Introductory ring theory: ring homomorphism (polynomials with coefficients in $\Bbb Q$) Show that there is no ring homomorphism $h: \Bbb Q [x]/(x^2-2) \to \Bbb Q[x]/(x^2-3)$
I know that $h$ must satisfy:
(i) $h(1)=1$
(ii )$h(a+b)=h(a)+h(b)$
(iii) $h(ab)=h(a)h(b)$
But don't really know how to proceed...
 A: The key is to see that if a map preserves addition and multiplication then it preserves relations - that is, if an element satisfies a relation in the domain, then its image must satisfy a corresponding relation in the range. In particular, equations like $ x\cdot x=1+1$ are relations. What happens when you apply a ring homomorphism to both sides of that equation? It says the image of $x$ satisfies the same equation! But does any element of $\Bbb Q[y]/(y^2-3)$ satisfy that equation? (Prove your answer, of course.)
A: If there exists the ring homomorphism $h$, then by composing $h$ with the canonical map $\mathbb{Q}[x]\to\mathbb{Q}[x]/(x^2-2)$, there is a ring homomorphism
$$
f\colon \mathbb{Q}[x]\to \mathbb{Q}[x]/(x^2-3)
$$
such that $x^2-2\in \ker f$. The homomorphism $f$ is determined by $f(x)$, because $\mathbb{Q}[x]/(x^2-2)$ is a ring containing (a copy of) $\mathbb{Q}$, so $f(a)=a$, for all $a\in\mathbb{Q}$ (prove it).
Say $f(x)=p(x)+(x^2-3)$, with $p(x)\in\mathbb{Q}[x]$; by using division with remainder, we can assume $p(x)=\alpha x+\beta$.
The fact that $x^2-2\in\ker f$ means then that
$$
(\alpha x+\beta)^2-2=\alpha^2x^2+2\alpha\beta x+\beta^2-2
$$
is divisible by $x^2-3$. Now
$$
\alpha^2x^2+2\alpha\beta x+\beta^2-2=
\alpha^2(x^2-3)+2\alpha\beta x+\beta^2-2+3\alpha^2
$$
so the condition is
$$
\begin{cases}
2\alpha\beta=0\\
3\alpha^2+\beta^2-2=0
\end{cases}
$$
However, $\alpha=0$ means $\beta^2=2$ and $\beta=0$ means $3\alpha^2=2$, but both equations have no solution in $\mathbb{Q}$.
