Do the rings $\mathbb{Z}[x]$ or $\mathbb{Q}[x]$ have a quotient isomorphic to the field with 9 elements?

This is a question from an old algebra qualifying exam.

(a) Prove or disprove the ring $\mathbb{Z}[x]$ has a quotient isomorphic to the field with 9 elements.

(b) Prove or disprove the ring $\mathbb{Q}[x]$ has a quotient isomorphic to the field with 9 elements.

I know you need some maximal ideal I in the ring $R = \mathbb{Z}[x]$ or $R = \mathbb{Q}[x]$ in order for the quotient $R/I$ to be a field. I also know that $3\mathbb{Z}[x]$ is a maximal ideal in $\mathbb{Z}[x]$. I also know that the only maximal ideal in $\mathbb{Q}$ is $(0)$. So, I think you can only produce a quotient isomorphic to a field with 9 elements using the ring $\mathbb{Z}[x]$. That is the only information I have so far. Is this the right direction? Do you have a suggestion for an ideal I should consider?

Please note that I have only learned about rings up to irreducibility criteria. We have not covered field extensions, nor are we expected to know it for this exam.

Thanks!

Solution 1:

Equivalently, we ask whether there is a surjective ring homomorphism from these rings to the field with 9 elements (we will denote this field as $F$).

Recall that as a group, the field with 9 elements is isomorphic to $\mathbb{Z}_3^2$. One of the two generators may be taken to be $1$. Let another generator of the additive group of $F$ be $k$.

Then there is a unique ring homomorphism $f : \mathbb{Z}[X] \to F$ such that $f(X) = k$, by the universal property of $\mathbb{Z}[X]$. This ring homomorphism’s image is a subgroup of $F$ containing both $1$ and $k$, and thus is surjective.

We could also have approached this more generally by noting that the multiplicative group of any finite field is cyclic and sending $X$ to a generator of the multiplicative group.

On the other hand, any ring homomorphism $\mathbb{Q}[X] \to F$ induces a ring homomorphism $f : \mathbb{Q} \to F$. But there can be no ring homomorphism between fields of differing characteristics.

Solution 2:

In (a), consider the ideal $(3, x^2 + 1)$ in $\mathbb{Z}[x]$. The quotient ring $$\frac{\mathbb{Z}[x]}{(3,x^2 + 1)} \simeq \frac{\mathbb{F_3}[x]}{(x^2 + 1)}$$ Since $x^2 + 1$ is irreducible over $\mathbb{F}_3$, the quotient ring is a field. Every element of $\mathbb{F}_3[x]/(x^2 + 1)$ has a unique representative of the form $a + bx$ where $a,b\in \mathbb{F}_3$. It follows that this field has nine elements.

As for (b), there is no such quotient. Otherwise, let $K$ be the quotient. Composing the canonical projection map $\mathbb{Q}[x] \to K$ with the inclusion map $\mathbb{Q}\to \mathbb{Q}[x]$, we obtain a nontrivial ring homomorphism $\phi:\mathbb{Q} \to K$. If $\phi(q) = 0$ for some nonzero $q\in \mathbb{Q}$, then $1 = \phi(1) = \phi(q)\phi(q^{-1}) = 0$, a contradiction. Hence $\phi$ is injective. It follows from the first isomorphism theorem that $K$ contains a subring isomorphic to $\mathbb{Q}$, contradicting the fact that $K$ is finite.