Consider the polynomial ring $\Bbb Z[x,y]$. Let $m,n$ be positive integers, and consider the ideal $(x^m, y^n)$. I want to show that $(x^m,y^n)$ is a $(x,y)$-primary ideal. (This is an exercise in Hungerford's Algebra.)
There is a following theorem which seems useful:
Theorem. Let $Q,P$ be ideals in a commutative ring $R$ with identity. Then $Q$ is $P$-primary if and only if the following statements hold: (1) $Q \subset P \subset $ Rad$(Q)$; (2) if $ab \in Q, a \notin Q$, then $b \in P$.
With $Q=(x^m,y^n)$ and $P=(x,y)$, (1) is easily shown, using the fact that Rad$(Q)=\{r \in R:r^k \in Q$ for some $k \geq 1 \}$. But I got stuck showing (2). How do I have to proceed?
It thus suffices to show that if $a\notin Q$ and $b\notin P$, then $ab\notin Q.$ To do this, we will use the following characterizations
$a\in Q$ iff all the terms of $a$ are of the form $x^ry^s$ where $r,s>0$, $n\mid r$ or $m\mid s$
$b\in P$ iff $b$ does not contain a constant term.
Then, if $a\notin Q$, it contains a minimal term of the form $cx^ry^s$ where either $n$ does not divide $r$ and $m$ does not divide $s$. Since $b$ contains a constant term $d\in\mathbb{Z}$, and $x^ry^s$ is chosen to be minimal in it's non-dividing power, the only term in the product $ab$ with power $x^ry^s$ must be $cdx^ry^s$, and thus $ab\notin Q$