About the Dimension of a complete local ring

Question

Let $k$ be a field, and let $A$ be a local, noetherian, complete k-algebra 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 $n$ ?

Answer 1

Not necessarily. See Examples of common false beliefs in mathematics and the answer by JSE and the reply to his example by Simon Wadsley.

Answer 2

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 $V=\sum_{i=0}^{d-1}t^iP_i(ct^d)$, 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.)

Answer 3

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:

1. They have the same residue field $k$ (this still works if the residue field extension is finite).
2. $A\otimes_B k$ is finite-dimensional 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 $k[[x,y]]\to k[[t]]$.