Let $k$ be a field and let $A$ be a finitely generated $k$-algebra. By Noether normalization one can find algebraically independent elements $x_1, \dots, x_n \in A$ such that $k [x_1, \dots, x_n] \hookrightarrow A$ is an integral extension. Can it happen that one of the $x_i$'s is a zero divisor in $A$? In particular I am interested in the case $n = 1$.
Edit: I found a counterexample for $n = 2$, it is taken from this paper, Example 3.6. Let $R = k [x_0, \dots, x_3]$ and let $$I = ( x_0^2 - 3 x_0 x_1 + 5 x_0 x_3, x_0 x_1 - 3 x_1^2 + 5 x_1 x_3, x_0 x_2 - 3 x_1 x_2, 2 x_0 x_3 - x_1 x_3, x_1^2 - x_1 x_2 - 2 x_1 x_3) \subset R.$$ Then $k [x_2, x_3]$ is a Noether normalization of $R / I$ and $x_2$ is a zero divisor in $R / I$.
Let $R=k[x,y]/(x^2,xy)$ and take $k[x,y]\subset R$. This is an integral extension (since $x$ and $y$ satisfies the monic polynomial $t^2-(x+y)t$) and $(x+y)x=0$.