How to show that the ideal $(X^{3},XY,Y^{n})$ of $K[X,Y]$ is primary? I'm working on a problem in Sharp's Steps in Commutative Algebra, to be precise exercise 4.28 which states the following:
Let $K$ be a field and $R = K[X,Y]$ be the polynomial ring in the indeterminates $X$ and $Y$ over $K$.
Let $I = (X^{3},XY)$.
(i) Show that the ideal $J_{n}:= (X^{3},XY,Y^{n})$ is primary for each $n \in \mathbb{N}$.
(ii) Show that $I = (X)\cap (X^{3},Y)$ is a minimal primary decomposition of $I$.
(iii) Construct infinitely many different primary decompositions of $I$.
I don't have trouble sorting (ii) out as I found this quite easy. 
I do however struggle to show that $(X^{3},XY,Y^{n})$ is primary. I have tried doing it directly and also to look at the quotient ring $R/J_{n}$ to show that every zero divisor is nilpotent. I also have not been able to explicitly find the radical of $J_{n}$ in general. Any help will be appreciated, thanks in advance.
 A: The way I thought about this question goes like this:
i) Look at the quotient $R/J_n = K[X,Y,Z]/(X^3, XY,Y^n)$. It looks like $K$ adjoin elements $a,b$ with $a^3 = ab = b^n = 0$. Now $1, a,a^2,b,b^2,...,b^{n-1}$ form a $K$-basis for this algebra. Given an element in $R/J_n$ , either it has its constant coefficient (that of the basis element $1$) nonzero, in which case it's not a zerodivisor; or it has only terms in the powers of $a$ and $b$, which means it will be nilpotent. So all zerodivisors in the quotient are indeed nilpotent. 
ii) You were okay with this bit: intersection of two primary ideals, neither contained in the other, different radicals. Cool cool.
iii) Just use part i). Given any $n$, $I = (X^3,XY) = (X)\cap(X^3,Y) = (X)\cap(X^3,Y)\cap (X^3,XY,Y)\cap...\cap(X^3,XY,Y^n)       $ just intersecting with as many $J_n$ as you fancy, because they all contain $(X^3,Y)$ so can't affect what the intersection actually is. This gives infinitely-many primary decompositions, but of course they're not minimal. Hope I understood that part of the question right!
As for finding the radical of $J_n$, I'm looking for the smallest prime ideal $P$ containing $J_n$. So, the quotient $R/P$ must be an integral domain, in which all elements of $J_n$ from $R$ are zero. So, $X^3 = Y^n = 0$ implies $\bar{X}$ and $\bar{Y}$ are both zerodivisors, and so need to be zero. So, $P$ contains $X$ and $Y$. But of course, $R/(X,Y)$ is an integral domain - isomorphic to $K$ - so $P = (X,Y)$ must be the radical.
Hope this was helpful, in spite of the messiness.
A: Perhaps you are overthinking things, because finding this radical is not hard. The ideal contains $x^3$ and $y^n$, and therefore its radical, by definition, must contain $x$ and $y$. Can you see what the radical must be now?
Applying this lemma you will also see why the ideal is primary, and you are done. 
A: you say "I also have not been able to explicitly find the radical of $J_n$". in fact if you solve this problem, (i) will be a result of Proposition 4.9 of the book. I prove this for $n\ge3$ (for $n\lt3 $ it is easy and similar). note that $(X,Y)^n\subset J_n\subset (X,Y)$. Taking radicals we have $(X,Y)\subset \sqrt {J_n}\subset (X,Y)$. so $\sqrt {J_n}= (X,Y)$.  
(iii) $I=(X)\cap (X^{3},XY,Y^{n})$
