Show whether two quotient rings are isomorphic

[Mr Davis 97]

Homework Statement

Sorry for the multiple postings. I actually solved the other problems, so I have this last one:

Are $\mathbb{Z}_3[x] / \langle x^2 + 2x + 1 \rangle$ and $\mathbb{Z}_3[x] / \langle x^2 + x + 2 \rangle$ isomorphic

Homework Equations

The Attempt at a Solution

It seems like they wouldn't be, because one ideal is maximal while the other isn't, but I am not sure if this is correct.

 

[fresh_42]

Homework Statement

Sorry for the multiple postings. I actually solved the other problems, so I have this last one:

Are $\mathbb{Z}_3[x] / \langle x^2 + 2x + 1 \rangle$ and $\mathbb{Z}_3[x] / \langle x^2 + x + 2 \rangle$ isomorphic

Homework Equations

The Attempt at a Solution

It seems like they wouldn't be, because one ideal is maximal while the other isn't, but I am not sure if this is correct.

Which is maximal? Have you factored the polynomials?

 

[Mr Davis 97]

Which is maximal? Have you factored the polynomials?

$x^2 + 2x + 1 = (x+1)^2$ but $x^2 + x + 2$ can't be factored because it has no roots in the field, hence it is maximal

 

[fresh_42]

Yes, you're right. I thought it was $x^2+x-2$. So one quotient ring is a field and the other one not even an integral domain.

 

[Mr Davis 97]

Yes, you're right. I thought it was $x^2+x-2$. So one quotient ring is a field and the other one not even an integral domain.

Also, one more thing. How does one list out the elements of quotient rings such as these? Also, is there a general way to count the number of elements in $\mathbb{Z}_n[x] / \langle p(x) \rangle$, where $p$ is an mth degree polynomial?

 

[fresh_42]

Yes, the equivalence classes of ${R}[x]/\langle p(x) \rangle$ are all of the form $[r(x)]_{p(x)} = r(x) + \{s(x)\cdot p(x)\,\vert \,s(x) \in R[x]\}$ which means, one can push all monomials of degree $m := \operatorname{deg}p$ and higher into the ideal. However, it gets a little bit more complicated, if the highest coefficient isn't a unit (and I would have to do the math first). Let's therefore assume the highest coefficient is $1$ and $p(x) = x^m +\ldots + p_1x+p_0$. Then $[x^m]_{p(x)}= -[p_{m-1}x^{m-1}]_{p(x)} - \ldots - [p_{1}x]_{p(x)} - [p_{0}]_{p(x)}$ which means all elements $x^n$ with $n \geq m$ can be reduced to lower powers. Now multiply the possible coefficients with the degree and you have the number of elements. Or better: $\{[1]_{p(x)}, \, \ldots \,, [x^{m-1}]_{p(x)} \}$ forms a basis of the $R-$ module (or vector space in case $R$ is a field). Of course if $R$ isn't finite, you will still get a finite dimension, but infinitely many elements.

 

[Mr Davis 97]

Yes, the equivalence classes of ${R}[x]/\langle p(x) \rangle$ are all of the form $[r(x)]_{p(x)} = r(x) + \{s(x)\cdot p(x)\,\vert \,s(x) \in R[x]\}$ which means, one can push all monomials of degree $m := \operatorname{deg}p$ and higher into the ideal. However, it gets a little bit more complicated, if the highest coefficient isn't a unit (and I would have to do the math first). Let's therefore assume the highest coefficient is $1$ and $p(x) = x^m +\ldots + p_1x+p_0$. Then $[x^m]_{p(x)}= -[p_{m-1}x^{m-1}]_{p(x)} - \ldots - [p_{1}x]_{p(x)} - [p_{0}]_{p(x)}$ which means all elements $x^n$ with $n \geq m$ can be reduced to lower powers. Now multiply the possible coefficients with the degree and you have the number of elements. Or better: $\{[1]_{p(x)}, \, \ldots \,, [x^{m-1}]_{p(x)} \}$ forms a basis of the $R-$ module (or vector space in case $R$ is a field). Of course if $R$ isn't finite, you will still get a finite dimension, but infinitely many elements.

So in this case, for both of the quotient rings, would we have $3^2 = 9$ elements?

 

[fresh_42]

So in this case, for both of the quotient rings, would we have $3^2 = 9$ elements?

I get six.

 

[Mr Davis 97]

I get six.

I'm trying to reconcile your work with comment I found on https://math.stackexchange.com/ques...nts-of-the-quotient-ring-of-a-polynomial-ring. In the comment why does he use exponentiation?

 

[fresh_42]

I'm trying to reconcile your work with comment I found on https://math.stackexchange.com/ques...nts-of-the-quotient-ring-of-a-polynomial-ring. In the comment why does he use exponentiation?

Because (for short without the annoying $[.]_{p(x)}$) we have a basis $\{1,x,x^2,x^3,\ldots , x^{m-1}\}$. These are all monomials of lower degree than $m$. All higher can be reduced with subtractions by $p(x)$ to a lower one. And $p(x)=0$ in this quotient ring, so the subtractions do no harm to the ring element. This means $m$ basis vectors (or elements) and each can carry $n$ elements.

And I have to correct myself. Your $9$ was correct, I mistakenly added the $n$'s instead of multiplying them. I should get some sleep ...