Corollary (?) of projective Nullstellensatz

Question

In a course on field invariants (in particular the diophantine dimension) the following claim is made.

For an algebraically closed field $K$, every system of $k$ forms in $n > k$ variables has a non-zero solution by the projective Nullstellensatz. Hence, the diophantine dimension of $K$ is $0$.

I can see why the diophantine dimension follows from the first claim, but after digging up my notes on algebraic geometry, I found this formulation of the projective Nullstellensatz (a bit altered to avoid introducting unneeded notation):

Let $K$ be an algebraically closed field. A system of forms in $n$ variables over $K$ has no nonzero solution if and only if there exists some $N \in \mathbb{N}$ such that the ideal $I$ generated by the forms contains all forms of degree at least $N$.

How does this theorem (which does not mention the number of forms in the system!) imply the first one?

Answer 1

A sketch of the solution a colleague of mine eventually found:

Suppose the ideal $I$ is generated by forms $Q_1, \ldots, Q_k$. We know that $\sqrt{I} \subseteq (X_1, \ldots, X_n)$ where $\sqrt{I}$ denotes the radical of $I$. If we can show that the inclusion is strict, then by the homogeneous Nullstellensatz we would have that $I$ is isotropic.

Suppose for the sake of a contradiction that $\sqrt{I} = (X_1, \ldots, X_n)$. Then $\sqrt{I}$ is a minimal prime ideal containing $I$, but by Krull's Höhensatz we must have that $\sqrt{I}$ cannot have height $> k$.

Answer 2

When we say all of forms of degree $N$, it includes $x_1^N,x_2^N,\ldots, x_n^N$. The only common zero for them is $(0,0,\ldots,0)$ which is not part of the projective space and defines empty set as the zero locus.

Now if the number of equations $k$ is less than the the number of variables, $n$, each of them can cut down the dimension by $1$, ending up with a variety of dimension $n-k$ or more which would be an infinite set.