Let $A$ be a (nonzero) commutative ring with $1$.
Suppose $f: A^n \longrightarrow A^m$ is an injective homomorphism of $A$ -modules. (This implies $n\leq m$.)
Then is it true that the induced map on symmetric algebras
$$\operatorname{Sym}A^n \longrightarrow \operatorname{Sym}A^m$$
is injective?
In other words, if $\varphi: A[x_1,...,x_n]\longrightarrow A[y_1,...,y_m]$ is a homomorphism of graded $A$-algebras, and if $\varphi(x_1),...,\varphi(x_n)$ are linearly independent over $A$, then is $\varphi$ injective?
Thanks.
Remark:
This is true for $n=m$, as then $\operatorname{det}f$ is a non-zero divisor;
also, it is so for $n=1$!!
It seems to be true for $A$ reduced.