TITLE: Why is $\langle x^2, xy,y^3\rangle$ primary in $k[x,y,z]$?
QUESTION [7 upvotes]: Can someone tell me a quick reason as to why $\langle x^2, xy,y^3\rangle$ is primary in $k[x,y,z]$? 

I'm trying to read a solution and I don't get this.

REPLY [10 votes]: Use these two properties of primary ideals:
$1$. If the radical of an ideal is a maximal ideal, then the ideal is primary. (In your case consider the ideal $\langle x^2, xy, y^3\rangle$ as an ideal of $k[x,y]$ and note that its radical is $\langle x, y\rangle$ which is maximal.)
$2$. If $\mathfrak q\subset R$ is primary (with radical $\mathfrak p$), then $\mathfrak q[t]\subset R[t]$ is primary (with radical $\mathfrak p[t]$). Here $\mathfrak q[t]$ stands for the extension of the ideal $\mathfrak q$ to the polynomial ring $R[t]$, that is, the set of polynomials having all coefficients in $\mathfrak q$. 
The ideal  $\langle x^2, xy, y^3\rangle$ in $k[x,y,z]$ is the extension of the primary ideal $\langle x^2, xy, y^3\rangle$ of $k[x,y]$ to $k[x,y][z]=k[x,y,z]$.
Edit. The second property relies on the following result that deserves to be proved here: 

Let $R$ be a commutative ring and $f\in R[X]$. Then $f$ is a zerodivisor if and only if there exists $a\in R$, $a\neq 0$ such that $af=0$.

If $f$ is a zerodivizor, then there exists $g\in R[X]$, $g\neq 0$ with $fg=0$. Choose $g$ of minimal degree with this property. Set $g(X)=b_0+b_1X+\cdots+b_mX^m$, $b_m\neq 0$. From $fg=0$ it follows that $a_nb_m=0$. Then the degree of $a_ng$ is less than $m$, $(a_ng)f=0$, and thus $a_ng=0$. In particular, $a_nb_{m-1}=0$ and looking at the coefficient of $X^{m+n-1}$ in $fg$ we get that $a_{n-1}b_m=0$. Then the degree of $a_{n-1}g$ is less than $m$, $(a_{n-1}g)f=0$, and therefore $a_{n-1}g=0$. Similar arguments show that $a_ig=0$
for all $0\leq i\leq n$. This implies $a_ib_m=0$ for all $i$, so $b_mf=0$, qed.