Avoidance Lemma
In commutative algebra, the well-known prime avoidance lemma states that if an ideal $I$ is contained in a union of finitely many prime ideals $P_i$, then it is effectively contained in $P_i$ for some $i$.
或许让你没想到的是, 这个引理实际上还有一个线性空间的版本:
$$
\DeclareMathOperator{\id}{id}
\DeclareMathOperator{\Hom}{Hom}
\DeclarePairedDelimiter{\abs}{\lvert}{\rvert}
\DeclarePairedDelimiter{\norm}{\lVert}{\rVert}
$$
Proposition
Let $F$ be an infinite field. Any $F$-vector space $V$
cannot be covered by a finite union of its proper subspace.
我们证明一个更强的结论:
Proposition
If $U_1, \cdots, U_n$ are proper subspaces of a $F$-vector space $V$
with $\abs{F} \ge n$, then $V \ne U_1 \cup \cdots \cup U_n$.
Proof.
Let $U = U_1 \cup \cdots \cup U_n$ and assume that it is irredundant,
i.e. no $U_i$ lies in the union of the others.
Choose $u, v$ such that $v \notin U_1, u \in U_1$
and $u \notin U_i$ for all $i > 1$.
We consider the "generic" line $L = v + uF$.
We claim that $L \cap U_1 = \varnothing$. If not, there exists some $c \in F$ such that $u \in U_1, v + cu \in U_1$, so $v = (v + cu) - cu \in U_1$, contradiction.
We claim that $\abs{L \cap U_i} \le 1$ for all $i > 1$. Indeed, if $v + cu, v + du \in U_i$, then $(c - d)u \in U_i$ and $c = d$.
Therefore $\abs{L \cap U} \le n - 1 < \abs{F} = \abs{L}$, so $L$ contains at least one point not in $U$. $\square$
We claim that $L \cap U_1 = \varnothing$. If not, there exists some $c \in F$ such that $u \in U_1, v + cu \in U_1$, so $v = (v + cu) - cu \in U_1$, contradiction.
We claim that $\abs{L \cap U_i} \le 1$ for all $i > 1$. Indeed, if $v + cu, v + du \in U_i$, then $(c - d)u \in U_i$ and $c = d$.
Therefore $\abs{L \cap U} \le n - 1 < \abs{F} = \abs{L}$, so $L$ contains at least one point not in $U$. $\square$