If $p_1,\ldots,p_n$ are polynomials over a field $F$, not all of which are $0$, there is a unique monic polynomial $d$ in $F[x]$ such that
(a) $d$ is in the ideal generated by $p_1, \ldots, p_n$;
(b) $d$ divides each of the polynomials $p_i$.
Any polynomial satisfying (a) and (b) necessarily satisfies
(c) $d$ is divisible by every polynomial which divides each of the polynomials $p_1,\ldots,p_n$.
I want to prove this theorem, but I don't understand the proof in my linear algebra text. Also, I want to prove "$d$ is the monic generator of the ideal".
Actually, the text starts the proof identifying $d$ as the monic generator without any proof. Even if the proof is same here but the verification that $d$ is the monic generator is in it, I can understand the proof.
The way to prove this depends a bit on what one knows. The comment suggests the in the context of OP the following is known:
If this is known one can chose $d$ as the monic generator of the ideal generated by the $p_i$, so $(d)= (p_1, \dots p_n)$.
The rest is then quite direct: a) is true essentially by definition of $d$, b) is true as $d$ divides each element of $(d)$ which contains all $p_i$. Finally if there is a polynomial $f$ such that $f\mid p_i$ for each $i$ then $(d)=(p_1, \dots p_n) \subset (f)$, so $d \in (f)$ and $f \mid d$.
If the fact I mention above is not know, one can defined $d$ as a monic polynomial in the ideal generated by $p_i$ with minimal degree. Using Euclidean division one can then show that this polynomial is unique and generates the ideal.