Question about principal generator of max ideals in $\mathbb{R}[X,Y]/(X^2+Y^2+1)$ [Liu Exercise 2.1.3] Exercise 2.1.3 from Qing Liu's (excellent) algebraic geometry text.

Let $k = \mathbb{R}$ be the field of real numbers. Let $A = k[X,Y]/(X^2+Y^2+1)$. We wish to describe Spec $A$. Let $x,y$ be the respective images of $X,Y$ in $A$.
(a) Let $\mathfrak{m}$ be a maximal ideal of $A$. Show that there exist $a,b,c,d \in k$ such that $x^2+ax+b, y^2+cy+d \in \mathfrak{m}$. Using the relation $x^2+y^2+1=0$, show that $\mathfrak{m}$ contains an element $f=\alpha x+\beta y+ \gamma$ with $(\alpha,\beta) \neq (0,0)$. Deduce from this that $\mathfrak{m} = fA$.

If such an $f$ exists, then we must have $fA = \mathfrak{m}$ as
$$\frac{A}{fA} \cong \frac{k[X,Y]}{(X^2+Y^2+1,\alpha X + \beta Y + \gamma)} \cong \frac{k[X]}{((\alpha^2+1)X^2 -2 \alpha\gamma X + \gamma^2 +1)},$$ and that last polynomial is irreducible.
To show that such an $f$ exists, one can show that there are only several cases of $\mathfrak{m}$ to consider, one of which is $\mathfrak{m} = (x^2+b, y^2+d)$, where $b,d > 0$. In this case, we can't simply add the two generating polynomials to obtain $f$ as we could in the unstated cases. As it turns out, $x^2+y^2+1 = 0 \Rightarrow b+d=1$ and $y^2+d = -x^2-(1-d)$, so $\mathfrak{m} = (x^2+b)$ is principal (no surprise as $A$ is a UFD and a Dedekind domain). Furthermore, it factors as $x^2+b = (\sqrt{d}x+\sqrt{b}y)(\sqrt{d}x-\sqrt{b}y)$, and, since $\mathfrak{m}$ is prime, one of those linear factors is contained in, and therefore generates, $\mathfrak{m}$.
My problem: This is one of those, "even if it's true, I have to see it with my own eyes" deals. Choosing $b=3/4, d=1/4$, we get that $\mathfrak{m} = (x^2+3/4)$ and I cannot for the life of me find a $g(x,y)$ such that $g(x,y)\cdot(x^2+3/4) = 1/2x \pm \sqrt{3}/2y$. I can see how it might be possible by higher order terms and mixed quadratic terms canceling, and the identity $1 = -x^2-y^2$ canceling out pure quadratic terms, but I'm definitely not confident. I think Groebner bases are useful here, but if so I don't know enough to use them, and Sage hasn't been too helpful. In fact, if, say, $(x^2+3/4) = (1/2x - \sqrt{3}/2y)$, then wouldn't they be associates and $1/2x + \sqrt{3}/2y$ necessarily a unit? I don't think either of the factors is a unit, though. At this point I'm just spinning my own wheels and could use some help finding an error in the proof (read: sketch) above, or help finding a $g(x,y)$ for the example  above.
 A: According to Zariski's Nullstellensatz (Qing Liu's Corollary 1.12, page 30) the extension $\mathbb R\subset A/\mathfrak m$ is finite, hence of degree $2$ (degree $1$ is excluded because $X^2+Y^2+1=0$ has no real solution).
Hence there indeed exist  $a,b,c,d \in \mathbb R$ such that $x^2+ax+b, y^2+cy+d\in \mathfrak{m}$.
Adding together these expressions we get $x^2+y^2+ax+b+cy+d\in \mathfrak m$, hence $-1+ax+b+cy+d\in \mathfrak m$ and (with obvious  notations) we get the required relation $f=\alpha x+\beta y+ \gamma \in \mathfrak m$.
Since  $fA\subset \mathfrak m$ to show equality it suffices to prove that $fA\subset A$ is maximal or that $A/fA=\mathbb R[X,Y]/(X^2+Y^2+1, \alpha X+\beta Y+ \gamma)$ is a field.
Assuming $\beta\neq0$ we replace $Y$ by $lX+m$ and obtain  $$A/fA=\mathbb R[X]/(X^2+(lX+m)^2+1)$$ which is obviously a field.
And so we have proved that $\mathfrak m=fA$ is indeed a principal ideal.
Optional Remark
It follows from Qing Liu's exercise that $A$ is a PID and hence factorial [=UFD].
The complexified algebra  $A_\mathbb C = \mathbb C[X,Y]/(X^2+Y^2+1)$ happens to also  be factorial, but beware that this is not obvious a priori.
For example the $\mathbb Q$-algebra
$B= \mathbb Q[X,Y]/(X^2+2Y^2+1)$ is a UFD but the extended $\mathbb Q(i)$-algebra $B_{\mathbb Q(i)}= \mathbb Q(i)[X,Y]/(X^2+2Y^2+1)$ is not a UFD.
On the other hand the the  algebra $B=\mathbb R[x,y]/(x^2+y^2-1)$ is not a UFD although the extended  $\mathbb C$-algebra
$B_\mathbb C=\mathbb C[x,y]/(x^2+y^2-1)$ is a UFD (https://mathoverflow.net/a/5596/450)
Moral: Factoriality is not necessarily preserved under field extension, nor can it be deduced from factoriality after a field extension.
A: Since I've accepted an answer, I thought I would post an answer that addressed my issue directly, in case someone else hits the same snag. I assumed, without checking, that the maximal ideals in $\mathbb{R}[X,Y]$ are correspondence with those of the form $(x^2+ax+b,y^2+cx+d)$, but this is not true! Maximal ideals are in correspondence with the kernels of surjective maps from $\mathbb{R}[X,Y] \to \mathbb{R}$ or $\mathbb{C}$. Since, in the latter case, $[\mathbb{C}:\mathbb{R}]=2$, a map from $\mathbb{R}[X,Y] \to \mathbb{C}$ sending $X \mapsto z, Y \mapsto w$ will be surjective iff $z$ and $w$ are linearly independent over $\mathbb{R}$. Therefore, along with others, ideals of the form $(X^2+b, Y^2+d)$ (like those considered above) are not maximal.
See also
https://math.stackexchange.com/a/2844259/39413, or https://math.stackexchange.com/a/2781509/39413, or more generally
https://mathoverflow.net/a/26503/69608
