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?
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:
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 $\mathfrak a$.