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

Question

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.

Answer 1

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.

Answer 2

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 $f \in (x^2, xy)$.