# Dimension of zero locus of polynomial without constant terms

Let $k$ be an algebraically closed field. Let $\mathbb A^n_k$ be the affine $n$ space over $k$. Let $0<r<n$ be an integer and suppose that $f_1,\ldots,f_r$ are polynomials in $k[x_1,\ldots,x_n]$ such that each $f_i$ has no constant term. Let $S$ denote the zero locus of $\{f_1,\ldots,f_r\}$. Show that $S$ has dimension at least $n-r$.

This is Theorem 2 from here, for which there is no proof provided. My knowledge of Algebraic Geometry is pretty limited, and basically I would try to show that the Krull dimension of $k[x_1,\ldots,x_n]/(f_1,\ldots,f_r)$ is at least $n-r$. For that I would try to find a surjection $k[x_1,\ldots,x_n]/(f_1,\ldots,f_r)\rightarrow k[y_1,\ldots,y_{n-r}]$, where $y_i$ are indeterminates. But this seems difficult to do without knowing what $S$ actually is. Another idea would seem to consider the projective $n$-space and realize $S$ as an affine chart of a projective algebraic set. But I don't know how are the dimensions of $S$ and its projectivization related and neither I see where such a method would fail if $f_i$ could be allowed to have constant terms.

For $r =1$, this is a consequence of a theorem of commutative algebra, namely the Krull's principal ideal theorem, which say that if $R$ is a noetherian ring and $x \in R$ then $\dim R/(x) \geq \dim R - 1$. You can use induction for $r > 1$.