Bosch exercise 1.3.2 on prime structure coordinate ring I have heard a commutative Algebra course and am preparing for Alg. geometry lectures next term with the textbook of Bosch “Algebraic geometry and comm. Algebra“.
I just did Exercise 1.3.2 of this book, stated below, but I find my solution a little clumsy and am wondering if it is perfectly correct argumented as well.. So feedback on this is most welcome! Thx!!
Ex: Consider the polynomial ring $K[X,Y]$ over a field $K$ and set $R=K[X,Y]/(X-XY^2,Y^3)$. Writing $\bar{X},\bar{Y}$ for the residue classes of $X,Y$ in $R$, show that Nil$(R)=(\bar{X},\bar{Y})$ is the only prime ideal in $R$ and that the reduced ring $R/Nil(R)\cong K$.
My solution try: Let $I:=(X-XY^2,Y^3)$, i.e. $R=K[X,Y]/I$. Then as a K-algebra $R=K[\bar{X},\bar{Y}]$ and $R/(\bar{X},\bar{Y})\cong K$, which proves that $(\bar{X},\bar{Y})$ is maximal, thus prime in R and the statement on the reduced ring when the nilradical was shown.
(Alternatively maximality of $(\bar{X},\bar{Y})$ is seen, using $(X,Y)$ maximal in $K[X,Y]$ and $I\subset (X,Y)$ and that the canonical projection $K[X,Y]\to R$ is an inclusion preserving correspondence of ideals in R to ideals in $K[X,Y]$ containing $I$.)
Now $Y^3\in I \implies (\bar{Y})^3=0$ in $R$, i.e. $\bar{Y}\in Nil(R)$. Thus $\overline{XY^2}\in(\bar{Y})\subset Nil(R)$ and thus $\bar{X}=\overline{X-XY^2}+\overline{XY^2} \in Nil(R)$ (since $X-XY^2 \in I \implies \overline{X-XY^2}=0$).
Thus $(\bar{X},\bar{Y})\subset Nil(R)$ and maximal in $R$ $\implies (\bar{X},\bar{Y})=Nil(R)$, since $Nil(R)\neq R$ due to R containing $K$. Remains to show it is the only prime ideal:
Let $\mathfrak{p}\in Spec(R)$. Then $\bar{Y}^3=0\in \mathfrak{p} \implies \bar{Y}\in\mathfrak{p}$ and $\bar{X}\in Nil(R)\implies \bar{X}^n=0$ for an $n \implies \bar{X}\in \mathfrak{p}$. Thus $(\bar{X},\bar{Y})\subset\mathfrak{p}\implies (\bar{X},\bar{Y})=\mathfrak{p}$ as $\mathfrak{p}$ is proper and $(\bar{X},\bar{Y})$ maximal.
What do you think? Are there more elegant alternatives?
 A: Your solution is essentially exactly how I would have solved this! Let me present a [very slightly] streamlined version of the argument as I would make it.
First, I would note that in any ring $R,$ maximality of the nilradical $\sqrt{(0)}$ implies that $\sqrt{(0)}$ is the unique prime ideal in that ring, as $\sqrt{(0)}\subseteq\mathfrak{p}$ for any prime $\mathfrak{p}\subseteq R.$ (This is essentially what you observed in the last paragraph of your argument, but it works in any ring. In fact, we can say more: $$\sqrt{(0)} = \bigcap_{\mathfrak{p}\textrm{ prime}}\mathfrak{p}.)$$
Second, I would observe that $(\overline{X},\overline{Y})\subseteq R$ is maximal by computing the quotient $R/(\overline{X},\overline{Y})\cong K[X,Y]/(X,Y)\cong K.$
Finally, we need only prove that $(\overline{X},\overline{Y})\subseteq\sqrt{(0)}.$ We see immediately that $\overline{Y}^3 = 0$ by definition of $R,$ and $\overline{X} = \overline{XY}^2$ implies that $\overline{X}^2 = 0$ as well. Thus $(\overline{X},\overline{Y})\subseteq\sqrt{(0)},$ and we are finished.
