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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0Do you ask for a homomorphism or an isomorphism? – 2012-09-09
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0I meant it as stated, homomorphism, but an answer for the isomorphism case is welcome too. – 2012-09-09
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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, I thought there'd be a simple answer, thanks a lot. – 2012-09-09
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6Or take $R=\mathbb C$ and the complex conjugation as ring automorphism. – 2012-09-09
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0@QiL Since you solved the case when we ask for an isomorphism, you can post your comment as an answer. – 2012-09-09
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1To 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. – 2012-09-09
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1how 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. – 2012-09-09
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1Dear @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. – 2012-09-09
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0(Of course, you can still fix this by declaring that $1=0$ in the target ring...) – 2012-09-09
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0@AaronMazel-Gee But this would force the target ring to be the zero ring, and then we'd have to take the source ring to be the zero ring, and then we'd have the identity map! – 2012-11-09
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0@AlexKruckman, It does force the target to be the zero ring, but not the source. Every ring maps to the zero ring uniquely, but the zero ring cannot map to any non-zero ring. The zero ring is the final object in the category of rings. – 2012-11-10
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0@KeenanKidwell I'm well aware of all that. But the question is about a map from a ring to itself! – 2012-11-10
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0@Alex Ah, I forgot about that. My apologies. – 2012-11-10
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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.