# Is $\mathbb{Z}[x]/(x^2-2)$ isomorphic to $\mathbb{Z}[x]/(x^2-3)$?

Is $\mathbb{Z}[x]/(x^2-2)$ isomorphic to $\mathbb{Z}[x]/(x^2-3)$?

It seems that an isomorphism can hardly be defined. But I cannot find any property that $\mathbb{Z}[x]/(x^2-2)$ has but $\mathbb{Z}[x]/(x^2-3)$ not or vice versa, so I cannot say those to quotient ring can not be isomorphic.

• is there any element in the second ring the square of which is $2$? Dec 20, 2017 at 3:04
• in the first ring $2$ and $x^2$ are in the same coset. I.e. $2=x^2$ in the quotient. This is not true in the second ring as $2$ and $3$ will be in different cosets. Dec 20, 2017 at 3:07
• But I don't mean those two quotient ring are the same. In the second ring, $3=x^2$, what if there is a morphism that map $2$ to $3$ and $x$ to $x$?
– HeHe
Dec 20, 2017 at 3:11
• In the first ring, the $x$ is a 'dummy variable', with the property that its square is $2$ - namely, the letter $x$ is unimportant - relabel it $y$ in the first ring. In the second ring, the square of the (different) $x$ is $3$. Now, an arbitrary element in the second ring is $z=ax+b$, with $a$ and $b$ integers. Suppose $z^2=2$ - can you get a contradiction? (don't forget $x$ and $1$ are linearly independent (over the integers), and $x^2=3$ in the second ring!) Dec 20, 2017 at 3:19
• @Hehe - I edited the previous comment a hundred times - sorry. And I am identifying '$x$' with the image(s) in the quotient ring(s). Dec 20, 2017 at 3:24

Suppose that $\phi : \Bbb Z[x]/(x^2 - 2) \to \Bbb Z[y]/(y^2 - 3)$ were a ring isomomorphism. Suppose that $p \in \Bbb Z[x]/(x^2 - 2)$ is such that $\phi(p) = y$ (such a $p$ must exist if $\phi$ is onto). Then we must have $$\phi(p^2 - 3) = \phi(p^2) - \phi(3 \cdot 1) = \phi(p)^2 - 3 \phi(1) \\ = \phi(p)^2 - 3 = y^2 - 3 = 0$$ So that $p^2 - 3 \in \ker \phi$. However, since $\phi$ is an isomorphism (i.e. since it is one-to-one), this implies that $p^2 - 3 = 0$, which is to say that $p^2 = 3$.
So, which element $p \in \Bbb Z[x]/(x^2 - 2)$ satisfies $p^2 = 3$?