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I have a problem when I am proving the following question:

Let $f:A \to A'$ be a surjective homomorphism of rings, and assume that $A$ is local, $A' \neq 0$, show that $A'$ is local.

I have found an answer in this question. However in my proof, suppose that $\mathfrak{m}$ is the unique maximal ideal in $A$, and I have proved that $f(\mathfrak{m})$ is an ideal and contains any other ideals. To finish this proof, I should prove $f(\mathfrak{m})\neq A'$. But I can't prove $f(1)\notin f(\mathfrak{m})$.

For example, let $g:\mathbb{Z}\oplus \mathbb{Z} \to \mathbb{Z}$ be the natural projection onto its first coordinate. We know $\mathbb{Z}\oplus 2\mathbb{Z}$ is a maximal ideal in $A$, but $g((1,1)) \in g(\mathbb{Z}\oplus 2\mathbb{Z})$.So it's not true for a general ring.

Is there any direct way to prove $f(1)\notin f(\mathfrak{m})$ for a local ring?

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    How are you using the fact that the homomorphism is surjective? – Mark Bennet Nov 17 '17 at 08:25
  • I use the fact to prove that $f(\mathfrak m)$ is an ideal of $A'$. –  Nov 17 '17 at 08:49
  • If your definition of local ring assume commutativity and identity, then you also have to show $A’$ is commutative and $f(1_A)$ is multiplicative identity of $A’$. These are easy to prove. – user264745 Jan 05 '24 at 20:30

2 Answers2

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Note that $f(1) = 1$, or more explicitly $f(1_A) = 1_{A'}$. Using the hypothesis that $A' \neq 0$, we have $1_{A'} \neq 0$. Therefore $1_A \notin \ker(f)$.

Therefore $\ker(f)$ is a proper ideal of $A$. Using the hypothesis that $A$ is a local ring, we have $\ker(f) \subseteq \mathfrak{m}$.

For any $x \in \mathfrak{m}$, we have $1-x \notin \mathfrak{m}$ (prove this by contrapositive: $x,1-x \in \mathfrak{m} \implies 1 \in \mathfrak{m}$). Therefore $1-x \notin \ker(f)$, so $f(1-x) \neq 0$, so $f(x) \neq f(1)$. This shows that $f(1) \notin f(\mathfrak{m})$.

Zach Teitler
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Since $f$ is surjective, the fundamental theorem of homomorphisms states that

\begin{align} \psi\colon A/\ker f&\to A'\\ x+\ker f&\mapsto f(x) \end{align}

is a isomorphism. The isomorphism theorems shows that there is a one-to-one correspondence between the set of ideals of $A'$ and the set of ideals of $A$ containing $\ker f$. Note that $\mathfrak m+\ker f$ is a maximal ideal for $A/\ker f$. Thence $\psi(\mathfrak m+\ker f)=f(\mathfrak m)$ is a maximal ideal of $A'$. Therefore $f(1)\notin f(\mathfrak m)$.

josephz
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  • I know this proof is right, but I'm looking for a 'direct' way. Could you please prove $f(1) \notin f(\mathfrak m)$ from the definition. I want to understand this more profoundly. –  Nov 17 '17 at 09:19