Show that the ideal generated by $x^2-2$ is maximal

Question

Let $A = \mathbb{Q}[x]$. Show that the ideal generated by $x^2-2$ is maximal.

I think it is sufficient to show that $\mathbb{Q}[x]/(x^2-2) \cong \mathbb{Q}\sqrt{2}$, where $\mathbb{Q}\sqrt{2} = \{a + b \sqrt{2} : a,b \in \mathbb{Q} \}$

I'm stuck on this problem.

Is anyone could help for this question?

Answer 1

Note that $\mathbb{Q}$ is a field, so $\mathbb{Q}[x]$ is a PID. In a PID $P$, if $p\in P$ then $\langle p\rangle$ is maximal iff $p$ is irreducible in $P$.

Now simply observe that $x^2-2$ has no roots in $\mathbb{Q}$, thus $x^2-2$ is irreducible in $\mathbb{Q}[x]$, so $\langle x^2-2\rangle$ is maximal.

Answer 2

Recall that an ideal $I$ in a ring $R$ is maximal if and only if $R/I$ is a field. To show that $\Bbb Q[x]/(x^2-2)$ is a field, you need to define a ring homomorphism $\phi : \Bbb Q [x]/(x^2-2) \rightarrow \Bbb Q(\sqrt2)$ by $\phi(a \overline x + b): = a\sqrt2 + b$, where $\overline x$ denotes the image of $x$ in the quotient ring, and check this is an isomorphism. Since the latter space is a field, you will be done.