Clarification on a solution for Atiyah-Macdonald Chapter 1 Q.3. The question is supposed to be a generalisation of Chapter 1 Q.2 to multivariable polynomials. However, I am specifically referring to the statement: $f$ is a zero-divisor iff there exists $a\in A$ s.t $a\not=0$ and $a*f=0$.
I saw a solution online: 
However, what I do not understand is how in the second line he states "exercise 1.2 allows us to assume that $g\in A[x_1,....,x_n]$" since exercise 1.2 only holds for 1 zero divisor and not ideals. Can someone care to explain how this works?
I have also seen other proofs which essentially use the same idea. i.e if there exists polynomial $f$ s.t $f*h_k=0$ for all k(where $h_k$ are polynomials themselves), then there exists $a\in A$ s.t $a\not=0$ and $a*h_k=0$ for all k. Again they are using the statement in exercise 1.2 for multiple zero divisors.
Any ideas?
 A: Fortunately, K. Conrad salvages the situation in this expository paper of his. In particular, see Theorem 3.3 therein. I reproduce the proof here for completeness:

Theorem. Let $A$ be a commutative ring with identity and $r\ge 1$. Let $\mathfrak a$ be an ideal of $A[x_1, \ldots, x_r]$ such that some $g\in A[x_1, \ldots, x_r]\setminus\{0\}$ annihilates $\mathfrak a$. Then there exists an $x\in A\setminus\{0\}$ that annihilates $\mathfrak a$.

Proof.
Base step for $r = 1$: Let $g$ be a nonzero polynomial of the least degree that annihilates $\mathfrak a$. If $\deg g = 0$, we are done. So suppose $\deg g\ge 1$. Hence, write $g = b_0 + \cdots + b_m x^m$ for $m\ge 1$ with $b_m\ne 0$. Now, since $b_m$ is a nonzero constant, it can't annihilate $\mathfrak a$ and hence take an $f\in\mathfrak a$ such that $b_m f\ne 0$. Write $f = a_0 + \cdots + a_n x^n$. Note that $g a_n$, if nonzero, will be an annihilator of $\mathfrak a$ of degree $ < \deg g$ (with $b_m a_n$ being $0$), giving the desired contradiction. But we can't be sure that $g a_n$ is indeed zero. So we take a bit convoluted route: Since $b_m f\ne 0$, we are guaranteed that some $b_m a_i\ne 0$ which in turn guarantees that $g a_i\ne 0$. Hence, we can take a largest integer $i_0$ such that $g a_{i_0}\ne 0$. Now observe that $0 = gf = (b_0 + \cdots + b_m x^m)(a_0 + \cdots + a_{i_0} x^{i_0})$ so that $b_m a_{i_0} = 0$. Thus we have constructed something exactly as we wanted, and we are done!
Inductive step: Suppose we have proven for $r\ge 1$. Now we prove for $r + 1$. Let $\mathfrak a$ be an ideal of $A[x_1, \ldots, x_{r + 1}]$ that is annihilated by some $g\in A[x_1, \ldots, x_{r + 1}]\setminus\{0\}$. Since we can identify $A[x_1, \ldots, x_{r + 1}]$ with $A[x_1, \ldots, x_r][x_{r + 1}]$, by the above reasoning, we can take $g$ to be in $A[x_1, \ldots, x_r]$. Hence $g$ annihilates all the (polynomial-)coefficients of the polynomials present in $\mathfrak a$, and hence annihilates the ideal $\mathfrak b$ generated by these coefficients. Now, being in the ring $A[x_1, \ldots, x_r]$, we use the inductive hypothesis, and find an $x\in A\setminus\{0\}$ that annihilates $\mathfrak b$, and hence $\mathfrak a$.
