Let $\Omega$ be an algebraically closed field, $k$ a subfield of $\Omega$, $I$ an ideal of $\Omega[X_1, ... , X_n]$, and $I_k = I \cap k[X_1, ... , X_n]$. Then $I_k$ is an ideal of $k[X_1, ... , X_n]$. There is a well defined homomorphism of $\Omega$-modules: $$I_k \otimes_k \Omega \rightarrow I$$ given by $f(X) \otimes c \mapsto c f(X)$, which I believe should be injective.
Here is what I have so far as a proof. Since $\Omega$ is a flat $k$-module, the inclusion $I_k \subseteq I$ gives that the natural map $$I_k \otimes_k \Omega \rightarrow I \otimes_k \Omega$$ is an injective homomorphism of $\Omega$ modules. I would like to say then, that $I \otimes_k \Omega \cong I \otimes_{\Omega} \Omega$, or at least the former injects into the latter, since then $I \otimes_{\Omega} \Omega \cong I$. Is this correct?
Consider the following commutative diagram of $\Omega$-modules with exact rows.
$$\require{AMScd} \begin{CD} 0 @>>> I_k \otimes_k \Omega @>>> k[x_1,\ldots,x_n] \otimes_k \Omega @>>> (k[x_1,\ldots,x_n]/I_k) \otimes_k \Omega @>>> 0 \\ \ @VVV @VV\sim V @VVV \ \\ 0 @>>> I @>>> \Omega[x_1,\ldots,x_n] @>>> \Omega[x_1,\ldots,x_n]/I @>>> 0. \end{CD} $$