Exercise 1.3 in Miles Reid's Commutative Algebra

Question

1.3 Let $A = k[X,Y]/(XY)$. Show that any element of $A$ has a unique representation in the form $$a+f(X)X+g(Y)Y\quad\text{with}\quad a\in k, f\in k[X],g\in k[Y].$$ How do you multiply two such elements?

Prove that $A$ has exactly two minimal prime ideals. If possible, find ideals $I,J,K$ to contradict each of the following statements:
(a) $IJ = I\cap J$;
(b) $(I+J)(I\cap J) = IJ$;
(c) $I\cap(J+K) = (I\cap J)+(I\cap K)$.
Image.

I have the following questions:

1. I suppose that $(x)$ and $(y)$ are two minimal prime ideals in $A$. How can I show that they are minimal and that there are no other minimal prime ideals. (How can I show that a prime ideal is minimal in general?)
2. I know that for any ring $R$ we have $(I+J)(I \cap J) \subseteq IJ$. For which $R$ it becomes an equality?
3. Is $(c)$ always true? I cannot find any counterexample

Answer 1

This is a solution to the first part 1. Let $P$ be a prime ideal then since $xy=0$ we have $xy\in P$ so $x\in P$ or $y\in P$. This shows minimality, it remains to show for example that $(x)$ is prime.

Let $f_1(x)x+g_1(y)$ and $f_2(x)x+g_2(y)$ be two elements where $g_i(y)$ is a polynomial in $y$, maybe with constant term. If $$(f_1(x)x+g_1(y))(f_2(x)x+g_2(y))\in (x)$$ then $$g_1(y)g_2(y)\in (x)$$ and this means that $g_1(y)g_2(y)=0$ and so $g_1(y)=0$ or $g_2(y)=0$.

In the former case, $$f_1(x)x+g_1(y)=f_1(x)x\in(x)$$ Thus $(x)$ is prime.

Answer 2

for the rest:
(2): Prüfer domain are what you are looking for; see here.
(3): You can find example for $I \cap (J+K) \neq I \cap J + I \cap K$ at the site; see here.