How many zero divisors in $\Bbb Z_5[x]/\langle x^3+2x+2 \rangle$

Question

How many zero divisors in $\Bbb Z_5[x]/\langle x^3+2x+2\rangle$

My attempt: in $\Bbb Z_5[x]$ we can factor $x^3+2x+2$ as $(x-1)^2(x-3)$, so the polynomial decomposes into $3$ first degree polynomials so maybe we could prove with the first isomorphism theorem that the quotient ring is isomorphic to $\Bbb Z_5 \times \Bbb Z_5 \times \Bbb Z_5$. In $\Bbb Z_5 \times \Bbb Z_5 \times \Bbb Z_5$ we have $60$ zero divisors (in accordance with the amount of zeros each entry in the vector $(a,b,c)$ can get) so I thought the answer is $60$ but that's wrong. Any ideas how to solve this question?

Answer 1

I suppose by $\mathbb{Z}_5$ you mean $\mathbb{Z}/5\mathbb{Z}$. As $x^3+2x+2=(x-1)^2(x-3)$ and the (principal) ideals $(x-1)^2$ and $(x-3)$ are coprime, Chinese Remainder Theorem implies \begin{align*}\mathbb{Z}_5[x]/(x^3+2x+2)&\cong\mathbb{Z}_5[x]/(x-1)^2\times\mathbb{Z}_5[x]/(x-3)\\&\cong\mathbb{Z}_5[x]/(x^2)\times \mathbb{Z}_5\end{align*} Note that the ideals $(x)$ and $(x)$ are not coprime, so one CANNOT use Chinese Remainder Theorem to infer $\mathbb{Z}_5[x]/(x^2)\cong\mathbb{Z}_5[x]/(x)\times\mathbb{Z}_5[x]/x$; that is WRONG!

To count zero divisors in $\mathbb{Z}_5[x]/(x^2)$ note that the sum of a unit and zero divisor is a unit. It readily follows that the zero divisors are precisely $ax$ for $a\in\mathbb{Z}_5$; there are 5 of them. In $\mathbb{Z}_5$, there is only one zero divisor, namely $0$.

An element in a product of rings is a zero divisor iff at least one component is a zero divisor. Note that this way, one has counted those pairs where both entries are zero divisors twice, so one must subtract this number.

To conclude, in $\mathbb{Z}_5[x]/(x^2)\times\mathbb{Z}_5$ there are $5\cdot5+25\cdot 1-5\cdot 1=45$ zero divisors.

Finally, let me mention that in $\mathbb{Z}_5\times\mathbb{Z}_5\times\mathbb{Z}_5$ there are (by inclusion-exclusion) $25+25+25-5-5-5+1=61$ zero divisors, not 60 as claimes by the OP.

Answer 2

The variant/application of the first isomorphism theorem you want is actually the Chinese remainder theorem. Let $R=\mathbb Z_5[x]$, $I=((x-1)^2)$, and $J=(x-3)$ Consider the map $R\to R/I \times R/J$. The kernel of this map is $I\cap J=((x-1)^2(x-3))$, and because the ideals are coprime, $I+J=(1)$, we have that the map will be surjective. Therefore we get an isomorphism $R/(I\cap J)\to R/I \times R/J$.

Before we try to count the number of zero divisors in each factor, we need to understand how zero divisors in the factors give rise to zero divisors in the product. If either factor is a zero divisor, then the result is a zero divisor, because if $x\neq 0$ and $xz=0$, then $(x,y)(z,0)=0$. Inclusion/exclusion tells us that the number of zero divisors of $R/I\times R/J$, which we will denote by $z(R/I\times R/J)$, will be $z(R/I)|R/J|+z(R/J)|R/I|-z(R/I)z(R/J)$, the number of ordered pairs where the first element is a zero divisor, plus the number of ordered pairs where the second element is a zero divisor, minus the pairs where both are.

Since $R/J\cong \mathbb Z_5$, there is only one zero divisor, namely $0$. So how many zero divisors are in $\mathbb Z_5/((x-1)^2)$? There are $5$, namely all the multiples of $x-1$. So how many total zero divisors are there? $5\cdot 5+1\cdot 25-5\cdot 1=25+25-5=45$.