$\mathbb{Z}_5 [x]/\langle x^2-2\rangle$ and $\mathbb{Z}_5 [x]/\langle x^2-3\rangle$

Question

Show that $\mathbb{Z}_5 [x]/\langle x^2-2\rangle $ and $\mathbb{Z}_5 [x]/\langle x^2-3\rangle$ are not isomorphic

Answer 1

In addition to the comments, it is also quite easy to build an explicit isomorphism between these two rings. Indeed, define ring homomorphisms $f \colon \mathbb{Z}[x]/\langle x^2-2\rangle$ and $g \colon \mathbb{Z}[x]/\langle x^2-3\rangle$ by the conditions: $$ \begin{array}{rcl} f\left([x]_{\langle x^2-2\rangle}\right) &=& [2x]_{\langle x^2-3\rangle}, \\ g\left([x]_{\langle x^2-3\rangle}\right) &=& [3x]_{\langle x^2-2\rangle}. \end{array} $$ By $[p(x)]_{\langle q(x) \rangle}$ I denoted the class of $p(x)$ modulo $q(x)$. Of course, the fact that these conditions do define two ring homomorphisms needs checking, which I will omit.

It is quite easy to see that $f\circ g$ and $g \circ f$ are identity maps, so $f$ is an isomorphism.

Answer 2

To simplify a bit the verifications needed in the isomorphism as constructed in the answer by Dan Shved, you could proceed as follows. Let $\alpha\in\Bbb Z_5[x]/\langle x^2-2\rangle$ and $\beta\in\Bbb Z_5[x]/\langle x^2-3\rangle$ be the respective images of $x$. You can think of $\alpha$ as $\sqrt2$ and of $\beta$ as $\sqrt3=\sqrt{-2}$. Neither $2$ nor $3$ has a square root in $\Bbb Z_5$ which shows that the quotients here are both fields; indeed both quadratic extensions of the finite field $\Bbb Z_5$, which is enough to know the are abstractly isomorphic. Since $\Bbb Z_5$ does contain squares root of $-1$ (or $4$), namely $2$ and $3$, you can easily locate the square roots of $2=(-1)\times(-2)$ in the extension $\Bbb Z_5[x]/\langle x^2-3\rangle$: they are $2\beta$ and $3\beta$.

Now consider the ring morphism $f:\Bbb Z_5[x]\to\Bbb Z_5[x]/\langle x^2-3\rangle$ that sends $x$ to $2\beta$; it is obviously surjective. Its kernel contains $x^2-2$ (because $(2\beta)^2=2$), and no polynomials of degree${}<2$ (since $1$ and $2\beta$ are linearly independent over $\Bbb Z_5$), so the kernel is precisely the ideal $\langle x^2-2\rangle$; then by the first isomorphism theorem $f$ induces an isomorphism $$ \overline f:\Bbb Z_5[x]/\langle x^2-2\rangle\to \Bbb Z_5[x]/\langle x^2-3\rangle. $$ One has $\overline f(\alpha)=2\beta$ (and also $\overline f(3\alpha)=\beta$), so $\overline f$ is just the isomorphism of the other answer.