determine the cardinality and all inverrtible elements of $\mathbb{Z}[x]/(4,x^2)$.

Question

I've got to determine the cardinality and all inverrtible elements of $\mathbb{Z}[x]/(4,x^2)$. I know that in this quotient ring $4=0$ and $x^2=0$. So $x^2=4$. The only elements would be $\{1,2,3, x,2x,3x,1+x,2+x,3+x\}$. So $9$ elements. Is that the correct presentation?

Answer 1

You are not arriving at the correct answer. Take $q \in \mathbb{Z}[X]/(4,X^2)$. Then $q = \pi(p)$ is the image of some $p \in \mathbb{Z}$. We can always write

$$ p = a + bX + X^2h $$

with $h \in \mathbb{Z}[X]$ (possibly zero, that causes no problems).

Hence $q = \pi(a+bX +X^2h) = \pi(a) + \pi(b)\pi(X).$ As you say, if $a \in \mathbb{Z}$ then $\pi(a)$ is the residue modulo $4$ of $a$. Hence $\mathbb{Z}[X]/(4,X^2) = \{a+b[X] : a,b \in \mathbb{Z}_4\}.$ You can easily check that none of these are equal. Thus the quotient has $16$ elements. As for invertible elements, now that you have that presentation, $1 = (a+b[X])(c+d[X]) = ac+(ad+bc)[X]+bd[X^2] = ac + (ad+bc)[X]$ if and only if $ac = 1$ and $ad+bc = 0$. In particular, $a$ has to be either $1$ or $3$ and $a = c$. So then we need $0 = ad+bc = a(b+c)$, which happens if $b = -d$ since we have taken $a$ invertible. I'll leave you to determine all possible cases.