How many zero divisors are in a finite ring?

Question

I'd like to know how many zero divisors there are in the ring $\mathbb{Z}_5[x]/(x^3-2).$

Is it sufficient to note that for $p(x) = x^3-2,$ $p(3) = 0$ and $p(5)=0?$

I haven't been able to write $x^3-2$ as a product of two polynomials. If it could be written as a product, and those products would be $0$ for some $x$, would they then be zero divisors?

Answer 1

Hint: $x^3-2 = 0$ if $x=\sqrt[3]{2}$, this means that $x^3 - 2 = (x-\sqrt[3]{2})q(x)$ where $\deg(q(x)) = 2$.

First of all, is there a $\sqrt[3]{2}$ in $\mathbb{Z}_5$? Since $2 = 27 = 3^3$, yes.

Now, $x-3 \neq 0$ and $q(x) \neq 0$ in the ring, but their product is $0$. So, what can you say about them?

Answer 2

Hint. Note that as $p(3) = 0$, we can divide $x^3 -2$ by $x-3$ using polynomial division, giving $$ (x^3 - 2):(x-3) = x^2 + 3x + 4 \in \def\F{\mathbf F}\F_5[x] $$ No we check whether $x^2 + 3x + 4$ is irreducible over $\F_5$, for a quadratic polynomial it suffices to check whether it has roots in $\F_5$, we have \begin{align*} 0^2 + 0 + 4 &= 4 \ne 0\\ 1^2 + 3 + 4 &= 3 \ne 0\\ 2^2 + 1 + 4 &= 4 \ne 0\\ 3^2 + 4 + 4 &= 2 \ne 0\\ 4^2 + 2 + 4 &= 2 \ne 0 \end{align*} So we have $$ (x^3 - 2) = (x-3) \cap (x^2 + 3x + 4) $$ and both ideals are coprime. This gives $$ \F_5[x]/(x^3 - 2) \cong \F_5[x]/(x-3) \times \F_5[x]/(x^2 + 3x + 4) $$ Now $\F_5[x]/(x-3) \cong \F_5$ via $x \mapsto 3$ and $\F_5[x]/(x^2 + 3x + 4) \cong \F_{25}$, the field with $25$ elements. That is the ring $R = \F_5[x]/(x^3 - 2)$ in question is the direct product of two fields, $$ R \cong \F_5 \times \F_{25} $$ Can you now determine the zero divisors?

Answer 3

Let $f\in\mathbb Z_5[X]$ such that its residue class modulo $X^3-2$ is a zero-divisor.

Then there is $g\in\mathbb Z_5[X]$, $g\notin(X^3-2)$ such that $fg\in(X^3-2)$.
Since $X^3-2=(X-3)(X^2+3X+4)$, and $X-3$ and $X^2+3X+4$ are irreducible, we get $$X-3\mid f \quad \text{or} \quad X^2+3X+4\mid f$$ so the zero-divisors are of the form $(X-3)(aX+b)$, respectively $c(X^2+3X+4)$ with $a,b,c\in\mathbb Z_5$.

Hence the number of zero-divisors is $25+5-1=29$. (Here I've used that the elements of $\mathbb Z_5[X]/(X^3-2)$ are residue classes of polynomials of degree $\le2$.)