2 + x + 1, y^2+y+1, x^2 + xy + y^2).

Last edited: Jan 2, 2014
4. Jan 2, 2014

### Mandelbroth

Oh! Sorry. I didn't even notice that I didn't follow your notation.

Yes. I just don't like the shorthand because it looks weird to me. $\mathbb{Z}/2\mathbb{Z}$ is the quotient of the integers with the ideal generated by 2. We could also write it as $\mathbb{Z}/(2)$.

I'm going to see if I can lead you without giving away the answer. You're almost sitting on it. From the First Isomorphism Theorem, what do we know about the kernel of $\phi$?

Also, it isn't obvious to me why multiplication works for $\psi$. Could you explain your thinking to me?

Last edited: Jan 2, 2014
5. Jan 2, 2014

### philbabbage

We know that if the images of phi and psi are isomorphic, then the kernels of each are equivalent. To me, this should imply that the number of canonical elements in each quotient are the same. In which case, I could argue that those polynomials less than z^2 + z + z are renamings like below:

$z + 1 \to y$
$z \to x$
$1 \to 1$
$0 \to 0$

which would essentially be the claim that y + 1 = x and x + 1 = y, which would work in $\mathbb{Z}_2$, and so perhaps this is reasonable .. ? I'll try evaluating this and see what I get.

Ahh, I used $z+1$ above. Thats also $z^{-1}$:

$z^2 + z + 1 = 0$
$z^2 = z + 1$
$z = 1 + z^{-1}$
$z^{-1} = z + 1$

I'm thinking about how to improve my writing for $\psi$. The idea though is that this is really just evaluating the polynomials, and that it doesn't matter whether you multiply two polynomials and evaluate them, or if you evaluate them and then multiply their results.

Umm, hopefully this notation isn't too bad. The z's being used as indeterminates hasn't burned me, but I can see how that can be weird.

6. Jan 2, 2014

### Mandelbroth

Your argument for $\psi$ feels uncomfortable to me. I don't know why. I'm pretty sure it's right, though.

What do we know that the kernel of $\phi$ is? We have a homomorphism $\phi$ from one ring, $R$, to a quotient ring, $R/I$, and that there is some kernel $\operatorname{ker}\phi$ such that $R/\operatorname{ker}\phi\cong R/I$. What do you think $\operatorname{ker}\phi$ is?

7. Jan 2, 2014

### philbabbage

Yeaah, it's definitely awkward and needs some work. I'm still thinking about it: it has to be right, but the improvement isn't obvious to me.

The kernel of phi is everything in $(\mathbb{Z}/2\mathbb{Z})[x,y]$ that is zero in $R/I$. This is the same as all multiples of $x^2+x+1, y^2+y+1, y^2+xy+x^2$, right?

So then if I remap these for $x = z, y = z + 1$:

$(z)^2 + z + 1 = 0$

$(z+1)^2 + (z + 1) + 1 = z + (z + 1) + 1 = 0$ (err, i guess thats the same?)

$(z)^2 + 1 + (z+1)^2 = (z + 1) + 1 + z = 0$

:x Wait, but then that means that I've actually found the isomorphism, the remapping of z -> x, z + 1 -> y, 1 -> 1 and 0 -> 0.

8. Jan 3, 2014

### philbabbage

Is that actually sufficient for showing that the kernels are the same?

9. Jan 4, 2014

### Mandelbroth

Do you think it's not? :tongue: