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How many ring automorphisms are there of $\mathbb{Z}/n\mathbb{Z}$? Calling such rings $A_n$, how many ring automorphisms are there of $\prod_1^n A_{n_j}$?

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    Since you are fairly new, I wanted to remind you of a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level.2012-06-25
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    Do you require ring homomorphisms of unital ring to map $1$ to $1$?2012-06-25
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    This is not homework, this is merely a question I had today that I couldn't come up with the answer to in a short time. I am curious about a full classification of such ring automorphisms. I wasn't thinking that $1$ had to map to $1$. If $1$ is the additive generator of $A_n$ (to use my notation), I suppose it could map to any other additive generator.2012-06-25
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    If you don't require homomorphisms of unital rings to map $1$ to $1$, then you should say so explicitly: just like a lot of people consider "rings" to automatically include a $1$, also a lot of people consider homomorphisms of rings to necessarily map the multiplicative identity to the multiplicative identity.2012-06-25
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    My hypothesis is that any sensible way of mapping 1 to another additive generator can be used to construct the ring automorphisms, so some expression involving the phi function would be enough to complete the problem2012-06-25
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    I will keep that in mind from now on, thanks.2012-06-25
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    For the second problem, it will depend on the $n_j$; the Chinese Remainder Theorem and the universal property of the product easily reduces the problem into the situation in which each $n_j$ is a power of the same prime, so that we have $\mathbb{Z}/p^{a_1}\mathbb{Z}\times\cdots\times\mathbb{Z}/p^{a_r}\mathbb{Z}$, with $a_1\leq\cdots\leq a_r$ (assuming finitely many factors, at any rate...). They are pretty restrictive, since the only idempotents have all coordinates equal to $0$ or $1$, and I think that other than swaps of coordinates if they happen to be equal, you won't have any...2012-06-25
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    I'm not sure what you mean by your "hypothesis", unless you mean "No, I don't require my ring homomorphisms to map the multiplicative identity to the multiplicative identity." Because, as it happens, mapping $1$ to $a$ will only give you a ring homomorphism if the image of $1\times 1$ is the same as the image of $1$; that is, if and only if $a^2\equiv a\pmod{n}$; and if you want this to be bijective, you need $\gcd(a,n)=1$, which immediately tells you that *no*, it is not the case that "any sensible way" will give you a ring homomorphism.2012-06-25

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For $\mathbb{Z}/n\mathbb{Z}$, the only ring automorphism is the identity. For finite products, the only nontrivial automorphisms come from exchanging prime-power order direct factors.

A group homomorphism of $\mathbb{Z}/n\mathbb{Z}$ to itself is completely determined by the image of $1+n\mathbb{Z}$. In order to be a group automorphism, you must map it to $a+n\mathbb{Z}$ with $\gcd(a,n)=1$. If $1\mapsto a$, then it is easy to verify that the homomorphism is just $r+n\mathbb{Z}\longmapsto ar+n\mathbb{Z}$.

But in order to be a ring isomorphism, we also require that it respects products. Therefore, the image of $bc$ must be the product of the images of $b$ and $c$; since $1^2=1$, we must have that $a^2\equiv a\pmod{n}$. And since $\gcd(a,n)=1$, this means that $a\equiv 1\pmod{n}$, so the map is the identity.

For the second problem, I'm assuming a finite number of direct factors. The Chinese Remainder Theorem establishes a ring isomorphism between $\mathbb{Z}/n\mathbb{Z}$ and the direct product of the rings $\mathbb{Z}/p_i^{a_i}\mathbb{Z}$, where $n=p_1^{a_1}\cdots p_r^{a_r}$ is a factorization of $n$ into prime powers of distinct primes, so we lose no generality by assuming that each $n_j$ is a prime power. Moreover, we may assume all $n_j$ are powers of the same prime $p$, since maps between direct factors corresponding to distinct primes must be the zero map.

Now, the only nontrivial idempotents in $\mathbb{Z}/p^a\mathbb{Z}$ are $0$ and $1$; this can be established easily using congruences; alternatively, note that a nontrivial idempotent $e$ in a commutative ring $R$ with unity will yield a direct product decomposition $Re \times R(1-e)$, and since $\mathbb{Z}/p^a\mathbb{Z}$ is indecomposable, it follows that if $e$ is idempotent then $e=1$ or $e=0$.

Let $R = \mathbb{Z}/p^{a_1}\mathbb{Z}\times\cdots\times\mathbb{Z}/p^{a_r}\mathbb{Z}$, with $1\leq a_1\leq\cdots\leq a_r$. And we consider the automorphisms of $R$.

Now, as a group, the product is generated by the idempotents $\mathbf{e}_i$, where $\mathbf{e}_i$ has a $1$ in the $i$th coordinate and $0$s elsehwere. Their images must be idempotents, so they must be of the form $(b_1,\ldots,b_r)$ with $b_i$ idempotent, hence $b_i\in \{0,1\}$. In particular, if $a_i\lt a_j$, then the $j$th coordinate of the image of $\mathbf{e}_i$ must be $0$, by considering the additive order of $\mathbf{e}_i$.

Now, since $\mathbf{e}_i\mathbf{e}_j = (0,0,\ldots,0)$ if $i\neq j$ (the idempotents are "mutually orthogonal"), it follows that the support of the image of $\mathbf{e}_i$ (the coordinates with nonzero component) must be disjoint from the support of the image of $\mathbf{e}_j$ if $i\neq j$.

Let $t$ be the largest index such that $a_t=a_1$. Then since $\mathbf{e}_i$, $1\leq i\leq t$ must be mapped to some idempotent in $\mathbb{Z}/p^{a_1}\mathbb{Z}\times\cdots \times\mathbb{Z}/p^{a_t}\mathbb{Z}$, and different $\mathbf{e}_i$ must map to mutually orthogonal idempotents, it follows that any automorphism of $R$ must simply permute the $\mathbf{e}_i$, $1\leq i\leq t$. Looking at the next largest power of $p$ that occurs we see therefore that the same thing must occur, etc. Thus, if the ring is of the form $$\left(\frac{\mathbb{Z}}{p^{a_1}\mathbb{Z}}\right)^{n_1}\times\cdots\times \left(\frac{\mathbb{Z}}{p^{a_s}\mathbb{Z}}\right)^{n_s}$$ with $1\leq a_1\lt \cdots \lt a_s$, $n_i\geq 1$, then the only automorphisms correspond to permutations of the factors of $\mathbb{Z}/p^{a_i}\mathbb{Z}$ for each $i$. Hence there are $(n_1!)\cdots (n_s!)$ such automorphisms.

In the case of a single factor, we get $s=1$ and $n_1=1$, so we get the previous result as a special case.