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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 $IK[X]$ denotes the Ideal generated by $I$ in $K[X]$.

I got 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.

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.