Injective ring homomorphism $f: \Bbb Z/ n\Bbb Z \to A$ does not imply $A$ is an integral domain

Question

Let $A$ be a ring such that there exists an injective ring homomorphism $f: \Bbb Z/ n\Bbb Z \to A$ for $n \geq 2$.

I know that if $A$ is an integral domain, then $n=p$ where p is prime but is the converse true ? The answer is obviously yes if $f$ is surjective since $A$ would be isomorphic to $\Bbb Z /p \Bbb Z$ that is a field. I thought of taking a ring of the form $\Bbb Z/ m\Bbb Z$ where m is not a prime, but $f: \Bbb Z/ p\Bbb Z \to \Bbb Z/ m\Bbb Z $ exists if and only if $m$ divides $p$ which gives $m=p$ or $m=1$, both are not counterexamples.

While I am writing this, I thought of a ring homomorphism $f: \Bbb Z /p \Bbb Z \to (\Bbb Z /p \Bbb Z)\times (\Bbb Z /p \Bbb Z)$ that maps $\overline n \to (\overline n,\overline n)$ :

• $f(\overline{x}+\overline{y})=(\overline{x}+\overline{y},\overline{x}+\overline{y})=(\overline{x},\overline{x}) + (\overline{y},\overline{y})=f(\overline x) + f(\overline y)$
• $f(\overline x \overline y)=(\overline x \overline y,\overline x \overline y)=(\overline x ,\overline x)(\overline y,\overline y)$
• $f(\overline 1)=(\overline 1, \overline 1)$
• $f$ is injective because $\ker(f)=\overline 0$

However, $(\Bbb Z /p \Bbb Z)\times (\Bbb Z /p \Bbb Z)$ is not an integral domain. Is everything ok with my proof ?

Answer 1

What you did is fine. To give a big picture, the condition that there exists an injective ring homomorphism $f:\mathbb Z/p\mathbb Z\rightarrow A$ is equivalent to say $A$ is a $\mathbb F_p$-algebra. Such an algebra is definitely not necessarily a domain, e.g. $\mathbb F_p[x]/(x^2)$.