Ideal in polynomial ring extension 
Let $K \subset L$ be a field extension, $K[X]$ and $L[X]$ the corresponding polynomial rings (in one variable) and $I \subset K[X]$ an ideal. I want to show that $I=K[X] \cap IL[X]$, where $IL[X]$ denotes the ideal generated by $I$ in $L[X]$.  

I was told that while there are many ways to show this abstractly, there is supposed to be a very simple proof only  involving Linear Algebra.
I don't really know where to start here. The inclusion from left to right is trivial, but I haven't got much more.
Any help - even just a hint - would be appreciated.
 A: In general, if $F$ is a field, then $F[x]$ is a PID (hint: Euclidean Algorithm).
So suppose $I = (f)$ where $f \in K[x]$. To show the non-obvious inclusion, it suffices to show that if $fg \in K[x]$ for some $g \in L[x]$, then $g \in K[x]$. Write $f = \sum a_i x^i$ and $g = \sum b_j x^j$, where $a_i \in K$ and $b_j \in L$. Then $fg$ has constant term $a_0b_0 \in K$, so $b_0 \in K$. Then the linear term is $a_0b_1 + b_0 a_1 \in K$, so $b_1 \in K$. You can then proceed similarly.
There is a flaw I left in - what should you do if $a_0$ or $b_0$ is zero? (or $a_1, b_1$, etc.) I'll leave this to you to fix it.
This is a typical question you will see in descent theory. For example, Galois descent says if $V$ is a $K$-vector space and $L/K$ is Galois, then $(V \otimes_K L)^{\operatorname{Gal}(L/K)} = V$. (So if $L/K$ is Galois, then you can apply $V=I$ and you are home.) Bernard asked whether you know about faithfully flat modules because, lo and behold, there's something called faithfully flat descent.
