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Let $A$ be an arbitrary ring. In "Commutative Algebra" by Zariski and Samuel it is claimed that every continuous homomorphism $A[[Y_1,...,Y_m]] \to A[[X_1,...,X_n]]$ is a substiution homomorphism $Y_i \to f_i$, where $f_i \in (X_1,...,X_n)$. I think the statement is false. [I have replaced the proof with a more general one]

If $B$ is a $A$-algebra, which is complete with respect to the $I$-adic topology of an ideal $I \subseteq B$, then continuous homomorphisms $A[[Y_1,...,Y_m]] \to B$ correspond to $m$-tuples in $rad(I)$.

Proof: Clearly any such homomorphism $h$ is determined by the values $f_i = h(Y_i)$. If $h$ is continuous, then there is some $l \geq 1$ such that $(Y_1,...,Y_m)^l \subseteq h^{-1}(I)$. If $g_i$ is the image of $f_i$ in $R/I$, this means $(g_1,...,g_m)^l=0$. Such an $l$ exists iff all the $g_i$ are nilpotent, i.e. $f_i \in rad(I)$. Assume conversely that $f_i \in rad(I)$, say $f_i^r \in I$ for some $r$. Then for every $k \geq 1$, $A[Y_1,...,Y_m]/(Y_1,...,Y_m)^{kr} \to B/I^k, Y_i \mapsto f_i$ is well-defined and is compatible in $k$. In the limit we get the desired homomorphism $A[[Y_1,...,Y_m]] \to B$.

Thus in the example $B = A[[X_1,...,X_n]]$, the $f_i$ should have a nilpotent constant terms; they don't have to vanish.

Am I right? I'm a bit confused since I thought everything is right in such a book, even more such elementary considerations. I also wonder why the universal property of $A[[Y_1,...,Y_m]] \to B$ is only stated (at least in the literature I know) in the form, that every $m$-tuple in $I$ yields a homomorphism $A[[Y_1,...,Y_m]] \to B$, instead of really describing the hom functor as I did above.

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Is a proof given? –  Harry Gindi May 11 '10 at 23:53
With my notation: "The continuity of $h$ requires that high powers of $f_i$ belong to high powers of the ideal $(X_1,...,X_n)$. Hence $f_i \in (X_1,...,X_n)$." Well the first sentence is correct, but the conclusion is wrong, I think. –  Martin Brandenburg May 11 '10 at 23:55
As far as I understand, you are correct, and this is well known (so unless the book implicitly assumes A to have no nilpotents, this is a mistake). There is a related geometric phenomenon: if you consider the group of automorphisms of a formal disk $Spf k[[x]]$ (or the semigroup of endomorphisms, if you prefer), at first sight any automorphism preserves the marked point... but then you realize that this is only true for closed points of the group --- the group has non-reduced directions that no longer preserve the special point. –  t3suji May 12 '10 at 0:37
As it happens, the failure of the claim is a starting point for the Gelfand-Kazhdan "formal geometry" constructions of bundles with connection on manifolds. Without the nilpotent constant terms in the substitutions, we wouldn't get connections. –  S. Carnahan May 12 '10 at 1:21
Very interesting answers ... erm comments. I would like to accept something as an answer. :-) –  Martin Brandenburg May 12 '10 at 7:59
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