Let $k$ be a field, and let $A$ be a local, noetherian, complete kalgebra with residue field $k$. Suppose that there are elements $t_1,\dots,t_n$ in the maximal ideal of $A$ such that the map $k[[X_1,\dots,X_n]] \rightarrow A$ that sends $X_i$ to $t_i$ for all $i$ is injective. Is the dimension of $A$ greater or equal than $n$ ?
3 Answers
Not necessarily. See Examples of common false beliefs in mathematics and the answer by JSE and the reply to his example by Simon Wadsley.

$\begingroup$ The example given there is essentially that if $f\in vk[[v]]$ is transcendental over $k(v)$, then $x\mapsto u$, $y\mapsto uv$, $z\mapsto uf(v)$ defines a continuous injective $k$algebra embedding of $k[[x,y,z]]$ into $k[[u,v]]$. Note that this answers negatively the question for all $n\ge 3$. This leaves the case $n=2$ open (the case $n\le 1$ being trivial). $\endgroup$– YCorOct 4, 2014 at 8:32

1$\begingroup$ Is the case n=1 that trivial, YCor ? $\endgroup$ Oct 4, 2014 at 10:21

$\begingroup$ Case $n=1$: a negative answer would mean that there is an injective continuous $k$algebra homomorphism from $k[[X]]$ into a 0dimensional $A$, which would be of finite length over $k$. Do I miss something? $\endgroup$– YCorOct 4, 2014 at 16:02

$\begingroup$ No, ok. I made a mistake: I thought you meant that the case where dim A=1 is trivial. And I do not know whether k[[x,y]] can be embedded in k[[t]]. $\endgroup$ Oct 4, 2014 at 18:10

An extended comment: if $k$ has characteristic zero then there is no continuous embedding of local rings $k[[u,v]]\to k[[t]]$. Indeed, suppose it maps $(u,v)$ to $(U,V)$. Let $d$ be the valuation of $U$, that is, $u\in t^dk[[t]]\smallsetminus t^{d+1}k[[t]]$. Write $U=ct^d(1+w)$ with $w\in tk[[t]]$. Using the usual power series for $(1+X)^{1/n}$, we can write $(1+w)=(1+W)^n$ for some $W\in tk[[t]]$. Thus $U=x(t(1+W))^{d}$. There exists a continuous $k$algebra automorphism of the local ring mapping $t$ to $t(1+W)$ (just because $t(1+W)$ has valuation 1). Hence after conjugation by this automorphism, we can assume that $w=0$, that is, $U=ct^d$.
Then we decompose $V$ according to the value modulo $d$ of the exponents: we write $V=\sum_{i=0}^{d1}t^iP_i(ct^d)$, with $P_i(T)\in k[[T]]$. Thus $V$ belongs to a finite extension of $k((ct^d))$, and hence $V$ is algebraic over $k((ct^d))$. Hence there exists a nonzero $Q(X)\in k[[ct^d]][X]$ such that $Q(V)=0$. If we write $Q(X)=R(ct^d,X)$, we deduce $R(U,V)=0$, and $R$ is a nonzero element of $k[[X]][Y]\subset k[[X,Y]]$. (Of course the special form of $R$ is due to the fact we used an automorphism to assume $U$ is a monomial.)

$\begingroup$ The argument would extend to characteristic $p$ if it were true that for all $x\in tk[[t]]\smallsetminus\{0\}$, the extension $k((x))\subset k((t))$ is finite. I don't know in general (it's true if $x$ has valuation $<p$, or if $x\in k[t]$). $\endgroup$– YCorOct 4, 2014 at 20:53

$\begingroup$ (line 3: "$u\in$" should be "$U\in$"; I'll edit later) $\endgroup$– YCorOct 5, 2014 at 11:15
Let me treat the remaining case $n=2$. More generally, assume $\varphi:(B,\mathfrak{m}_B)\to (A,\mathfrak{m}_A)$ is a local homomorphism of complete noetherian local rings with the following properties:
 They have the same residue field $k$ (this still works if the residue field extension is finite).
 $A\otimes_B k$ is finitedimensional over $k$. Equivalently, $\mathfrak{m}_B\,A$ is $\mathfrak{m}_A$primary.
(Note that condition 2 is automatic if $\varphi$ is an injection $k[[x,y]]\to k[[t]]$).
Claim: $A$ is a finite $B$module.
Indeed, our assumptions imply that $A$ is $\mathfrak{m}_B$adically complete and separated as a $B$module, so a standard limit argument shows that $A$ is generated by any sequence whose image in $B/\mathfrak{m}_A B$ is generating.QED
In other words, the morphism $\mathrm{Spec}(A)\to \mathrm{Spec}(B)$ is finite. This implies $\dim(A)\leq\dim(B)$, with equality if $\varphi$ is injective. In particular, there is no injection $k[[x,y]]\to k[[t]]$.

$\begingroup$ If your claim is true, then the consequence about power series rings seems to have nothing to do with $n=2$. How to reconcile this with the existing counterexamples for $n\ge 3$? $\endgroup$ Nov 22, 2014 at 10:26

$\begingroup$ All I prove is that if $n\geq2$ then $\dim A\geq2$. If $\dim A\leq1$, this forces property 2. $\endgroup$ Nov 22, 2014 at 10:59