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F is a field $\Rightarrow$ $F\left[ x\right]$ is a principal ideal ring. How to proof? Second question F have to be field?(why)

FMath
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2 Answers2

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In $F[x]$ you have division with remainder. This shows that each nonzero ideal $I$ in $F[X]$ is generated by the monic polynomial $f$ in $I$ with smallest degree. Division of a polynomial $f$ by a polynomial $g$ requires the leading coefficient of $g$ to be invertible. This should answer your 2nd question.

Wuestenfux
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$F$ is field $\implies$ $F[x]$ is E.D $\implies$$F[x]$ is PID