Let $K$ be an algebraically closed field and $A:=K[x,y,z]/(xyz-z^2)$. Prove that $(\overline{x})$ and $(\overline{y})$ are primary ideals of $A$ and that $(\overline{z})$ is prime. Find the radical ideals of $(\overline{x})$ and $(\overline{y})$.
I don't even know how to start with this one. Trying use the simple definitions of prime and primary feels tricky to me because of the quotients involved. Is there some simple, by-the-book way to do this?
The definitions work fine. For $ x, y $, you can argue as follows:
$$ (K[x, y, z]/(xyz - z^2))/(\bar x) \cong K[x, y, z]/(xyz - z^2, x) \cong K[x, y, z]/(x, z^2) \cong K[y, z]/(z^2) $$
and every zero divisor in this ring is nilpotent, since $ (z^2) $ is a primary ideal of $ K[y, z] $. It follows that both $ (\bar x) $ and $ (\bar y) $ are primary ideals in $ A $. (We checked it only for $ x $, but the same argument shows it for $ y $ as well.) For $ z $, we have
$$ (K[x, y, z]/(xyz - z^2))/(\bar z) \cong K[x, y, z]/(xyz - z^2, z) \cong K[x, y, z]/(z) \cong K[x, y] $$
which is a domain, thus $ (\bar z) $ is a prime ideal of $ A $. To compute the radicals, we use the Nullstellensatz, and the fact that the radical of an ideal $ I + J $ in $ R/J $ is the ideal of $ R/J $ corresponding to $ \textrm{rad}(I + J) $ under correspondence. We have that
$$ \textrm{rad}((x, xyz - z^2)) = I(V(x, xyz - z^2)) = (x, z) $$ $$ \textrm{rad}((y, xyz - z^2)) = I(V(y, xyz - z^2)) = (y, z) $$
so that the radicals of $ (\bar x) $ and $ (\bar y) $ are $ (\bar x, \bar z) $ and $ (\bar y, \bar z) $ respectively.