Definition of an algebraic set being "defined over $k$" in terms of tensor product

Question

Let $\Omega$ be an algebraically closed field, $I$ a radical ideal, and $k$ a subfield of $\Omega$. If $I = \mathcal I(A)$ for some closed set $A \subseteq \Omega^n$, then $I_k := I \cap k[X_1, ... , X_n]$ is an ideal of $k[X_1, ... , X_n]$. We say that $A$ is defined over $k$ if the homomorphism of $\Omega$-modules: $$I_k \otimes_k \Omega \rightarrow I$$ defined on generators by $f(X) \otimes c \mapsto cf(x)$ is an isomorphism. When $A$ is not necessarily defined over $k$, what is the nature of the homomorphism I just mentioned? Is it injective? Also, is $k$ allowed to be a subring (even a subring without $1$) of $\Omega$ rather than a subfield? What is the significance of this definition?

Answer 1

Being defined over $k$ basically means that you can write down generators for the ideal with coefficients in $k$.

To see how this is encoded by the definition you've given, take $\Omega = \mathbb{C}$, $k = \mathbb{R}$, $n = 1$, and $$I = (x-i) < \mathbb{C}[x].$$ Of course any polynomial in $\mathbb{R}[x]$ divisible by $x-i$ is also divisible by $x+i$, and hence by $(x+i)(x-i) = x^2 + 1$. So $$I_\mathbb{R} = (x-i) \cap \mathbb{R}[x] = (x^2+1) < \mathbb{R}[x].$$ When we try to take $I_\mathbb{R}$ back to $\mathbb{C}[x]$ by tensoring with $\mathbb{C}$, of course, we get $$I_\mathbb{R} \otimes_\mathbb{R} \mathbb{C} \mapsto (x^2+1) \subsetneq (x-i)$$ That is, since we couldn't write down generators for $I$ with coefficients in $\mathbb{R}$, when we restricted it to $\mathbb{R}[x]$ we "lost part of the ideal," which remained lost when we brought it back to $\mathbb{C}[x]$ by tensoring with $\mathbb{C}$. This ideal is defined over $\mathbb{R}$.

On the other hand, if you'd taken $J = (x-7) < \mathbb{C}[x]$, then you'd get $$J_\mathbb{R} = (x-7) < \mathbb{R}[x],$$ and so $$J_\mathbb{R} \otimes_\mathbb{R} \mathbb{C} \mapsto (x-7) = (x-7);$$ this ideal is defined over $\mathbb{R}$.