Is a particular element of a particular ring a nonzerodivisor?

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.

-

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 $I$.
An additional note on embedded points: $C$ has no embedded points since it is a projective hypersurface, and the locus defined by $I$ is flat over $C$, hence maps associated points to associated points, hence has no embedded points by a dimension count. (All the fibers over $C$ are isomorphic to $\mathbb{P}^1$.) – Charles Staats May 28 '11 at 20:19