Why does passing to the reduced ring not change the number of primes ideals? I'm reading a note of Hochster's, and I don't follow something. He writes as the Corollary on page 9,

Let $K\subseteq S$, where $K$ if a field, and $S$ is a finitely-dimensional $K$-vector space of dimension $\leq n$. Passing to $S_\mathrm{red}$ only decreases its $K$-vector space dimension, while the number of prime ideals does not change.

The note doesn't define $S_\mathrm{red}$ is. What exactly is $S_\mathrm{red}$, and why does the number of prime ideals not change upon passing to it? 

Given a ring $A$, its reduction $A^{\text{red}}$ is the quotient $A/\mathrm{nil}(A)$, where $\mathrm{nil}(A)$ is the nilradical
$$\mathrm{nil}(A)=\{x\in A\mid x^n=0 \text{ for some }n\geq 0\}$$
In general, a reduced ring is one whose nilradical is trivial, and as you would expect, $A^{\text{red}}$ is a reduced ring (and in an appropriate sense, the natural way of producing a reduced ring from $A$).
It is a basic theorem of commutative algebra that any prime ideal $P\subset A$ contains the nilradical; in other words,
$$\{\text{prime ideals of }A\}=\{\text{prime ideals of $A$ containing $\mathrm{nil}(A)$}\}$$
The fourth (a.k.a., "lattice") isomorphism theorem says that the quotient map $A\to A/\mathrm{nil}(A)$ creates a bijection
$$\{\text{ideals of $A$ containing $\mathrm{nil}(A)$}\}\longleftrightarrow\{\text{ideals of }A/\mathrm{nil}(A)\}$$
This bijection in fact restricts to a bijection
$$\{\text{prime ideals of $A$ containing $\mathrm{nil}(A)$}\}\longleftrightarrow\{\text{prime ideals of }A/\mathrm{nil}(A)\}$$
Combining, there is a bijection
$$\{\text{prime ideals of }A\}\longleftrightarrow\{\text{prime ideals of }A/\mathrm{nil}(A)\}$$
Thus, they have the "same number" of prime ideals.
