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Suppose we have an injective homomorphism of power series rings

$$k[[x_1,...,x_m]]\to k[[y_1,...,y_n]]$$

can $x_i$ map to a unit in $k[[y_1,...,y_n]]$?

Of course in general, for injective homomorphisms of rings, non-units can become units in larger rings. For example, $\mathbb{Z}\to \mathbb{Q}$. I am also assuming that $k$ is fixed under the above embedding.

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any hypotesis on $m,n$? $m\le n$, $n\le m$, $m<n$, $n<m$? – Fosco Loregian Mar 9 '12 at 0:31
You'll have to know something about $k$. See this. – Dylan Moreland Mar 9 '12 at 0:59
@DylanMoreland: Sorry, I am unable to make a connection between my question and your link. Let's assume $k$ is algebraically closed for simplicity. – Merlin Mar 9 '12 at 1:15
@tetrapharmakon: I haven't thought of it, so I will leave it open. – Merlin Mar 9 '12 at 1:15
up vote 2 down vote accepted

The answer is no (injective or not) at least if you consider homomorphisms of $k$-algebras. Suppose you have such a map $k[[x_1,\dots, x_n]]\to k[[y_1,\dots, y_m]]$. Composing with the evaluation at $(0,\dots, 0)\in k^m$, you get a map $$ \phi: k[[x_1,\dots, x_n]]\to k $$ and you want to prove that $\phi(x_i)=0$ for all $i\le n$. Suppose for instance that $\phi(x_1)=\lambda\in k^*$. Then $\phi(x_1-\lambda)=0$. But $x_1-\lambda$ is an unit in $k[[x_1,\dots, x_n]]$ and $$ 0 = \phi(x_1-\lambda) \phi(1/(x_1-\lambda))=\phi(1).$$ Contradiction.

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Thank you, that's a neat trick. – Merlin Mar 9 '12 at 15:12
Very clever. Well done! – user5137 Mar 9 '12 at 16:22

I don't think this can be done without addressing the issue of infinite sums in $k$. If we have such a mapping $\varphi:k[[x_1,\cdots,x_m]]\rightarrow k[[y_1,\cdots,y_n]]$ with

$$\varphi(x_i)=a+g(y_1,\cdots,y_m)\in k[[y_1,\cdots,y_m]]$$ where $a\in K$, $a\not=0$, and $g(0,0,\cdots,0)=0$, then let $$f=1+x_i + x_i^2 + x_i^3 + \cdots$$

What is $\varphi(f)$? If it's something even close to obvious, then it's the power series that one obtains by substituting $\varphi(x_i)$ into each term. However, this gives us a constant term of $$1+a+a^2 + a^3 + \cdots$$ which is something that we know about if, say, $k\in \{\mathbb{Q},\mathbb{R},\mathbb{C}\}$, but I'm not aware of any kind of study of infinite series in general fields (off the top of my head, I don't think that a series whose terms aren't eventually zero could possibly converge in a finite field).

And even if, say $k=\mathbb{R}$ and, say $a=\frac{1}{2}$, then the constant term of $\varphi(f)$ would be 2, but then consider what $\varphi$ would do to the power series $$1+3 x_i + 9x_i^2 + 27 x_i^3 + \cdots + 3^n x_i^n + \cdots$$ In this case, the series formed by the constant term diverges. What then?

The issue here is why we have the word "formal" in the phrase "rings of formal power series": in a general algebraic setting, we don't care about convergence, and we don't plug anything into our power series except zero.

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I am gathering you are saying that such homomorphisms, if they exist cannot be substitution homomorphisms. Zariski-Samuel restrict to power series without constant terms while defining substitution homomorphisms. Although I am not sure whether there are no other homomorphisms between power series rings, which is what led me to this question. – Merlin Mar 9 '12 at 1:17
@Merlin: this is actually a good question. – user18119 Mar 9 '12 at 9:21
Yes, upon further thinking and after asking for feedback as to whether I was missing something obvious, this only works for homomorphisms that respect infinite sums. This clearly need not be the case. – user5137 Mar 9 '12 at 16:23

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