How to construct the subring generated by a set, T? I'm trying to find a constructive way of describing the subring generated by some subset, T, of a ring R.  I think I could describe it as all finite sums of finite products of elements of T, but I have no idea how to write that as a set or how to prove that it is equal to the subring generated by T.
 A: Let $S$ be a subset of a commutative ring $R$. By the universal property of polynomial rings, there exists a (unique) morphism of $\mathbb{Z}-$algebras $\phi:\mathbb{Z}[x_{s}: s\in S]\rightarrow R$ such that $\phi(x_{s})=s$ for all $s\in S$. Then $\mathbb{Z}[S]$, the image of this morphism, is the subring of $R$ which is generated by $S$. In fact, it is easy to see that $\mathbb{Z}[S]$ is the intersection of all subrings of $R$ that are contain $S$. One can observe that $a\in\mathbb{Z}[S]$ if and only if $a=f(s_{1},...,s_{n})$ where $f(x_{s_{1}},...,x_{s_{n}})\in\mathbb{Z}[x_{s}: s\in S]$ and $s_{i}\in S$ for all $i$.
A: Your guess is correct: the subring of $R$ generated by $T \subseteq R$ is the set $S$ of all finite sums of finite products of elements of $T$, i.e. $S = \{ \sum_{\text{finite}} \left( \prod_{\text{finite}} t_i \right) \mid t_i \in T \}$. To see this, check that $S$ is a subring (closure under multiplication follows from distributivity, the empty product is $1$ if $R$ has a $1$, and the empty sum is $0$), and also that any other subring $S'$ of $R$ containing $T$ also contains $S$.
If $R$ is commutative with unity, another way to think about $S$ is by giving it a presentation as a $\mathbb{Z}$-algebra. If $x_t$ is a set of indeterminates, one for each element of $T$, then $\mathbb{Z}[T] := \mathbb{Z}[x_t \mid t \in T]$ is the free commutative ring on the set $T$. There is a universal map $\mathbb{Z}[T] \to R$ sending $x_t \mapsto t$, and if $I$ is the kernel, then $\mathbb{Z}[T]/I$ is isomorphic to the subring $S$ defined above. In this view, the subring $S$ is often written as $\mathbb{Z}[t \mid t \in T]$
If $R$ is not commutative but still has a $1$, then one would need to take the free noncommutative algebra over $\mathbb{Z}$ instead. Finally, although the universal construction may seem rather abstract, note that the kernel $I$ above can be easily described as the set of relations, under operations of $R$, satisfied by elements of $t$.
