Is the ideal $\langle x^2 + 1\rangle$ maximal in $\mathbb{Z}_3[x]$?

Question

Is the ideal $\langle x^2 + 1\rangle$ maximal in $\mathbb{Z}_3[x]$?

I am going about this by trying to prove that $\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}$ is a field. I can prove commutative and unity, but I need some help on how I should show that every nonzero element is a unit.

Answer 1

OK, let us show explicitly that every element in the ring $R=\Bbb F_3[x]/(x^2+1)$ is a unit. (This is of course a field, the extension of degree two of the field $\Bbb F_3$ with three elements, since the ideal $(x^2+1)$ is generated by an irreducible = prime element. The ring $R$ is by general means isomorphic to the one field $\Bbb F_9$ with nine elements - taken up to isomorphism. But well, the OP wants to show that every element is a unit.)

The non-zero elements in $\Bbb F_3$, identified with (classes of) constant polynomials are units. Let us consider all "other" (classes of) polynomials $f$. These have degree one. After possibly norming the principal coefficient (with units $\pm 1$ in $\Bbb F_3$) we can and do consider the monic case only. There are only three cases:

• For $f$ being (the class of) $x+1$ we have $(x+1)(x-1)=x^2-1\equiv -1-1=-2=1$.
• For $f$ being (the class of) $x-1$ we have the same computation.
• For $f=x$ consider $x(-x)=-x^2\equiv -(-1)=1$.

So every non-zero element in $R$ is a unit.

$\square$

Answer 2

$x^2 + 1$ has no roots in $\Bbb{Z}_3 := \Bbb{Z}/3\Bbb{Z}$, so the ring of equivalence classes $$\frac{\Bbb{Z}_3[x]}{\langle x^2 + 1 \rangle} := \{ \overline{a + bx}: a, b \in \Bbb{Z}_3 \}$$ is a finite integral domain, which means it's a field.

In general, if what you really want to know is whether or not an ideal $M \subset R$ of the ring $R$ is maximal, it's very inefficient (not to say impractical, or sometimes impossible) to directly check that every nonzero equivalence class $\overline{a} = a + M \in R/M$ has a multiplicative inverse in $R/M$ by hand-computation.
It's better to use some heavier machinery--in this case, the fact that $R = \Bbb{Z}_3$ is a field, so $\Bbb{Z}_3[x]$ is a Euclidean domain, which means every ideal is principal and the maximal ideals are precisely those generated by irreducible polynomials (since non-constant factors of $p(x) \in \Bbb{F}[x]$ become zero divisors in the quotient ring $\frac{\Bbb{F}[x]}{\langle p(x) \rangle}$, which is finite iff $\Bbb{F}$ is finite).

I was a little sloppy in just saying "$x^2 + 1$ has no roots", actually, because having no roots is a necessary but not a sufficient condition to be an irreducible polynomial. For instance, $x^4 + 1$ has no roots in $\Bbb{Z}_5$, but it factors over $\Bbb{Z}_5$ as $(x^2+2)(x^2+3)$. In this specific case though, $M = \langle x^2 + 1 \rangle$, so it's sufficient to say the polynomial has no roots, because the only possible factors of a quadratic like $x^2 + 1$ are linear.