# Prove $(0) = (x)\cap (xz^{n-1} + \lambda y^n)$ in $R=\frac{k[x,y,z]}{(x^2,xy)}$

Studying for my algebra final and doing some practice problems, and I can't seem to understand this one...

Full problem:

Let $$k$$ be a field, and $$R=\frac{k[x,y,z]}{(x^2,xy)}$$. For $$n\in\mathbb{N}, \lambda\in k^*,$$ show $$(0) = (x)\cap (xz^{n-1} + \lambda y^n)$$.

I was going to do a proof by contradiction, but I'm having trouble understanding the form of elements of $$R$$. I feel like if I understood that better, it would be fairly easy to show that no nonzero element of R can be in the right hand side of the equation provided.

Any help and hints are greatly appreciated, thanks in advance!

edit: So my next thought is to find a minimal primary decomposition for $$(x^2, xy)$$, as it's my understanding that that decomposition is the same as the primary decomposition for $$(0)$$ in $$R$$. It's a monomial ideal, so you can reduce it to $$(x)\cap(x^2,y)$$. I'm not really sure where to go from here though.

Let $$\pi: k[x,y,z] \rightarrow k[x,y,z]/(x^2, xy)$$ be the canonical projection. Note that an ideal $$J \subseteq k[x,y,z]/(x^2, xy)$$ is $$0$$ iff $$\pi^{-1}(J) \subseteq (x^2, xy)$$. Thus we don't really need to work in the factor ring at all. Rather, it suffices to show that $$f \in (x) \cap (xz^{n-1} + \lambda y^n) \subseteq k[x,y,z]$$ implies $$f \in (x^2, xy)$$.

To do this, write $$f = g(xz^{n-1} + \lambda y^n) = hx$$ for some $$g,h \in k[x,y,z]$$. Since $$x$$ is a prime element in $$k[x,y,z]$$, necessarily $$x$$ divides $$g$$ or $$xz^{n-1} + \lambda y^n$$. The latter is clearly impossible, so $$x$$ divides $$g$$, and it follows quickly that $$f \in (x^2, xy)$$.

Note that we didn't need $$k$$ to be a field, it could have been any commutative ring.

• Thanks so much, this was really helpful! – tenzs Dec 9 '18 at 5:43

For a geometric perspective, we think of $$R$$ as the ring of polynomial functions on $$X = \operatorname{Spec} R$$ which we can think of as $$\{(a,b,c) \in k^3 : a^2 = ab = 0\}$$. This is contained inside the plane $$\{a = 0\}$$. But because of the square, there is a non-reduced structure. Namely, if $$b = 0$$ then $$a^2 = 0$$ is the only remaining equation and we think of this as having a doubled line $$\{a = 0, b = 0\}$$ inside $$X$$. Therefore $$X$$ is the plane $$\{a = 0\}$$ where the line $$\{a = 0, b = 0\}$$ is doubled.

To say that a function $$f \in R$$ is zero now means that $$f$$ should vanish if we substitute $$x = 0$$ and should vanish "doubly" if we substitute $$x = 0, y = 0$$.

So now let's say we have a function (i.e. polynomial) $$f \in (x)\cap (xz^{n-1} + \lambda y^n)$$. Then we can factor $$f$$ as $$f = xg = (xz^{n - 1} + \lambda y^n)h$$. If we substitute $$x = 0$$ then we can see that $$f = xg = 0$$. So indeed $$f$$ vanishes on the plane $$\{a = 0\}$$ in $$k^3$$.

On the other hand, still keeping $$x = 0$$, we have $$0 = f = \lambda y^n h$$. Now if we substitute $$y = 0$$ we can see that somehow $$f$$ is vanishing a second time. To make this more precise, the fact that $$\lambda y^n h(0,y,z) = 0$$ means $$h$$ factors as $$h = xh_0$$ so that $$f = (x z^{n - 1} + \lambda y^{n-1})x h_0$$ and now setting $$y = 0$$ we get $$f = x^2z^{n - 1}$$ which indeed vanishes doubly on the line $$\{a = 0, b = 0\}$$.

Ignoring the geometric perspective, you can still see that we have shown that $$f \in (x) \cap (xz^{n - 1} \lambda y^n)$$ (inside $$k[x,y,z]$$) implies $$f \in (x^2, xy)$$.