I've been stuck on one step of a proof, and am hoping someone will have a helpful hint/explanation :)
Here is the setting:we have two Lie groups $K \hookrightarrow T$ acting on a manifold $M$. We can define the equivariant cohomology of $M$ (with respect to the $T$-action, say), as $H_T^\bullet(M) := H^\bullet(\dfrac{M \times ET}{T})$, and similarly for $H_K^\bullet$, where we choose $EK = ET$. ($H^\bullet$ represents the cohomology ring for whichever cohomology theory we're working with).
So in our case, $H_T^\bullet(\text{pt}) = \mathbb{C}[x_1,...,x_n]$. We associate the Lie algebras $\mathfrak {t}, \mathfrak{k}$ to $T$ and $K$ respectively (after complexification, $\mathfrak{k} \subseteq \mathfrak{t} \cong \mathbb{C}^n$, I'll just use the same notation for the complexifications for clarity). We choose an $H_K^\bullet(\text{pt})$ module $P$, which is also a module over $H_T^\bullet(\text{pt})$ by restriction.
So we define the Zarisky topology on $\mathfrak{t}$ where the closed set induced by a polynomial $f$ is $V_f := \{y \in \mathbb{C}^m \mid f(y) = 0\}$, and the support of the ($H_T^\bullet(\text{pt}))$ module $P$ to be Supp$P := \bigcap\limits_f V_f$ running over all the $f$ such that $fP = 0$.
Then, I want to show that $\boxed{\text{Supp}P \subset \mathfrak{k}}$. The proof I'm reading has the following line:
"It is obvious that for any $f \in \mathbb{C}[x_1,...,x_n]$ with $f|_\mathfrak{k} = 0$ we have $f \cdot P = 0$ and Supp$P \subseteq V_f$. It follows that Supp$P \subseteq \mathfrak{k}$."
I don't understand that line
Say we do start with a function that is identically zero on $\mathfrak{k}$ (viewed as a subset of $\mathbb{C}^n$). Then, I'm pretty sure the fact $f \cdot P = 0$ just follows from how we defined our module structure, by restricting from $T$ to $K$. Then, I'm ok with the fact that Supp$P \subseteq V_f$, however, I don't see why that should imply Supp$P \subseteq \mathfrak{k}$.
The fact that $f|_\mathfrak{k} = 0$ does imply that $\mathfrak{k} \subseteq V_f$ (since we define $V_f$ as the zero set of $f$). So all this tells us, is that for any function that is zero on $\mathfrak{k}$, $\mathfrak{k} \subseteq V_f$.
But I don't see why that should imply Supp$P \subseteq \mathfrak{k}$.
My thinking: it would be enough to find a polynomial that is identically zero on $\mathfrak{k}$ and only on $\mathfrak{k}$, but I don't see why its existence would be guaranteed.
Note: Here is the proof I'm reading for anyone who wants more context. I twitched the notations a bit, writing $K = K_0$. Note that the last character is a typo, and should say $\mathfrak{k}_\mathbb{C}$ instead of $\mathfrak{t}_\mathbb{C}$.