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.


1 Answer 1


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. $$R=\Bbb F_5[(x_n)_{n\in\Bbb N}]/\langle (x_n^5-x_n)_{n\in\Bbb N})\rangle$$

  • $\begingroup$ So $A$ is a field and applying that to $R/f^-(Q)$ ( which is actually ID) gives $f^-(Q)$ as maximal . Right ? $\endgroup$ Jan 11, 2021 at 11:12
  • $\begingroup$ @user710290 Correct $\endgroup$
    – leoli1
    Jan 11, 2021 at 11:12
  • $\begingroup$ What is that eg ? Can you describe a bit ? Sorry for my lack of knowldege ! $\endgroup$ Jan 11, 2021 at 11:14
  • 1
    $\begingroup$ 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 $\endgroup$
    – leoli1
    Jan 11, 2021 at 11:20
  • $\begingroup$ This solution answers why 4th option is correct. math.stackexchange.com/questions/210940/… $\endgroup$ Feb 22, 2021 at 7:58

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