Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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? –  user69810 Apr 23 '13 at 4:30
1  
@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

1 Answer 1

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 $S\cup T$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.