Unique maximal ideal and ring epimorphism kernel with prime numbers equivalence

Question

Let $A$ be a ring. Let $f: \mathbf{Z} \to A$ be a surjective ring homomorphism. Prove that $A$ has a unique maximal ideal iff there exists $n\in \mathbf{N}$ and $p\in\mathbf{N}$ a prime number such that $\ker f =(p^{n})$.

$(p^{n})$ denotes the ideal generated by $p^{n}$, I'm not sure if that's the standard notation.

We had to prove that in a recent exam and nobody could prove it. I haven't found it online neither. Thank you!

Answer 1

I assume by 'ring epimorphism' you mean 'surjective ring homomorphism'. Since $f$ is surjective we have $A\cong \Bbb Z/\ker f$. We know that the maximal ideals of $\Bbb Z/\ker f$ correspond bijectively to the maximal ideals of $\Bbb Z$ containing $\ker f$, hence $A$ has a unique maximal ideal iff there is a unique maximal ideal of $\Bbb Z$ containing $\ker f$. Now use the fact that $\Bbb Z$ is a PID.

Answer 2

Hint By the first isomorphism theorem $A$ is isomorphic to $\mathbb Z / \mbox{Ker}(f)$.

$\Rightarrow$ Note that $\mbox{ker}(f)=(m)$ for some integer $m$. Assume by contradiction that $n$ is not a power of a prime. Since $n \neq 1$, there exist two primes $p \neq q$ which divide $m$. Show that $(p)/(m)$ and $(q)/(m)$ are maximal ideals in $\mathbb Z / \mbox{Ker}(f)$.

$\Leftarrow$: Show that $(p)/(p^n)$ is the only maximal ideal in $\mathbb Z/(p^n)$.

Second solution If you do not want to use the FIT, you can proceed as follows:

Since $\mathbb Z$ is a PID then $\mbox{ker}(f)= (m)$ for some integer $m$.

Claim $I$ is ideal in $A$ exactly when $I= f((k))$ for $k|m$.

Proof

$\Leftarrow$ follows from $f$ being surjective.

$\Rightarrow$ Let $I \subseteq A$ be any ideal. Then $f^{-1}(I)$ is an ideal in $\mathbb Z$ and hence $$ f^{-1}(I)=(k) $$ for some $k |m$.

[This comes from $(m) =\mbox{ker}(f) \subseteq f^{-1}(I) =(k)].

Since $f$ is onto, you get $$ I=f((k)) $$

Now, your problem follows immediately from the claim.

Answer 3

Use the Chinese remainder theorem. Every ideal in $\mathbb Z$ looks like $n\mathbb Z$ for a unique nonnegative integer $n$. If you write $n = p_1^{e_1} \cdots p_s^{e_s}$ for primes $p_i$, then $n\mathbb Z = p_1^{e_1}\mathbb Z \cap \cdots \cap p_s^{e_s}\mathbb Z$. The Chinese remainder theorem gives

$$\mathbb Z/n\mathbb Z \cong \mathbb Z/p_1^{e_1}\mathbb Z \times \cdots \times \mathbb Z/p_s^{e_s}\mathbb Z$$.

The rings $\mathbb Z/p_i^{e_i}\mathbb Z$ are local rings with unique maximal ideals $p_i\mathbb Z/p_i^{e_i}\mathbb Z$. Now, use the fact that if $R_1, ... , R_s$ are commutative rings with identity, the maximal ideals of

$$R = R_1 \times \cdots \times R_s$$

are all of the form

$$R_1 \times \cdots \times R_{i-1} \times \mathfrak m \times R_{i+1} \times \cdots \times R_s$$

for some $1 \leq i \leq s$, where $\mathfrak m$ is a maximal ideal of $R_i$. Thus $R$ cannot be a local ring unless $s=1$.