Proof that the ideal $(xy, xz)$ in $\mathbb{A}^3$ is radical but not prime

Question

The proof that $(xy, xz)$ is not prime seems easy. In particular, $xy \in (xy, xz)$, but neither $x$ nor $y$ is in $(xy, xz)$.

On the other hand,

I don't know how to prove that $(xy, xz)$ is radical.

My proposed setup is as follows. Let $f^n = g \cdot xy + h \cdot xz$, then it suffices to show that $f \in (xy, xz)$. I'm not sure how to finish the argument though, even though the result seems clear.

As for the title: $\mathbb{A}^3$ is not a ring. You mean its coordinate ring $k[x,y,z]$. — Martin Brandenburg, Sep 14 '13 at 20:14

Martin Brandenburg · Answer 1 · 2013-09-14T20:20:11.590

We have that $(xy,xz) = (x) \cap (y,z)$ is an intersection of prime ideals.

The inclusion $\subseteq$ is trivial, and for $\supseteq$ assume that $f \cdot x \in (y,z)$. Then $\overline{f} \cdot \overline{x}= 0$ in $k[x,y,z]/(y,z) \cong k[x]$. But here $x$ is regular, hence $\overline{f} = 0$ and therefore $f \in (y,z)$, i.e. $f \cdot x \in (xy,xz)$.

More generally, if $R$ is any commutative ring, $x \in R$ and $I \subseteq R$ is an ideal such that $\overline{x} \in R/I$ is regular, then $(x) \cdot I = (x) \cap I$.

score 3 · Answer 2 · answered Sep 14 '13 at 19:14

3

Hint/Sketch: $f\in (xy,xz)$ is equivalent to every monomial term of $f$ being divisible by $xy$ or $xz$ (write $f$ as a sum of monomials). Assume $f \notin(xy,xz)$, so there is some nonzero monomial term of $f$ not divisible by $xy$ or $xz$. It is straightforward to show $f^n$ must also have a similar nonzero term and you are done.

answered Sep 14 '13 at 19:14

PVAL-inactive

8,551

Somehow this approach seems quite general. What makes it work for these special monomials? Remark that it doesn't work with arbitrary monomials. Therefore, I think that some essential details are missing. – Martin Brandenburg Sep 14 '13 at 20:15
Well, that detail is pretty obvious as soon you try to fill in the «it is straightforward»! – Mariano Suárez-Álvarez Sep 14 '13 at 20:30

score 2 · Answer 3 · answered Sep 14 '13 at 20:33

2

There is a general result.

Call an ideal $I\subseteq k[x_1,\dots,x_n]$ monomial if it is generated by monomials.

Then:

The radical of a monomial ideal $I$ is also a monomial ideal, and it is generated by the monomials obtained from those generating $I$ by making them square-free.

For example, the radical of $(x^2yz^3,y^2z^6w^3,x^2w)$ is $(xyz,yzw,xw)$.

If you complete all the details in PVAL's answer, then proving the above statement should not be too difficult.

answered Sep 14 '13 at 20:33

Mariano Suárez-Álvarez

135,076

(Or you could look at http://math.stackexchange.com/questions/125907/the-radical-of-a-monomial-ideal-is-also-monomial, but don't) – Mariano Suárez-Álvarez Sep 14 '13 at 20:34

score 1 · Answer 4 · 2013-09-15T19:15:04.660

For $f\in\sqrt{(xy,xz)}$ there exists $n\ge 1$ such that $f^n\in(xy,xz)$, and want to prove that $f \in (xy, xz)$.

We have $x\mid f^n$, so $x\mid f$ and there is $f_1\in K[x,y,z]$ such that $f=xf_1$. We get $x^{n-1}f_1^n\in(y,z)$.

Now let's see when a polynomial $p\in(y,z)$: by writing $p$ as a polynomial in $y,z$ with coefficients in $K[x]$ we can show easily that $p\in(y,z)$ iff $p(x,0,0)=0$.

Returning to $x^{n-1}f_1^n\in(y,z)$ we get $x^{n-1}f_1^n(x,0,0)=0$, so $f_1(x,0,0)=0$, that is $f_1\in (y,z)$ and thus we obtain $f\in(xy,xz)$.

Proof that the ideal $(xy, xz)$ in $\mathbb{A}^3$ is radical but not prime

4 Answers4