# Inverse image of a non-zero prime ideal under a surjective ring homomorphism

Let $$R, S$$ be commutative rings with unity . Let $$f:R\to S$$ be a surjective ring homomorphism

$$Q\subseteq S$$ be a non-zero pime ideal . Which of the following statements are true?

$$(a)f^-(Q)$$ is a non-zero prime ideal in $$R$$

$$(b)f^-(Q)$$ is a maximal ideal in $$R$$ if $$R$$ is a PID

$$(c)f^-(Q)$$ is a maximal ideal in $$R$$ if $$R$$ is a finite commutative ring with unity .

$$(d)f^-(Q)$$ is a maximal ideal in $$R$$ if $$x^5=x$$ for all $$x\in R$$

$$(a)$$ True.

Let $$r_1r_2 \in f^-(Q)$$

Then $$f(r_1r_2 )=f(r_1)f(r_2) \in Q$$. Since $$Q$$, is prime ideal, we have either of $$f(r_1),f(r_2)$$ belongs to $$Q$$ .

So either $$r_1 \in f^-(Q)$$ or $$r_2 \in f^-(Q)$$

Also $$f^-(Q)$$ is non-zero and proper since $$Q$$ is non-zero and $$f$$ is surjective.

This proves the result.

$$(b)$$ True since every non-zero prime ideal in a PID is maximal.

$$(c)$$ True.

$$f^-(Q)$$ is non-zero prime .

$$\Rightarrow R/f^-(Q)$$ is finite ID (since $$R$$ is finite) and hence a field .

$$\Rightarrow f^-(Q)$$ is maximal ideal.

$$(d)$$ I am not sure about this.

Can it be said that order (multiplicative) of every element is $$4$$ ?

Can it be said that the ring is finite (since order of every element is finite) and then considering the logic in $$(c)$$ ?

Can it be said that the ring is PID under the given conditions ?

Please help me complete by appropriate hints . Thank you.

Hint for (d): If you have an integral domain $$A$$ with the property $$x^5=x$$ for all $$x\in A$$, then for any $$x\ne0$$ we have $$x^4=1$$, so ...
Apply this to $$R/f^{-1}(Q)$$
I think $$R$$ might neither be finite nor a PID with the given assumptions, consider e.g. $$R=\Bbb F_5[(x_n)_{n\in\Bbb N}]/\langle (x_n^5-x_n)_{n\in\Bbb N})\rangle$$
• So $A$ is a field and applying that to $R/f^-(Q)$ ( which is actually ID) gives $f^-(Q)$ as maximal . Right ? Jan 11, 2021 at 11:12
• In the finite field $\Bbb F_5$ we have $x^5=x$ for all elements $x$ and in characteristic $5$ the map $z\mapsto z^5$ is a homomorphism. By taking the quotient of the polynomial ring $\Bbb F_5[y]/(y^5-y)$ I add some other element $y$ to the ring that satisfies $y^5=y$. If we do that infinitely many times we get a ring such that its elements still satisfy $x^5=x$ but the ring is no longer finite. And it isn't a PID since it has zero-divisors Jan 11, 2021 at 11:20