Let $M$ be a non-compact complex manifold of complex dimension $m$. As an example, one can let $M = \mathbb{C}^m$, with coordinates $z_1,\cdots,z_m$. "Usually", given a holomorphic $f$ function on $M$, you would expect $\{f=0\}$ to have codimension 1, i.e. to have dimension $m-1$. However, there are some degenerate cases:
A) trivially, if $f=0$, then $\{f=0\}$ is all of $M$. This case can be easily ruled out by requiring $f$ not to vanish identically.
B) there are examples where $\{f=0\}$ could be empty. This is the case if for instance $M = \mathbb{C}^m$, and we let $f(z_1,\cdots,z_m) = e^{z_1+\cdots+z_m}$.
Are there other possible degenerate cases than A) and B)? In other words, if $f$ does not vanish identically, but does vanish at some point in $M$, then is it necessarily the case that $\{f=0\}$ has dimension $m-1$?
Edit: If $\partial f \neq 0$ at some point $p$ in $\{f=0\}$, then the dimension of $\{f=0\}$ in a neighborhood of $p$ is indeed $m-1$, by the holomorphic version of the implicit function theorem. Hence in this case, there is a non-empty open subset of $M$ of dimension $m-1$. But then I just realized I have an issue. What is the definition of dimension for a singular analytic space? This is I am sure in Grauert's work. In the case where $\partial f \neq 0$ at some point $p$ in $\{f=0\}$, does it follow that the dimension of $\{f=0\}$ is $m-1$? It would then remain to consider the case where $\partial f$ vanishes identically on $\{f=0\}$.
If $M$ is a connected complex manifold of dimension $m$ and $f\in \mathcal O(M)$ is a non zero, non-invertible holomorphic function, then the zero set of $f$ always has codimension one: $$\operatorname {dim}Z(f)=m-1$$ Remarks
1) The function $f$ is invertible if and only if it nowhere vanishes.
Non-zero constants or your $e^{z_1+\cdots+z_m}$ are examples of invertible functions.
2) The result is valid for $M$ compact, but in a vacuous sense: every holomorphic $f$ is constant and thus zero or invertible!
3) The result is false for a non smooth complex analytic space.
For example, take $M=Z(z_1\cdot z_2)\subset \mathbb C^2$ (the union of the coordinate axes) and $f=z_2$.
That function is neither zero nor invertible, but nevertheless $Z(f)$ is of dimension $1$ although $M$ has dimension $1$ too.