# Can you have a ring homomorphism from a ring to itself which isn't the identity?

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.

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Do you ask for a homomorphism or an isomorphism? – Davide Giraudo Sep 9 '12 at 14:01
I meant it as stated, homomorphism, but an answer for the isomorphism case is welcome too. – Monty Gill Sep 9 '12 at 14:07
Galois theory is all about automorphisms of certain fields. – Karolis Juodelė Sep 9 '12 at 14:41

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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Ah, I thought there'd be a simple answer, thanks a lot. – Monty Gill Sep 9 '12 at 13:41
Or take $R=\mathbb C$ and the complex conjugation as ring automorphism. – user18119 Sep 9 '12 at 14:50
To play more with the polynomial example: Take the homomorphism $R[x]\to R[x]$ induced by $x\mapsto x^2$. This is injective but not surjective. – Hagen von Eitzen Sep 9 '12 at 16:51
how about $\phi(x)=0$ for all $x \in R$. I believe this qualifies as a homorphism since $\phi(x+y) = \phi(x)+\phi(y)$ and $\phi(xy) = \phi(x)\phi(y)$. Obviously far from an isomorphism, but it is a general, all be it silly, example. – James S. Cook Sep 9 '12 at 17:51
Dear @James, When dealing with rings with identity, usually (but not always) people insist that a homomorphism send $1$ to $1$. This is the case in, e.g., commutative algebra and algebraic geometry. – Keenan Kidwell Sep 9 '12 at 19:28

I post my comment as suggested by Davide.

Take $R=\mathbb C$ and the complex conjugation as ring automorphism.

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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 $\mathbb{R}.$

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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.

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