I know that if $F$ is a field then $F[x]$ is a PID, and I also know that if $F[x]$ is a Euclidean domain then it a PID and therefore $F$ is a field. However, can we say that if $F$ is a field then $F[x]$ is a Euclidean domain? If so, what would be a sketch of a proof? Otherwise, is there an easy example of $F[x]$ that is a PID but not Euclidean?
If $F$ is a field then $F[x]$ is euclidean domain
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abstract-algebra
polynomials
ring-theory
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6I thought the first proof everyone sees that $F[x]$ is a PID is by showing it is a Euclidean domain using the obvious division algorithm. – 2017-02-15
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1Do you know how to do the euclidean division in $\Bbb R[x]$ (or, say, $\Bbb Q[x],\,\Bbb C[x]$...)? Because it is exactly the same. – 2017-02-15
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0$F[x]$ is Euclidean with $d(f)={\rm deg}(f)$. This is usually explained in abstract algebra books dealing with Euclidean rings. – 2017-02-15