Let $R=K[[x,y,z]]$ be the formal power series ring over a field $K$. We set $M=K[[x,y]]$. Then $M$ has a structure of $R$-module by $f:R\longrightarrow M $ via $g(x,y,z)\rightsquigarrow g(x,y,0)$. In this way we can regard $M$ as an $R$-submodule because $K[[x,y]]\subset K[[x,y,z]]$. We say an $R$-module $M$ over an integral domain $R$ is torsion-free if zero is the only element of $M$ annihilated by some non-zero element of the ring $R$. It is obvious that every submodule of torsion-free module is torsion-free. By view of this definition $K[[x,y,z]]$ is torsion-free as itself-module since its integral domain and therefore $K[[x,y]]$ must be torsion-free $K[[x,y,z]]$-module. On the other hand $z$ is not zero in $K[[x,y,z]]$ and annihilates every member of $K[[x,y]]$. I can't figure out where is the mistake.
Torsion-free modules over formal power series rings
commutative-algebraformal-power-series
Best Answer
$M=K[[x,y]]$ is a subset of $R=K[[x,y,z]]$, but is not a submodule of $R$. One way to see this is to use your own argument.
Claim: There is no injective $R$-linear map from $M$ to $R$.
Proof. Suppose $\phi: M\to R$ is an injective $R$-linear map. Let $m\in M$ be a nonzero element. Then $\phi(m)\neq 0$ and hence $0\neq z\phi(m)=\phi(zm)=\phi(0)=0,$ a contradiction.