Showing an ideal is irreducible I am currently trying to show that the ideal $\langle x^3, y^5, z^2 \rangle \subset \mathbb{C}[x,y,z]$ is irreducible (i.e.:  it cannot be written as the intersection of two larger ideals $J$ and $K$).
Simply put, I just can't do it!  I'm guessing the strategy would be to assume that $I = J \cap K$ and then show that either $I=J$ or $I=k$, or alternatively, show that if $I$ is strictly contained in $J$ and $K$, then $J \cap K \neq I$.  But as I say, I can't formalize any sort of argument.
Any help would be greatly appreciated. :)
 A: Let $J$ be a strictly larger ideal.
Let $f\in J\setminus I$ be a polynomial with minimal number of monomials.
Then each of these monomials is of the form $a_{i,j,k}x^iy^jz^k$ with $a_{i,j,k}\ne 0$, $0\le i<3$, $0\le j<5$, $0\le k<2$.
If $f$ consists of at least two monomials $a_{i,j,k}x^iy^jz^k$ and $a_{i',j',k'}x^{i'}y^{j'}z^{k'}$ then the exponents cannot all be the same. If $i>i'$, say, then $f(x,y,z)\cdot x^{3-i}-a_{i,j,k}x^3y^jz^k\in J\setminus I$ has less monomials, contradicting minimiality. The same works for the other exponents. We conclude that $f$ has only one monomial and wlog. the coefficient is $1$, so $f(x,y,z)=x^iy^jz^k$.
Let $K$ be another ideal with $I\subsetneq K$ and such that $J\cap K=I$.
As above we find $g\in K$ with $g(x,y,z)=x^ry^sz^t$, $0\le r<3$, $0\le s<5$, $0\le t<2$.
Then the polnomial  $x^{\max(i,r)}y^{\max(j,s)}z^{\max(k,t)}$ is in $J$ and in $K$ but not in $I$, contradiction.

The same argument in a different formulation:
Assume $J$ is strictly larger than $I$. For $f\in J\setminus I$, the sequence  $f, xf, x^2f, x^3f$ ends in $I$, hence if we replace $f$ with the last in this sequence that is $\notin I$, we have $f\in J\setminus I$ with $xf\in I$. Repeating the process with $y$ and $z$, we find $f\in J\setminus I$ with $xf,yf,zf\in I$. It follows that $f(x,y,z)=ax^2y^4z$ and we conclude $x^2y^4z\in J$ for all ideals $J\supsetneq I$. Consequently, the intersection of two structly larger ideals also contains $x^2y^4z$.
