Q:

maximal ideal of a quasi-affine variety

Problem
Here $k$ is a fixed algebraically closed field. Let $X$ be an affine variety over $k$, $U\subseteq X$ is its open subset (quasi-affine variety), and $p\in U$.
Then, show that
$$
I_{U}(p) := \{f\in\mathcal{O}_{U}(U) = \mathcal{O}_{X}(U)\ |\ f(p) = 0\}
$$
, which is an ideal of $\mathcal{O}_{U}(U) = \mathcal{O}_{X}(U)$, is maximal.
What I have already proven
We have injective restriction maps
$$
k[X] \to \mathcal{O}_{X}(U) = \bigcap_{q\in U}k[X]_{I_{X}(q)} \to k[X]_{I_{X}(p)}.
$$
Here $k[X]_{I_{X}(p)}$ is a local ring with maximal ideal $\mathfrak{m} := I_{X}(p)k[X]_{I_{X}(p)}$. We have that $\mathcal{m}\cap \mathcal{O}_{X}(U) = I_{X}(U)$. In particular, $I_{X}(U)$ is prime, for contraction of prime ideal is prime. How to prove $I_{X}(U)$ is maximal?
Notation
(1) $k[X]$ is the coordinate ring of $X$.
(2) $k(X)$ is the rational function field of $X$, which is the fraction field of $k[X]$.
(3) $\mathcal{O}_{X}(U) = \bigcap_{q\in U}k[X]_{I_{X}(q)}$ is the set of regular functions on $U$.

A:

We have an onto homomorphism $\mu:\mathcal{O}_X(U)\longrightarrow k$  defined by $\mu(f)=f(p)$ for all $f\in \mathcal{O}_X(U)$. Then $\ker(\mu)=I_U(p)$ and hence it is maximal.

