# Show whether two quotient rings are isomorphic

1. Apr 25, 2017

### Mr Davis 97

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

2. Relevant equations

3. 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.

2. Apr 25, 2017

### Staff: Mentor

Which is maximal? Have you factored the polynomials?

3. Apr 25, 2017

### Mr Davis 97

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

4. Apr 25, 2017

### Staff: Mentor

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.

5. Apr 25, 2017

### Mr Davis 97

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?

6. Apr 25, 2017

### Staff: Mentor

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.

7. Apr 25, 2017

### Mr Davis 97

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

8. Apr 25, 2017

### Staff: Mentor

I get six.

9. Apr 25, 2017

### Mr Davis 97

10. Apr 25, 2017

### Staff: Mentor

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 ...