Question about Maps of power series rings 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.
 A: 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.
A: 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.
