4
$\begingroup$

$AB \subseteq A \cap B$ is clear. I have seen reverse inclusion proven thus,

Let $x \in A\cap B$. Since $A+B=R$, there exist $a \in A$, $b \in B$, such that $a+b=1$. Then $x= axa + axb + bxa + bxb$. Therefore, $A \cap B \subseteq AB$.

My problem: I cannot see why $bxa \in AB$.

Edit: If this does not hold for noncommutative rings, give a counterexample

$\endgroup$
5
  • $\begingroup$ Is the ring assumed to be commutative? $\endgroup$
    – Potato
    Apr 10, 2013 at 3:47
  • $\begingroup$ No, it explicitly said "any" ring. The proof is by PM Cohn, for the record $\endgroup$
    – Rachmaninoff
    Apr 10, 2013 at 3:54
  • $\begingroup$ For some people, all rings are commutative. In what text is this? $\endgroup$
    – Potato
    Apr 10, 2013 at 3:55
  • $\begingroup$ Intro to Ring Theory, PM Cohn pg 16, #7. $\endgroup$
    – Rachmaninoff
    Apr 10, 2013 at 3:58
  • $\begingroup$ The other reference I found this lemma in requires the ring to be commutative. I suspect this is necessary. $\endgroup$
    – Potato
    Apr 10, 2013 at 4:04

1 Answer 1

Reset to default
5
$\begingroup$

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 $A = \begin{pmatrix} 1 & 0\\ 0 & 0 \\ \end{pmatrix}$, $B = \begin{pmatrix} 0 & 0\\ 0 & 1 \\ \end{pmatrix}$, $I = (A), J = (B)$ (as left ideals). Now all matrices of $I$ are of the form $ \begin{pmatrix} a & 0 \\ b & 0 \end{pmatrix}$ and the matrices of $J$ are of the form $ \begin{pmatrix} 0 & c \\ 0 & d \end{pmatrix}, \forall a,b,c ,d \in \mathbb{R}$. Now $A+B = Id$, so $I+J = R$. Yet, $I \cap J = 0$ and if $C = \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} \in J$, $AC = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \in IJ$.

$\endgroup$
7
  • 1
    $\begingroup$ Do you happen to have a counterexample for the noncommutative case? $\endgroup$
    – Potato
    Apr 10, 2013 at 4:24
  • 1
    $\begingroup$ Here is an example I found: math.stackexchange.com/questions/47212/… Note that $IJ$ = 0, but $I \cap J$ are strictly triangular matrices with their second row being zero. Now if $IJ = JI$ the proof should still go through. $\endgroup$
    – srvasude
    Apr 10, 2013 at 4:24
  • 1
    $\begingroup$ $I+J \neq R$ in this example since $R$ contains the matrix with nonzero entry at $(2,2)$ $\endgroup$
    – Rachmaninoff
    Apr 10, 2013 at 4:37
  • $\begingroup$ I was not thinking straight, my mistake. Here is a counterexample which should work this time. Take again the ring of $2\times2$upper triangular matrices, except over $\mathbb{R}$. Take the ideal $I$ generated by the matrix $A$ with a 1 in the upper left corner, and zero everywhere else, and $J$ generated by the matrix $B$ with a 1 in the lower left corner, and zero everywhere else (left ideals). $\endgroup$
    – srvasude
    Apr 10, 2013 at 5:59
  • 2
    $\begingroup$ Now $A+B=1$ so $I+J=R$. However $I$ has matrices which have the second column zero, and $J$ has matrices which have the first column zero, so $I \cap J = 0$. Yet, $IJ$ is non zero (take $A$ and multiply by matrix with zero's in first column and 1's in second column and get a matrix that is zero except for a 1 in the upper right corner) $\endgroup$
    – srvasude
    Apr 10, 2013 at 6:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .