# a question on the minimal prime divisors of an ideal

This question is motivated by the second part of Step 1 in the proof of Theorem 14.14 in Matsumura's Commutative Ring Theory, p. 112.

Let $k$ be an infinite field and $Q$ a homogeneous ideal of $k[x]=k[x_1,\cdots,x_s]$. Suppose that $\operatorname{dim} k[x] / Q =d>0$. Let $V$ be the $k$-vector space generated by the elements $x_1,\cdots,x_s$ (i.e. the vector space of linear forms over $k$) and let $P_1,\cdots,P_t$ be the minimal prime divisors of $Q$. Matsumura says: "By the assumption that $d>0$ we have that $P_i \not\supset V$...".

Question: why is that true? If e.g. $P_1 \supset V$, then $P_1$ is equal to the maximal ideal generated by $x_1,\cdots,x_s$. Since $P_1$ is a minimal prime divisor of $Q$, this means that the coheight of $P_1$ in $k[x]/Q$ is zero. This is fine, as long as there exists some other $P_i$ with coheight $d$. Any comments?

Remark: if Matsumura's argument does not hold in the general setting that i present here, then i suppose the nature of $Q$ comes into play. Any corroborations of that will be appreciated.

If a proper ideal contains $V$ this must be $M=(X_1,\dots,X_n)$. Moreover, you know that $M$ is minimal over $Q$ and $\dim R/Q=d>0$, where $R=K[X_1,\dots,X_n]$. Since $Q$ is homogeneous $R/Q$ is a finitely generated graded $K$-algebra and its Krull dimension equals the height of the irrelevant maximal ideal (which is $M/Q$), a contradiction.