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Let $A$ be a commutative ring with an identity element $1$. An element $x$ in an $A$-module $M$ is called cyclic if $Ax=M$. An $A$-module which has a cyclic element is called cyclic $A$-module.

Let $M_{1},M_{2}$ be cyclic $A$-modules and $I_{1},I_{2}$ annihilators of $M_{1},M_{2}$ respectively: $$ I_{k}=\{ a \in A \mid ax=0\ (\forall x \in M_{k}) \} \ \ \ (k=1,2). $$ Prove that a necessary and sufficient condition for the direct sum $M_{1} \oplus M_{2}$ to be a cyclic $A$-module is that $I_{1}+I_{2}=A$.

My attempt:

"$\Longrightarrow$": We know that $$M_{1} \oplus M_{2} \simeq M_1\times M_2 \simeq A/I_1 \times A/I_2 \simeq A/(I_1 \cap I_2).$$ Since $A/(I_1 \cap I_2)$ is a cyclic $A$-module with a cyclic element $1+I_1 \cap I_2$, we must have $M_{1} \oplus M_{2}$ is cyclic. But it seems that condition $I_{1}+I_{2}=A$ is not necessary?

"$\Longleftarrow$": Assume that $M_{1} \oplus M_{2}$ is cyclic then $$M_{1} \oplus M_{2} \simeq A/\rm{Ann}(M_1 + M_2) \simeq A/(I_1 \cap I_2).$$ I know that if $M_{1} \oplus M_{2} \simeq A/I_1I_2$ then $I_1 \cap I_2 =I_1I_2$ forcing $I_{1}+I_{2}=A$. But how can I prove that?

Any help would be much appreciated.

user
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1 Answers1

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Assume $I+J \neq A$. Let $\mathfrak m$ be maximal ideal containing $I+J$.

We get a surjection $A/I \oplus A/J \twoheadrightarrow A/\mathfrak m \oplus A/\mathfrak m$.

If $A/I \oplus A/J$ was cyclic, we can compose this surjection with $A \twoheadrightarrow A/I \oplus A/J$ to get a surjection

$$A \twoheadrightarrow A/\mathfrak m \oplus A/\mathfrak m$$

After tensoring with $A/\mathfrak m =: k$, we get a surjection

$$k \to k^2,$$

which is absurd.

This shows: If $I+J \neq A$, $A/I \oplus A/J$ is not cyclic. Precisely, what you wanted.

MooS
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  • It isn't quite clear to me that $k \to k^2$ cannot be a surjection? Can you explain a little bit? – user May 06 '16 at 06:26
  • basic linear algebra tells you that there is no surjective linear map $k \to k^2$. – MooS May 06 '16 at 10:13