These are problems bothering me.
Let $S$ and $T$ are subrings of a ring $R$.
And $I$ and $J$ are ideals of the $R$.
Question 1
Let product of subrings like ideal product $IJ$ which means
$ST = \{ s_1t_1+s_2t_2+...+s_nt_n \vert s_k \in S , t_k \in T, n \in \mathbb{N}\}$
This $ST$ might be not subring of the $R$ so I've tried to find the counterexample of this but I failed.
Questuion 2
When ideal case, $IJ$ is the subring of $I \cap J$
Then generally, if $ST$ is a subring of $R$, is $ST$ subring of the $S\cap T$?
Please help me :(
An example is $S =k[x]$ and T = $k[y]$ inside of the noncommutative polynomial ring $k\langle x,y\rangle$. So for instance $xyxy \not\in ST$, but its is a valid product of two elements in $ST$.
The issue comes from noncommutativity; if $S$ and $T$ commute in the larger ring then $ST$ will be a ring. (Or more generally if you can rewrite $ts = s’t’$, or in other words if $TS \subseteq ST$.)
No we don’t have $ST \subset S \cap T$, this property for ideals requires absorptivity, which subrings in general lack. For instance in the example above $S\cap T = k$ is much smaller than $ST$. And izf your subrings contain identity then $S\cup T \subset ST$ so there is no way its contained in $S\cap T$.