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I want to show that if $K$ is a field then $K[x]$ is principal

Here is what I did:

I took 2 polynomial p(x) and q(x). Since we're on a field we can do a euclidean division of polynomials:

$q(x) = p(x)a(x) + r(x)$ with $deg(r) < deg(p)$

Now if $r= 0$ we obviously have $p \in (q)$

But I don't know how to show that it is principal if $r \neq 0$

Also I would like to show that the assertion is false if we replace "$K$ is a field" by "$K$ is a principal domain and an integral domain". I can't figure out how to do that

Sylvester Stallone
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    This and variants have been asked before. Check for example these: http://math.stackexchange.com/q/1214667/133781 http://math.stackexchange.com/q/2014146/133781 and http://math.stackexchange.com/q/873088/133781 – Xam Mar 12 '17 at 16:02

1 Answers1

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Hint: Let $0 \neq I \subsetneq K[x]$ be an ideal. Let $0\neq f\in I$ such that

$$ deg(f)= \min\{ deg(g) \ : \ 0\neq g\in I \}.$$

Use your argument above to show $I=(f)$.

For your second question I'd suggest that you show that the ideal $(2, x)\subseteq \mathbb{Z}[x]$ is not principal.

Severin Schraven
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  • Why do you need the condition: $deg(f)=min{deg(g) : g \in I}$? – user33 Nov 26 '24 at 12:46
  • @user33 Pick any $0\neq h\in I$. Then by the Euclidean algorithm we get that $$h=pf+r$$ with $\mathrm{deg}(r)<\mathrm{deg}(f)$. However, because $f,h\in I$ we get that $r=h-pf\in I$. But $f$ has minimal degree in $I\setminus {0}$, thus, $r=0$. – Severin Schraven Nov 26 '24 at 13:56