# 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!

• $3\mathbb{Z}[x]$ isn't a maximal ideal in $\mathbb{Z}[x]$: the quotient is $\mathbb{F}_3[x]$ which isn't at all a field. Commented Jan 9, 2022 at 21:30
• @Mindlack, so what would a maximal ideal look like in $\mathbb{Z}[x]$? Is it possible to find a quotient that produces a field with 9 elements? Commented Jan 9, 2022 at 21:32
• You could, for example, try to find a quotient of $\mathbb{F}_3[x]$ that has nine elements. Commented Jan 9, 2022 at 21:33
• There is nothing with $\mathbb Q[x]$ because if $F\to R$ is a non-zero commutative ring homomorphism, then it is $1-1.$ In particular, for $\mathbb Q\to\mathbb Q[x]\to F$ is not possible if $F$ is finite. Commented Jan 9, 2022 at 21:42

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.

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.

• What is the universal property of $\mathbb{Z}[X]$ again? And also, is the generator $1$ of $\mathbb{Z}_2^3$ is in the image of $f$ because $f(1)=1$?
– user637978
Commented Jan 9, 2022 at 22:26
• @NobelCat The universal property of $\mathbb{Z}[X]$ is that for all rings $R$ and all $r \in R$, there is a unique ring homomorphism $f : \mathbb{Z}[X] \to R$ such that $f(X) = r$. And $1$ is indeed in the image of $f$ because $f(1) = 1$. Commented Jan 9, 2022 at 22:28