Questions on the proof of Theorem 14.14 in Matsumura's Commutative Ring Theory.

Question

I have three questions on the proof of 14.14. For the full proof, please refer this google book link.

The theorem states that

Let $(A, \mathfrak{m})$ be a $d$ -dimensional Noetherian local ring, and suppose that $A / \mathrm{m}$ is an infinite field; let $\mathfrak{q}=\left(u_{1}, \ldots, u_{s}\right)$ be an $\mathrm{m}$ -primary ideal. Then if $y_{i}=\sum a_{i j} u_{j}$ for $1 \leqslant i \leqslant d$ are $d$ 'sufficiently general' linear combinations of $u_{1}, \ldots, u_{s}$, the ideal $b=\left(y_{1}, \ldots, y_{d}\right)$ is a reduction of $\mathfrak{q}$ and $\left\{y_{1}, \ldots, y_{d}\right\}$ is a system of parameters of $A$.

To do this, let $k:=A/\mathfrak{m}$, which is infinite by the given condition. Then, let $Q$ be an ideal generated by all homogeneous polynomials $\varphi(x)$ in $A[X_{1},\cdots, X_{s}]$ such that $\overline{\varphi}(u_{1},\cdots, u_{s}) \in \mathfrak{q}^{n}\mathfrak{m}$ where $\overline{\varphi}(x) \in k[X_{1},\cdots,X_{s}]$ by canonical homomorphism. With some calculation in the book, we can calculate that $\dim k[X]/Q=d$.

My question starts at this point. Now, let $V=\sum_{i=1}^{s}kX_{i}$, a sub-$k$-vector space of $k[X]$ consisting of all linear forms. Let $P_{i}$, $i=1,\cdots, t$ be minimal prime divisors of $Q$. Then, $P_{i} \not\supset V$, otherwise $P_{i}=(X_{1},\cdots, X_{s})$, thus $d=\dim k[X]/Q = \operatorname{ht}(P_{i})=0$, contradiction.

Question1: Why the condition $k$ is infinite lead us to conclude that $V \neq \bigcup_{i=1}^{t}\left(V \cap P_{i}\right)$

Anyway, from this condition, we may choose $l_{1}(X) \in V$ not belong to any $P_{i}$. Then, we may repeat this choice to get $(Q,l_{1},\cdots, l_{d})$.

Question2: Why $(Q,l_{1},\cdots, l_{d})$ is $(X_{1},\cdots, X_{s})$-primary?

Answer 1

Question 1: A finite-dimensional vector space over an infinite field cannot be written as a finite union of proper subspaces. See here.

Question 2: My commutative algebra is rusty, so this may be roundabout. By the minimality condition, adding each $l_1, \ldots, l_d$ decreases the Krull dimension by $1$. Hence $\dim k[X]/(Q, l_1, \ldots, l_d) = 0$. By standard results (summarized here), this quotient is finite-dimensional over $k$, so there exists $M$ such that $X_i^M \in (Q, l_1, \ldots, l_d)$ for all $i$. But that says $(X_1, \ldots, X_s)^{sM} \subset (Q, l_1, \ldots, l_d) \subset (X_1, \ldots, X_s)$. It follows that $(Q, l_1, \ldots, l_d)$ is $(X_1, \ldots, X_s)$-primary (e.g. see Rotman's Advanced Modern Algebra, Prop. 5.111).