Points on the "conic" $\mathbb{R}[x, y] / (x^2 + y^2 + 1) $ On exercise 2.1.3 of Qing Liu's algebraic geometry book, which I'm self studying together together with Vakil's notes, he asks us to prove the variety defined by $x^2 + y^2 + 1$ over the real field looks like the projective plane minus a point.
The later more abstract parts of the exercise I was able to do. However the first step, where he asks us to show that the maximal ideals in the quotient ring must contain single variable quadratic polynomials $x^2+ax+b$ and $x^2+cx+d$, I haven't been able to.
I suspect we can use the fact that quotient by maximal ideals of fg algebras are field extensions of the base field, but this didn't lead anywhere yet... Any hints? 
More generally, I feel I'm missing some feeling for the computational part of the subject. The abstract exercises about sheaves and whatnot go fine, but I'm lost with the tricks for multivariate polynomial algebra. Is there a general philosophy here? For instance, if instead of $\mathbb{R}[x,y]$ we had $\mathbb{Q}[x,y]$ how would one begin? 
 A: Fix a maximal ideal $m\subset R = \Bbb R[x,y]/(x^2+y^2+1)$. Clearly $R/m\cong \Bbb C$ (if this is not clear, you should think about it more or post a comment asking for more explanation). Consider the image of $x$ under this isomorphism. It's an element of $\Bbb C$, a finite algebraic extension of $\Bbb R$, so it has to have a minimal polynomial. Since $\Bbb C$ is a degree two extension of $\Bbb R$, the degree of this polynomial is bounded above by 2. So the image of $x$ satisfies a quadratic polynomial, hence the maximal ideal must contain a quadratic polynomial in $x$.
A: Let $I=(x^2+y^2+1)$ and suppose $I\subset M$, where $M$ is a maximal ideal of $\mathbb{R}[x,y]$.

Claim:$\;f\in M$ for some $f\in\mathbb{R}[x]$ with $\deg(f)=2$.

Here's a pedestrian proof . . .

Choose $g\in M{\setminus}I$ such that the degree of $g$ in the variable $y$ is least, say $k$, and among those with degree $k$ in $y$, with least degree in $x$, say $n$.

We can't have $k\ge 2$ else, by polynomial long division, we could divide $g$ by the polynomial $y^2+x^2+1$, regarded as a monic quadratic polynomial in $y$, and the remainder would be in $M{\setminus}I$ (since $g\not\in I$), and would have degree in $y$ at most $1$.

Next suppose $k=1$.

Thus, write $g=sy + t$, where $s,t\in\mathbb{R}[x]$, and $s\ne 0$.

Then 
\begin{align*}
&
\!\!\!\!
\begin{cases}
s^2(y^2+x^2+1)\in M\\[4pt]
s^2y^2-t^2=(sy+t)(sy-t)\in M\\
\end{cases}
\\[4pt]
\implies\;&s^2x^2+s^2+t^2\in M\\[4pt]
\end{align*}
Since $s\ne 0$, $s$ has nonzero range, hence $s^2x^2+s^2+t^2$ is a nonzero element of $\mathbb{R}[x]$.

But it can't be in $I$ since any nonzero multiple of $x^2+y^2+1$ has degree at least $2$ in $y$.

Hence $s^2x^2+s^2+t^2\in M{\setminus}I$, contrary to the assumption that $k=1$.

It follows that $k=0$, hence $g\in\mathbb{R}[x]$.

Note that $g$ is nonzero since $g\in M{\setminus}I$, and $g$ can't be a nonzero constant since $g\in M$.

Hence $n\ge 1$.

We can't have $n > 2$, else $g$ would factor nontrivially in $\mathbb{R}[x]$, and since $M$ is a prime ideal, at least one of the factors of $g$ would be in $M$, contrary to the minimality of $n$.

Hence $n=1$ or $n=2$.

If $n=1$, let $f=g^2$, and if $n=2$, let $f=g$. In either case, we're done.

This completes the proof.
