$R$ is an algebra over an infinite field. If $\exists$ ideals s.t. $J\subseteq \bigcup_{k=1}^nI_k$ then $J\subseteq I_k$ for some $k$

Solution 1:

Here's a solution, it has nothing to do with commutative algebra, this is purely linear algebra. Let's proceed by induction. The case $n=1$, is obvious. I've treated the case $n=2$, in the comment above.

Assume that we have proved the result for $1,..., N-1$, and take $E$ a $k$-vector space, $W_1,...,W_N$ sub vector spaces of $E$ and $V$ a sub vector space contained in $W_1\cup...\cup W_N$.

If $V\subset W_1$ we're done, and if $V\subset W_2\cup...\cup W_N$ we're done too by the induction hypothesis.

So we may assume that we can find $x\in V$ s.t $x\in W_1$, $x\notin W_2\cup...\cup W_N$ and $y\in W_2\cup...\cup W_N$ and $y\notin W_1$.

As $k$ is infinite, the elements $x+ty$ are all in $V$ and there a infintely many of them, so 2 of them must lie in a $W_k$.

Wa have 2 case, if $x+ty$ and $x+sy$ both lie in $W_1$, with $s\neq t$, then $y$ lies in $W_1$ which we excluded. If $x+ty$ and $x+sy$ both lie in $W_k$ for some $k>1$ and $s\neq t$, then certainly $s\neq 0$ and $t\neq 0$ otherwise $x$ would lie in $W_k$, thus $x/s+y$ and $x/t+y$ lie in $W_k$ thus $x.(1/s-1/t)$ lies in $W_k$, which again is excluded.