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

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

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Unless I'm mistaken both $x^2-2$ and $x^2-3$ are irreducible over $\mathbb Z_5[x]$. In particular both rings are isomorphic to the field with $25$ elements. –  JSchlather Nov 19 '12 at 8:21
Looking through your questions, it appears that they are all homework questions, but none of them have been marked as such. Please read the FAQ and follow the rules of the site. You should mark homework questions as homework, and you should say something about what you've already tried. –  Noah Snyder Nov 19 '12 at 11:25

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.

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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.