Let $R$ be a commutative ring. Denote by $X\ast Y=\{xy\mid x\in X,y\in Y\}$ the complex product of subsets. I want to show that given subsets $X,Y\subseteq R$ the following ideals are equal: $$(X)(Y)=(X\ast Y)\,.$$ By defnition, $(X)(Y)=((X)\ast(Y))$, so I have to show that
- $X\ast Y\subseteq((X)\ast(Y))$,
- $(X)\ast(Y)\subseteq(X\ast Y)$.
The first point is trivial, because $X\subseteq(X)$ and $Y\subseteq(Y)$ implies $$X\ast Y\subseteq(X)\ast(Y)\subseteq((X)\ast(Y))\,.$$
As for the second point, let $a=\sum_i r_ix_i\in(X)$ and $b=\sum_js_jy_j\in (Y)$. Then $$ab=\sum_{i,j}r_ix_is_jy_j=\sum_{i,j}r_is_jx_iy_j\in(X\ast Y)\,,$$ which proves 2..
I have two questions:
- Is my proof of the above assertion correct?
- Does the statement still hold if $R$ is not commutative and we replace ideal by left-, right- or two-sided ideal? I think I used commutativitiy in the second step.
Edit: I think I have a counterexample for 2.: Let $R$ be any ring and let $R\langle x,y,z\rangle$ be the $R$-algebra freely generated by $x,y,z$. Then I think $xyz\in(x)(z)$ but $xyz\notin(xz)$. This should work when $()$ is the generated left ideal aswell as if it is the generated two-sided ideal. Is this correct?
Yes, your proof looks right.
But I'm confused about your remark "by definition $(X)(Y) = ((X)(Y))$". It's not a definition, these sets just turn out to be equal because $R$ is commutative.
By definition, $(XY)$ is the ideal generated by $XY$. So an element in $(XY)$ is a linear combination $\sum_{i=1}^n r_i x_i y_i$.
$(X)(Y)$ on the other hand, consists of elements of the form $\left ( \sum_{i=1}^n r_i x_i \right ) \left ( \sum_{j=1}^m r'_j y_j \right )$.
Regarding your example: You need to assume that $R$ is not commutative. Also, I assume you meant finitely generated rather than freely. Then: yes, indeed since $yz \in (z)$, $xyz \in (x)(z)$ but $xyz \notin (xz)$ since $R$ is assumed to be non-commutative.