Let $A$ be the ring $\Bbbk[\alpha\_0, \alpha\_1, \alpha\_2, x\_0, x\_1, x\_2]$ (where $\Bbbk$ is an infinite field, algebraically closed if it matters). Let $g \in \Bbbk[\alpha\_0, \alpha\_1, \alpha\_2]$ be a homogeneous polynomial of degree at least one, such that $\alpha\_0$ does not divide $g$. Let
$$I = (g, \sum\_i \alpha\_i x\_i).$$



> 
> Is $\alpha\_0$ necessarily a nonzerodivisor in $A/I$?
> 
> 
> 


A little bit of motivation: I've shown that $I$ has a certain nice property that I would like to carry over to its localization $I[\alpha\_0^{-1}] \subset A[\alpha\_0^{-1}]$. This will hold automatically if $I[\alpha\_0^{-1}] \cap A = I$, which is true iff $\alpha\_0$ is a nonzerodivisor in $A/I$. (Each of these is clearly equivalent to the statement: if $\alpha\_0 f \in I$, then $f \in I$.)


I can show that at least $\alpha\_0$ is not nilpotent, by showing that $(A/I)[\alpha\_0^{-1}]$ is not the zero ring.



Is a particular element of a particular ring a nonzerodivisor?
I think the answer is yes.


Let's try to think of the question geometrically. The polynomial $g$ defines a curve $C$ in $\mathbb{P}^2$ with coordinates $\alpha\_i$ (I'd prefer these to be $x$'s and your $x$'s to be $\alpha$'s, but we won't change your notation). The bihomogeneous polynomial $\sum \alpha\_ix\_i=0$ defines the codimension $1$ locus in $\mathbb{P}^2\times (\mathbb{P}^2)^\ast$ of pairs $(p,\ell)$ where $p$ is a point in $\mathbb{P}^2$ and $\ell$ is a line containing $p$. Then $I$ defines the codimension $2$ locus consisting of pairs $(p,\ell)$ with $p\in C$ and $p\in \ell$. The irreducible components of this locus are in bijective correspondence with the irreducible components of $C$, and a polynomial $f\in k[\alpha\_i]$ will be a zero divisor in $A/I$ if and only if $gcd(f,g)\neq 1$, so that $f$ vanishes on a component of the locus defined by $I$.


