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Just got a question about ideals of $F[x]$ being principal with $F$ being a field.

The proof basically starts assuming $g(x) \in N \subset F[x]$ where $N$ is an ideal and $g(x)$ is a minimal degree element. The rest is just deducing the what the division algorithm says.

But my question is that shouldn't we let $g(x)$ be arbitrary? Why are we choosing the minimal element?

Lemon
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  • Minimal in terms of degree. – Wuestenfux Jun 02 '17 at 08:39
  • Yes, but why are we choosing the minimal degree? – Lemon Jun 02 '17 at 08:39
  • @naturalstupidity, $g(x) = x(x)$. I am not sure where this is going. – Lemon Jun 02 '17 at 08:40
  • Well you aren't choosing the minimal degree, so you can't use the division algorithm. But that isn't my question. – Lemon Jun 02 '17 at 08:47
  • Oh you know what? I think I know what went wrong. The reason we are allowed to choose the minimal element is because every element bigger (in degree) than $g(x)$ can be reconstructed by the division algorithm. So we can assume without loss of generality to begin with the minimal element. – Lemon Jun 02 '17 at 08:55
  • I am not question its existence. – Lemon Jun 02 '17 at 09:04

2 Answers2

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You are trying to show the ideal $N$ is a principal ideal.

Let $g \in N$ have minimal degree.

The standard proof shows $N=(g)$, i.e., $N$ is the ideal generated by $g$.

If you choose $h \in N$ with $\deg(h)>\deg(g)$, you can form the principal ideal $(h)$, but it won't contain $g$, since any nonzero multiple of $h$ will have degree at least equal to $\deg(h)$.

Hence $N \ne (h)$.

The reason $g$ works is that when you use the division algorithm, the remainder, if nonzero, would be an element of $N$ with degree less than $\deg(g)$, which is impossible, since $\deg(g)$ is minimal (in $N$).

For the same reason $h$ fails, since when you use the division algorithm, the remainder, if nonzero, is an element of $N$ with degree less than $\deg(h)$, but $N$ has such elements (e.g., $g$), so there's no contradiction.

quasi
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  • Nah I already highlighted what was my original issue. It's literally the first statement "Let $g \in N$ have minimal degree". We do this because assuming non-minimal elements wouldn't help, but assuming the minimal lets us reconstruct the non-minimal elements. – Lemon Jun 02 '17 at 09:14
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    Perhaps you've now answered your own question, but my answer addresses your posted question (and is there for the benefit of others as well). – quasi Jun 02 '17 at 09:16
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The key idea behind the proof is as follows. In domains that enjoy division with remainder, ideals are closed under remainder, so by induction (Euclidean algorithm) they are also closed under gcd. Therefore the least (degree) element $g\in I$ necessarily divides every $f\in I$ since otherwise $\,\gcd(g,f)\in I$ is a smaller (degree) element in $I$, contra minimality of $g$.

Remark $ $ Presenting the descent as above by gcd (vs. mod or remainder) is not only clearer from a conceptual (divisibility) viewpoint, but it more easily leads the way to generalizations, e.g. for PIDs we can view ideal closure under gcds as reducing the number of prime factors of elements, yielding a gcd descent proof that nonzero ideals are generated by any nonzero member having the least number of prime factors.

Bill Dubuque
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