2022-04-30

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

nick1337

Expert2022-08-10Added 777 answers

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 $xa+xb=x$. Note that $xa\in AB$ because $x\in A\cap B$ (see note below) and similarly $xb\in AB$. Thus $x$ is the sum of two elements in $AB$. Thus, $x\in AB$, and so we have reverse inclusion.

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

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

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