Number of ideals in the factor ring $\mathbb{Q}[x]/\langle x^4+3x^2+2 \rangle$

Question

Let $R=\mathbb{Q}[x]/\langle x^4+3x^2+2 \rangle$. Then $R $ has

$(a)$ four maximal ideals

$(b)$ four prime ideals

$(c)$ exactly two maximal ideals.

$(d)$ exactly four ideals

My thoughts:

$$x^4+3x^2+2=(x^2+1)(x^2+2)$$

Now $x^2+1$ is irreducible over $\mathbb{Q}$ (since no roots).

Now I show $x^2+2$ is irreducible over $\mathbb{Q}(i)$.

On the other hand, it is reducible iff it has a root. So let $a+ib$, $a,b \in \mathbb{Q}$ be a root of $x^2+2$. Then

$a^2-b^2+2+(2ab)i=0$

$\Rightarrow a=0$ or $b=0$

In the respective cases

$b^2=2$ or $a^2+2=0$, impossible.

Basically, the above quotient ring is the extension field $\mathbb{Q}(i,√2i)$ and it is known that there are only two ideals in a field.

I am stumped to see the options!! I am still learning these things. Am I missing something obvious? Please provide me useful hints to proceed.

Thanks for your suggestions and time.

Answer 1

Observe the following fact: Let $R$ be a commutative ring and $I$ be an ideal of $R$. If $J\supseteq I$ is also an ideal of $R$, then $$J/I=\{a+I\mid a\in J\}$$ is also an ideal of $R$. This fact immediately gives two ideals of your ring $\mathbb{Q}[x]/\langle (x^2+1)(x^2+2)\rangle$. Moreover, every ring has two trivial ideals. Therefore we have four ideals of $\mathbb{Q}[x]/\langle (x^2+1)(x^2+2)\rangle$.

We need to check that there are no more ideals other than them: if $J$ is an ideal of $\mathbb{Q}[x]/\langle (x^2+1)(x^2+2)\rangle$ and $\pi:\mathbb{Q}[x]\to \mathbb{Q}[x]/\langle (x^2+1)(x^2+2)\rangle$ be the canonical epimorphism, then $\pi^{-1}[J]$ is an ideal containing $\langle (x^2+1)(x^2+2)\rangle$.

Moreover, the map $J\mapsto \pi^{-1}[J]$ is one-to-one for ideals (This is known as the corresponding theorem for rings.) Hence our problem reduces to counting ideals of $\mathbb{Q}[x]$ which contains $\langle (x^2+1)(x^2+2)\rangle$.

It remains to comprehend the maximal ideals of $\mathbb{Q}[x]/\langle (x^2+1)(x^2+2)\rangle$. We can check it directly, as we have only four ideals.

Answer 2

Hint:

Since you have a factorisation of the polynomial into a product of irreducible polynomials, you can apply the Chinese remainder theorem: the natural map \begin{align} \mathbf Qx]/(x^4+3x^2+2)&\longrightarrow \mathbf Q[x]/(x^2+1)\times\mathbf Q[x]/(x^2+2)\\ p(x)&\longmapsto \bigl(p(x)\bmod x^2+1,p(x)\bmod x^2+2\bigr) \end{align} is a ring isomorphism. Hence, you have to determine the number of ideals in the product of two fields.