Units in $A = \mathbb{Z}_3[x]/(x^2+1)$

Question

Let $A = \Bbb{Z}_3[x]/(x^2+1)$, the quotient ring by the ideal $(x^2+1)$. Which ones are units?

I did this question in a very boring way, merely listing all the possibilities and check. I cannot find an efficient way to find the unit. Is there any efficient way to do this?

Answer 1

Perhaps the most efficient elementary way of doing it in this case is to note that

1. everything nonzero in $\mathbb Z/3\mathbb Z$ is a unit
2. everything in $A$ can be written (uniquely) in the form $a+bx$ for $a,b\in\mathbb Z/3\mathbb Z$

and then ask what an inverse for $a+bx$ would look like in general. For this, since we are identifying $x^2+1\sim0$, we can think of $x$ as a square root of $-1$, so you can seek inspiration from the complex numbers. For complex numbers, the inverse of $a+bi$ is given by rationalising the denominator: $$ \frac1{a+bi} = \frac{a-bi}{(a+bi)(a-bi)} = \frac{a-bi}{a^2+b^2} = \frac a{a^2+b^2} + \frac{-b}{a^2+b^2}i $$ you can check that if the right-hand side is well-defined in $A$ (using $x$ instead of $i$), then it will also serve as an inverse. You can now ask when the right-hand side is defined, which is whenever $a^2+b^2\neq0\pmod3$, and this is true whenever $a$ and $b$ are not both zero.

Therefore, the units are $a+bx$ for $a$ and $b$ not both zero, and this constitutes all nonzero elements of $A$.

Answer 2

Certainly, @shibai's solution is the best way to think about things without using much machinery, but since you ask for an efficient way to solve the problem, I will add the following.

Observe that $f(x) = x^2 + 1$ is irreducible modulo $3.$ Considering that $f(x)$ is quadratic, this is equivalent to checking that $f(x)$ has no roots modulo $3.$ (Explicitly, by the Factor Theorem, a root will correspond to a linear factor. But one linear factor gives rise to another linear factor for degree reasons, and that gives a way to write $f(x)$ as a product of two polynomials of strictly lesser degree.) We have that $f(0) = 1,$ $f(1) = 2,$ and $f(2) = 5 \equiv 2 \operatorname{ (mod } 3),$ so $f(x)$ is irreducible modulo $3.$ Over a field $k,$ the polynomial ring $k[x]$ is a principal ideal domain (PID), hence every ideal of $k[x]$ can be generated by one polynomial. Further, one can prove that prime ideals in a PID are actually maximal, and an irreducible polynomial $f(x)$ in $k[x]$ is prime, hence the ideal $(f(x))$ is prime and so maximal. Last, the quotient of a ring by a maximal ideal is a field, so the quotient ring $\mathbb Z_3[x] / (x^2 + 1)$ is a field, hence any nonzero element is a unit. But the nonzero elements of $\mathbb Z_3[x] / (x^2 + 1)$ are the polynomials $ax + b$ such that either $a \neq 0$ or $b \neq 0.$