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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.

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    $\begingroup$ If you don't understand the proof it could be good to mention some details about the proof. It will not be productive if somebody might just give you the same proof here. $\endgroup$ – quid Jul 29 '15 at 10:08
  • $\begingroup$ 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. $\endgroup$ – cokecokecoke Jul 29 '15 at 10:15
  • $\begingroup$ Oh! i got it!!! $\endgroup$ – cokecokecoke Jul 29 '15 at 10:37
  • $\begingroup$ Thanks for the clarification. Even if you problem now seems resolved I will add some remarks on this specific point. $\endgroup$ – quid Jul 29 '15 at 11:18
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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:

Each (non-zero) ideal of $F[X]$ is generated by a unique monic polynomial (which is a bit stronger than $F[X]$ is a PID).

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.

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