I Let A and B be ideals of a ring. Prove that AB subsete A &nbsp;cap B

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I Let A and B be ideals of a ring. Prove that AB subsete A &amp;nbsp;cap B

nick1337 · Accepted Answer

I will modify your reverse inclusion statement. Let us suppose we have $a + b = 1$ for $a \in A$ and $b \in b$ . Then for $x \in A \cap B$ we have $x a + x b = x$ . Note that $x a \in A B$ because $x \in A \cap B$ (see note below) and similarly $x b \in A B$ . Thus $x$ is the sum of two elements in $A B$ . Thus, $x \in A B$ , and so we have reverse inclusion.

As mentioned by Potato above a Commutative Ring is needed here (I use it implicitly when I say $x a \in A B$ , because $x a$ is really in $B A$ , but by being in a commutative ring, $B A = A B$ ).

As a counterexample, take the ring of $2 \times 2$ upper triangular matrices over $R$ . Take $A = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ , $B = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})$ , $I = (A), J = (B)$ (as left ideals). Now all matrices of $I$ are of the form $(\begin{matrix} a & 0 \\ b & 0 \end{matrix})$ and the matrices of $J$ are of the form $(\begin{matrix} 0 & c \\ 0 & d \end{matrix}), \forall a, b, c, d \in R$ . Now $A + B = I d$ , so $I + J = R$ . Yet, $I \cap J = 0$ and if $C = (\begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix}) \in J$ , $A C = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) \in I J$ .

I Let A and B be ideals of

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