Why is $\langle x^2, xy,y^3\rangle$ primary in $k[x,y,z]$? 
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.
 A: More generally, if $R$ is an integral domain then $I=\left<x^2,xy,y^3\right>$ is a primary ideal of $R[x,y]$.  (Note that $k[x,y,z] \cong k[z][x,y]$, so in this case, $R=k[z]$.)
Proof:
Any element of $q\in R[x,y]$ can be written uniquely as:
$$q=a + bx + cy + dy^2 + m$$
where $m\in I$ and $a,b,c,d\in R$.
Claim (without proof): $\exists n>0; q^n\in I$ if and only if $a=0$.
Now assume $q_i = a_i + b_ix + c_iy + d_iy^2 + m_i$, as above, where $i=1,2$.
Then $q_1q_2\in I$ means that $a_1a_2=0$ and $a_1b_2+b_1a_2=0$ and $a_1c_2+c_1a_2=0$ and $a_1d_2+c_1c_2+d_1a_2 =0$.
Now, if $a_2=0$, then $q_2^n=0$ for some $n>0$ by our claim above and we are done. So assume $a_2\neq 0$.
Then $a_1=0$. Substituting, we get: $b_1a_2=0$ so, again since $a_2\neq 0$, $b_1=0$.
Substituting again, we get that $c_1a_2=0$ so $c_1=0$.
Finally, substituting again, we get that $d_1a_2=0$ so $d_1=0$.
So $a_1=b_1=c_1=d_1=0$ and hence $q_1 = m_1\in I$, and we are done.
A: 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.
