Prove that $A$ a UFD implies $A[T]$ a UFD. Hint: You know that $K[T]$ is a UFD, where $K=\text{Frac}(A)$. Use Gauss' lemma to compare factorisation in $K[T]$ and $A[T]$. Deduce that the polynomial rings $\Bbb Z[x_1,\ldots,x_n]$ and $k[x_1,\ldots,x_n]$ are UFDs.
First I prove that:
Let $f\in A[T]$ be a primitive polynomial then, $f$ is irreducible over $A[T]\iff$ $f$ is irreducible over $K[T]$, where $K=\text{Frac}(A)$.
Proof: $(\Rightarrow)$ Suppose $f$ is irreducible over $A[T]$, but reducible over $K[T]$. Then $f=gh$, with $g,h\in K[T]$, so write $f=af_0g_0$ where $a\in K$ and $f_0,g_0\in A[T]$ are primitive polynomials. Since $f,f_0,g_0\in A[T]$ we have that $a\in A$, and so $f$ is reducible over $A[T]$. $\Rightarrow\Leftarrow$
$(\Leftarrow)$ Suppose $f$ is irreducible over $K[T]$, but reducible over $A[T]$. Then $f=gh$, with $f,g\in A[T]$. WLOG assume $g$ is a unit, in $K[T]$, so that $f=\frac{p}{q}h$ for $p,q\in A$. Since $f$ is primitive, $\frac{p}{q}$ is a unit in $A$. $\Rightarrow\Leftarrow$
$\blacksquare$
Now for my attempt at the proof in question.
Let $f\in A[T]\subseteq K[T]$. $f$ may be factored, over $K[T]$, as $$f=up_1\cdot\ldots\cdot p_n$$
Let $q_i=a_ip_i$ such that $q_i\in A[T]$ is both primitive and irreducible. Then $$f=cq_1\cdot\ldots\cdot q_n$$ As $A$ is a UFD $c$ may be factored to yield $$f=(r_1\cdot\ldots\cdot r_m)q_1\cdot\ldots\cdot q_n$$ This is a factorisation of $f$. To show that it is unique, suppose that $$(r_1\cdot\ldots\cdot r_m)q_1\cdot\ldots\cdot q_n=(s_1\cdot\ldots\cdot s_k)t_1\cdot\ldots\cdot t_v$$
Since $A$ is a UFD, and the irreducible polynomials are primitive, $r_i,s_j$ are associates. To show that $n=v$ and that the irreducible polynomials $q_i,t_j$ are associates of each other consider $q_1\cdot\ldots\cdot q_n=t_1\cdot\ldots\cdot t_v$ over $K[T]$. Over $K[T]$ irreducible and prime elements are the same, so $q_i$ and $t_j$ are associates. This means that we can write $q_i=\frac{a}{b}t_j$ for $a,b\in A$. Since $q_i,t_j$ are primitive $\frac{a}{b}$ is a unit in $A$. This forces $n=t$, and so the factorisation is unique.
Is my proof correct?