By a ring I mean a ring with a multiplicative identity. To me, at this point, this sounds like a fairly simple question, but I haven't been able to come up with any such homomorphism, nor has searching Google for one proved fruitful.
Can you have a ring homomorphism from a ring to itself which isn't the identity?
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9[Galois theory](http://en.wikipedia.org/wiki/Galois_theory) is all about automorphisms of certain fields. – 2012-09-09
4 Answers
Sure, let $R$ be a commutative ring with identity and consider the map $R[x] \to R[x]$ determined by $p(x) \mapsto p(0)$.
To see a case where the map is an isomorphism, let $R = \Bbb Z[\sqrt{2}]$ and consider the map $a + b \sqrt{2} \mapsto a - b \sqrt{2}$. You should check that this is a homomorphism, and actually gives an isomorphism from $\Bbb Z[\sqrt{2}]$ to itself.
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0Ah, evidently I forgot about that too! @KeenanKidwell, just to clarify -- being terminal doesn't mean you can't map to any other object. (I know that you didn't quite say this, of course.) For instance, $\mbox{pt} \in \mbox{Top}$ is terminal; in fact, in certain contexts people *define* a "point" to be a map from the terminal object! – 2012-11-10
I post my comment as suggested by Davide.
Take $R=\mathbb C$ and the complex conjugation as ring automorphism.
You can consider for example $M_n(\mathbb{R}),$ the ring of $n\times n$ matrices over the real numbers (with $n>1$) and take the map $M_n(\mathbb{R})\ni A \mapsto PAP^{-1}\in M_n(\mathbb{R}),$ where $P\ne xI$ is an invertible matrix of $M_n(\mathbb{R})$ and $x\in \mathbb{R}.$
Of course you can take any ring $R$ instead of $\mathbb{R}.$
Recall that functors between categories preserve automorphisms. So for any faithful functor $F : \mathbf{Set} \rightarrow \mathbf{C}$, the object $F(\kappa)$ will have at least $\kappa!$-many distinct automorphisms, where I write $\kappa!$ for the number of set-theoretic automorphisms of a set with $\kappa$-many elements. For example, let $F : \mathbf{Set} \rightarrow \mathbf{CRing}$ denote the free functor. Then $F(2)$ has at least $2! \;(=2)$ different automorphisms (one for each permutation of the variables!), and $F(\aleph_0)$ has at least $\aleph_0!$-many (that is, $2^{\aleph_0}$-many; that is, continuum-many) distinct automorphisms.