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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).

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While $\mathbb{R}$ does not contain any nontrivial finite subrings (think about addition), there are other infinite rings that contain (nontrivial) finite subrings. Think about an infinite field with finite characteristic. – hardmath Sep 1 '11 at 14:41
up vote 6 down vote accepted

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.

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I agree that any subring of $\mathbb{R}$ must contain $\mathbb{Z}$ and so there are no finite nontrivial subrings $S \subset \mathbb{R}$. I just used $\mathbb{R}$ to express the trivial subring. – Alex Hirzel Sep 1 '11 at 19:03

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.

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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.,

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My strategy is usually to make a post quickly and edit things in later as I think of them; I didn't see you had posted the same example as an answer until I'd already added it in. You have my +1 :) – Zev Chonoles Sep 1 '11 at 14:53
@Zev, no hard feelings, I do the same thing myself. – Henning Makholm Sep 1 '11 at 14:57

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).

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Depending on whether one wants the multiplicative identity of the big ring to coincide with the one of the smaller one, this might not be considered a subring... – Sebastian Sep 1 '11 at 15:36

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