Is $\mathbb{F}_{3} [x] / (x^{2} + 2x + 1)$ isomorphic to $\mathbb{F}_{3} [x] / (x^{2} + 1)$ or $\mathbb{F}_{3} [x] / (x^{2} + 2)$?

Question

Is $\mathbb{F}_{3} [x] / (x^{2} + 2x + 1)$ isomorphic to $\mathbb{F}_{3} [x] / (x^{2} + 1)$ or $\mathbb{F}_{3} [x] / (x^{2} + 2)$?

I think $\mathbb{F}_{3} [x] / (x^{2} + 2x + 1) \not\cong \mathbb{F}_{3} [x] / (x^{2} + 1)$ because the former is not a field since ($x^{2} + 2x + 1$) is reducible, but the latter is a field since $(x^{2}+1)$ is irreducible. Is this correct?

I am not sure about the second ring.

Answer 1

Yes, you are correct about the first one. One way to show that also $\mathbb{F}_3[x]/(x^2+2x+1) \ncong \mathbb{F}_3[x]/(x^2+2)$ is to note that $(x+1)^2 = 0$ in $\mathbb{F}_3[x]/(x^2+2x+1)$ while the square of any non-zero element of $\mathbb{F}_3[x]/(x^2+2)$ is non-zero, so they can't be isomorphic.