I have recently come across the following exercise: given an algebraically closed (and of course commutative) field $K$ of characteristic not equal to $2$, describe all the graded prime ideals of the quotient $K[X, Y, Z]/\left(X^2+Y^2+Z^2\right)$.
The problem was encountered at the end of chapter 7 of the "Algèbre commutative" by Bourbaki, more precisely as problem 6. a) of $\S{3}$. I assume (with complete certainty) that the grading in question is the one induced by the natural $\mathbb{N}$-grading on the three indeterminate polynomial ring $K[X, Y, Z]$, so that the task of the problem amounts to finding all the graded prime ideals of $K[X, Y, Z]$ which include the principal ideal $\left(X^2+Y^2+Z^2\right)$.
The hypothesis that $\mathrm{char}K \neq 2$ plays the role of ensuring the homogenous polynomial $X^2+Y^2+Z^2$ is irreducible (equivalently, nonzero and prime, since we are working in a UFD). Other than drawing this conclusion I am afraid I haven't managed to make any progress in solving the problem and I have the distinct feeling I do not know what key observations I should be making to enable a classification of the required prime ideals. The generic prime ideal satisfying the conditions of the problem will be generated by irreducible homogenous polynomials (one of which the homogenous quadratic with respect to which the quotient is taken), however this is far from permitting a clear classification.
I would be very grateful even for an indication of what results/constructions/algebro-geometric considerations one should have in mind in order to solve such a problem.
Let $\mathfrak{p}$ be such a prime ideal – assume moreover that $\mathfrak{p}$ isn’t all the polynomials that vanish at zero. Then $\mathfrak{p}$ contains $X^2+Y^2+Z^2$ and a nonzero homogeneous polynomial $P(X,Y,Z)=A(X,Y)+B(X,Y)Z$.
If $B(X,Y) \neq 0$, then $\mathfrak{p}$ contains $(A(X,Y)-B(X,Y)Z)P(X,Y,Z) \in (A(X,Y)^2+(X^2+Y^2)B(X,Y)^2)+\mathfrak{p}$, so $\mathfrak{p}$ contains a nonzero polynomial $Q(X,Y)$, thus it contains a nonzero irreducible polynomial which depends on $X,Y$ only. As $K$ is algebraically closed, such a polynomial is proportionnal to (up to switching variables) $Y-\alpha X$. Write $1+\alpha^2=-\beta^2$.
Thus $\mathfrak{p}$ contains $(1+\alpha^2)X^2+Z^2=(Z-\beta X)(Z+\beta X)$, so it contains (up to renaming $\beta$) $Z-\beta X$.
Thus $\mathfrak{p}$ contains $Z-\beta X, Y-\alpha X$, and $P_{\alpha,\beta}=(Z-\beta X,Y-\alpha X)$ is prime and is contained in $\mathfrak{p}$.
Assume $\mathfrak{p}$ isn’t of the form above, so it contains some $P_{\alpha,\beta}$ as a proper subset. Note that $K[X,Y,Z]/P_{\alpha,\beta}$ is isomorphic to $K[X]$, and every integral proper quotient of $K[X]$ is a finite integral $K$-algebra with $K$ algebraically closed, so is isomorphic to $K$. Thus $K[X,Y,Z]/\mathfrak{p}=K$ so $\mathfrak{p}$ is the ideal of polynomials vanishing at zero, and we are done.
So $\mathfrak{p}$ is either $(X,Y,Z)$ or some $(Z-\alpha X,Z-\beta X)$, with $\alpha^2+\beta^2=-1$.