# Smallest Subring

Suppose that $S$ and $T$ are subrings of a ring $R$. Show that their ring-theoretic product $ST$ is a subring of $R$ that contains $S \cup T$, and is the smallest such subring.

I understand that $ST$ is a subring because its an additive subgroup, its closed under multiplication, and contains the multiplicative identity. However, how do I go about proving that its the smallest subgroup?

Thanks

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It's the smallest in the sense of inclusion. You must verify that if $B$ is a subring of $A$ containing $S \cup T$ then it must contain $ST$. –  Vinicius M. Apr 23 '13 at 4:12
Can you explain this a bit more? How would that suffice to prove that its the smallest? Wouldnt a union of $S$ and $T$ be pretty large? –  Ben Apr 23 '13 at 4:24
Which is larger, the sum of two things or the product of two things? –  John Douma Apr 23 '13 at 4:30
@Ben, its the smallest satisfying the property 'contains $S\cup T$'. See if this example is clearer, let $G$ be a group and $a \in G$, the subgroup generated by $a$ can be defined as $\{a^n \mid n \in \Bbb Z\}$, but it also can be defined as the smallest subgroup $H$ such that $a \in H$, in other words, its the subgroup $H$ such that if a subgroup $N$ contains $a$, then it contains $H$. You can also describe this as the intersection of all subgroups containg $a$. Repeat the idea replacing $a$ with $S \cup T$ –  Vinicius M. Apr 23 '13 at 5:14

To show that $ST$ is the smallest subring containing $S\cup T$ we must show two things.
1. $S\cup T\subset ST$
2. $S\cup T\subset R^{'} \Rightarrow ST\subset R^{'}$
If $s\in S$ then $s=s\cdot 1_R\in ST$ and if $t\in T$ then $t=1_R\cdot t\in ST$. Since this is true for all $s\in S, t\in T$, $S\cup T\subset ST$.
If $S\cup T\subset R^{'}$, then for any $st\in ST$, $st\in R^{'}$ because $R^{'}$ is a subring and is closed under multiplication.
Therefore, $ST\subset R^{'}$ and so is the smallest subring containing $S\cup T$.