3

Theorem: Given $\{f_i\}_{1 \leq i \leq n}$, $f_i \in \mathbb{K}[x]$. Then the monic generator $f$ of the ideal $\langle \{f_i\} \rangle$ is $f = \gcd \{ f_i \}$.

In other words: $\langle \{f_i\} \rangle = \langle \gcd \{f_i\} \rangle$

My try at a proof: Let $d = \gcd \{f_i\}$.

$\subseteq$) If $g \in \langle \{f_i\} \rangle$, $g = \sum_i q_i f_i$ for some $q_i \in \mathbb{K}[x]$. As $d | f_i \ \forall i$, $f_i = r_i d$ then $g = \sum_i q_i r_i d$, and that means that $g = d \sum_i q_i r_i$ that is $d | g$, $g \in \langle d \rangle$

$\supseteq$) Let $g \in \langle d\rangle$. We know that $d = \sum_i q_i f_i$. And that $g = d q$. Then $dq = g = \sum_i q_i q f_i$. Then $g \in \langle \{f_i\} \rangle$.

I would like to know if my proof is correct. What puzzles me most is that I hadn't used that d is the greatest common divisor, just that it divides the other polynomials.

quid
  • 42,135
dami
  • 486
  • I'm pretty sure you used the fact that $d$ is the gcd of the $f_{i}$s when you asserted $d = \sum_{i}q_{i}f_{i}$. Otherwise, I'm not sure what you meant by that line. – Alex Wertheim Dec 11 '14 at 18:27

2 Answers2

3

My impression is that the point is showing that $\langle d\rangle\subseteq\langle f_1,f_2,\dots,f_n\rangle$ which is equivalent to $$ d=\sum_{k=1}^n q_kf_k. $$ The impression follows from the fact that once you know this the entire proof is trivial. Let's take another path.

You should know that $\langle f_1,f_2,\dots,f_n\rangle=\langle g\rangle$ for a unique monic polynomial $g$. Let's show that $g$ is the (monic) greatest common divisor of $f_1,\dots,f_n$. (By general theory about $\mathbb{K}[x]$, $g$ is the monic polynomial of least degree in $\langle f_1,f_2,\dots,f_n\rangle$.)

First of all, from $f_k\in\langle g\rangle$ we deduce that $g$ divides $f_k$ $(k=1,2,\dots,n)$. Now, let $h$ be a polynomial that divides $f_k$ $(k=1,2,\dots,n)$ and let us show that $h$ divides $g$.

Since $g\in\langle f_1,f_2,\dots,f_n\rangle$, we can write $g=\sum_{k=1}^n q_kf_k$; also $f_k=p_kh$, so $$ g=\sum_{k=1}^n q_kp_kh=\biggl(\sum_{k=1}^n q_kp_k\biggr)h $$ and $h$ divides $g$ as requested.

Therefore $g$ is the greatest common divisor of the given polynomials.

egreg
  • 238,574
  • re "You should know" There is no indication given by OP that that they know the ring of polynomials is a PID. The point could be precisely to show this. By contrast they say explicitly that they know that the GCD is a combination of the $f_i$, which you reprove. Indeed, it is not implausible that this is indeed the case. For example, I teach routinely in introductory courses GCD of integers, and if time permits of polynomials, covering this latter fact, while not talking at all about rings, ideals or anything. – quid Dec 12 '14 at 13:38
  • @quid Without any precise statement of the problem, with the already known facts, it's difficult to establish what the OP already knows. My impression is that “We know that” wasn't really the case, but I may be wrong. – egreg Dec 12 '14 at 15:32
  • Given that the issue is high-lighted in a comment on my post on which OP commented, I consider it as strange that it is your impresssion that OP does not know this, and that you were even that sure about it as even not to leave open the possibility that your answer could be completely besides the point. The point might be to show what you claim (though there are reasons to doubt it), to assert "The point is" seems a bit strong. It is this assertivnes that I find problematic. – quid Dec 12 '14 at 16:15
  • @quid I changed the wording and added a motivation. I think you agree that when you know that the gcd is a linear combination, the proof is entirely trivial. – egreg Dec 12 '14 at 16:20
  • Thank you for the change. What is trivial or not depends on the context and the development of the ideas. It could also be that the fact that the gcd is a linear combination is a trivial consequence of the argument that showed that the notion of a GCD makes sense in the structure in the first place. Anyway, I think the information you recorded is a good complement to the one that was already present. – quid Dec 12 '14 at 16:26
  • 1
    I agree with both of you. I choosed quid answer because it answers exactly what I was asking, and I can only choose one answer, but the information egreg posted was actually very useful to me to fill the gaps in my understanding. I just didn't want to reply thanks because (I'm new to mse) it seems to be discoraged here to post thanks or +1 or something like that, but I'm thankful to both of you, and voted up for both answers. And yes, I knew polys over a field is a PID, but to remind me the proof does help (I'm still learning this stuff). – dami Dec 12 '14 at 18:46
  • I meant to remind me of the proof egreg posted. – dami Dec 12 '14 at 18:52
1

The proof is by and large correct; if I would write it myself I would present some things differently, but I would grade it as correct.

To adress your specific doubt.

The first part of your argument is a special case of the fact that the ideal generated by the $f_i$ is contained in the ideal generated by $h$ for $h$ being some divisor common to the $f_i$; thus here you really only need that it is a divisor common to the $f_i$.

The second part of your argument uses the facts that a greatest common divisor is an element of the ideal generated by the $f_i$, and only this property is important for this part (that this element is a common divisor is not used).

As mentioned in a comment, it is not true for each common divisor that it is an element of the ideal generated by the $f_i$; for this you need it is the greatest.

quid
  • 42,135
  • @user26857 OP wrote "We know that $d = \sum_i q_i f_i$" I thus assume they actually know this (from the course, or preceeding chapter of the book or whereever the question came up). If they in fact do not know this then there is a gap there. Yet the exact formulation "We know that" suggests that they are aware that this is not evident but something that needs jsutification. – quid Dec 11 '14 at 19:30
  • 1
    Thanks, that answers exactly what I was asking. – dami Dec 12 '14 at 04:09
  • @dami This is just an ideal-theoretic form of the well-known number theoretic fact that a common divisor that is linearly representable (Bezout) is necessarily greatest. This is discussed in many of my prior posts, e.g. here for numbers, and here for polynomials over a field. It will be helpful to first grasp the number-theoretic case. – Bill Dubuque Dec 18 '14 at 15:35