Prove that the two distinct bases of ideal $I$ will have identical affine variety I have a proposition here that states

If $f_1,\ldots,f_2$ and $g_1,\ldots,g_t$ are bases of the same ideal $I$ in some field $k$, so that $\langle f_1,\ldots,f_s\rangle=\langle g_1,\ldots,g_t\rangle$ then we have that $V(f_1,\ldots,f_s)=V(g_1,\ldots,g_t)$

Apparently, my lecturer left the proof as "very straightforward" and didn't touch on it. Unfortunately, I am unable to prove it. Here' s my attempt and question,

Let $(a_1,\ldots,a_n) \in V(f_1,\ldots,f_s)$. Then, we clearly have $V(f_1,\ldots,f_s) \subseteq V(I)=V(\langle g_1,\ldots,g_t\rangle)$ since if each $f_1,\ldots,f_s$ disappears for any such $(a_1,\ldots,a_n)$ then the span will consequently vanish.

My question is, I need to show $V(f_1,\ldots,f_s) \subseteq V(g_1,\ldots,,g_t)$ and vice versa but not $V(f_1,\ldots,f_s) \subseteq V(\langle g_1,\ldots,g_t\rangle)$. And I don't think $V(\langle g_1,\ldots,g_t\rangle)=V(g_1,\ldots,g_t)$ so to me, this isn't straightforward at all.
So how do I go about this? Please tell me what I need to do
 A: It seems that you are confused about the following statement:
Lemma. Let $k$ be an algebraically closed field. Let $f_1, \ldots, f_s \in k[X_1,\ldots,X_n]$, and set $I = (f_1, \ldots, f_s)$. Then $V(f_1, \ldots, f_s) = V(I)$.
Proof. Let $x \in \mathbb A^n$. If $x \in V(I)$, then $f(x) = 0$ for all $f \in I$. In particular, setting $f = f_i$ gives $x \in V(f_1,\ldots, f_s)$.
Conversely, if $x \in V(f_1,\ldots,f_s)$, then $f_i (x) = 0$ for all $i \in \{1,\ldots,s\}$. If $f \in I$, then there exist $a_i \in k[X_1,\ldots,X_n]$ such that $f = \sum a_i f_i$. Thus,
$$f(x) = \sum a_i f_i(x) = \sum 0 = 0.$$
$\square$
A: The polynomials that are zero on a given set, even the zero set of a given set of polynomials, form an ideal, so they must contain the ideal generated by the given set of polynomials.
On the other hand, if a collection of polynomials has a common zero $(x_1,\ldots,x_n)$ then any linear combination of those polynomials (with arbitrary polynomials as coefficients) also has that zero.
