Are there any nontrivial, finite subrings of an infinite ring? For example, $S\subset\mathbb{R}$ where $S=\{0\}$ is the trivial subring which is finite. Is there a nontrivial subring of an infinite ring (i.e. of $\mathbb{R}$ or not) that is non-infinite?
This came up as a student question in my 3000-level abstract algebra class discussion and is not a homework problem (though I am a student).
 A: A subring $S\subset\mathbb{R}$ is nontrivial when it contains some $c\neq0$, and therefore it must contain infinitely many elements: 
$$\ldots,-2c,c,0,c,2c,\ldots$$
(in fact, under the usual definition of subring, any subring must contain $1$, and so by the above argument any subring of $\mathbb{R}$ must already contain $\mathbb{Z}$, which is infinite. But I'm assuming you're not including this in your definition?)
However, there are many infinite rings having finite subrings. For example, let $R$ be any finite ring; then the polynomial ring $R[x]$ is infinite but has $R$ as a subring.
A: Zev's argument in fact shows that there cannot be any finite subrings in a ring of characteristic zero (when subrings are required to have the same multiplicative unit), and Niel's answer shows that there can be a non-trivial finite subrng in an infinite ring.
So, for an example of a finite subring of an infinite ring, let us look at rings of positive characteristic. For example, let $\mathbb{F}_p$ be the finite field of $p$ elements. Then it is a subring of the polynomial ring $\mathbb{F}_p(X)$, which is infinite. It is also a subring of the algebraic completion $\overline{\mathbb{F}_p}$, which is an infinite field.
There are also infinite rings which contain no non-trivial finite subrngs: $\mathbb{Z}$ is an example, since every additive subgroup is infinite. In fact, every integral domain of non-zero characteristic cannot contain any non-trivial finite subrngs. Let $R$ be such a ring, and suppose $S$ is a non-trivial additive subgroup. Then, $S$ contains a non-zero element $x$. But $R$ contains an isomorphic copy of $\mathbb{Z}$, and $R$ is an integral domain, so for any non-zero integer $n$, $n x \ne 0$. In particular, if $n \ne m$, then $n x \ne m x$, and both are in $S$ if $n$ and $m$ are integers. So $S$ is infinite.
A: As Zev points out, it's clear that $\mathbb{R}$ has no nontrivial finite subring.
On the other hand, there are also clearly some infinite rings that have non-trivial finite subrings -- for example $R[x]$ for any finite ring $R$, which is infinite and has $R$ itself as a subring.,
A: Yet another simple example: ℝ × (ℤ / 2ℤ) is an infinite ring, and has a finite subring which is isomorphic to ℤ / 2ℤ which is generated by (0,1).
