I'm working through Algebraic Geometry: A Problem Solving Approach by Garrity et al, and I have found myself stuck on Exercise 4.13.1, which is the section Points and Local Rings.
Let $V = V(x^2 + y^2 - 1) \subset \mathbb{A}^2(k).$ Let $p = (1, 0) \in V.$ Define $$\mathfrak{m}_p = \{f \in \mathcal{O}_V : f(p) = 0\}.$$
Show that $\mathfrak{m}_p$ is a maximal ideal in $\mathcal{O}_V.$
We're told that $\mathcal{O}_V$ is a subring of the function field $\mathcal{K}_V = \{\frac{f}{g} : f, g \in \mathcal{O}(V), g \neq 0\},$ where $\mathcal{O}(V) = k[x, y]/I(V).$
It was easy enough to show that $\mathfrak{m}_p$ is an ideal, but I'm getting caught up on how to show that it is maximal.
I know that maximal ideals in $k[x, y]$ are all of the form $I = \langle x - a_1, y - a_2 \rangle,$ so my idea was to show that $\mathfrak{m}_p = \langle x - 1, y \rangle,$ and then I know that maximal ideals in $k[x, y]/I(V)$ correspond to maximal ideal in $k[x, y]$ containing $I(V).$ This feels close to something correct, but I think I just don't understand the different spaces well enough to put it all together. Also, $\mathfrak{m}_p$ is an ideal in $\mathcal{O}_V,$ while $\langle x - 1, y\rangle$ is an ideal in $k[x, y],$ so setting them equal to each other doesn't make any sense.
Map $\mathcal{O}_V \to k$, $f \mapsto f(p)$. Show this is a ring homomorphism with kernel $\mathfrak{m}$.