Suppose that $k=\mathbf{R}$ and $A=\mathbf{R}[x, y] .$ Deduce that if $\mathfrak{m}$ is a maximal ideal of $A$, then $A / \mathfrak{m}$ is isomorphic to either $\mathbf{R}$ or $\mathbf{C}$. In the first case, note that $\mathfrak{m}=(x-a, y-b)$ for suitable $a, b \in \mathbf{R}$. Suppose now that $A / \mathfrak{m} \cong \mathbf{C} .$ The projection map $A \rightarrow A / \mathfrak{m}$ gives a surjective homomorphism:
$$
f: \mathbf{R}[x, y] \rightarrow \mathbf{C}
$$
where we are identifying $A / \mathfrak{m}$ with $\mathbf{C}$. Note that $\mathfrak{m}$ is the kernel of $f .$ Let $f(x)=\alpha…

Suppose that $\alpha$ and $\beta$ are both complex. Set $f_{x}=(x-\alpha)(x-\bar{\alpha})$ and define $f_{y}$ similarly. Note that $f_{x}, f_{y} \in \mathfrak{m}$ and thus $\left(f_{x}, f_{y}\right) \subset \mathfrak{m} .$ Let:
$$
B=\mathbf{R}[x, y] /\left(f_{x}, f_{y}\right)
$$
Show that the dimension of $B$ as a vector space over $\mathbf{R}$ is 4 . Write down an explicit basis of $B$. Deduce that $\mathfrak{m}$ strictly contains the ideal $\left(f_{x}, f_{y}\right)$. However, $\mathfrak{m}$ can be generated by adding one more polynomial $g$. Can you find such a polynomial (hint: use the …