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Let $R$ is an integral domain .

Show that If $R[X]$ is Euclidean domain then $R$ is a field .

I'll be waiting for your help. Thank you very much in advance!

Sara
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1 Answers1

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More generally:

If $R[X]$ is a PID, then $R$ is a field.

Indeed:

Take $a\in R$ with $a\ne0$. Consider the ideal $I=(a,X)$. Then $I=(f)$ for some $f$. Now, $a=fg$ implies that $f$ is a constant. Finally, $X=fh$ implies that $f$ is a unit. Thus, $I=R[X]$. Write $1=ap+Xq$ and evaluate at $X=0$. You get $1=ap(0)$, which implies that $a$ is a unit.

If $R[X]$ is an Euclidean domain using degree as the Euclidean function, there is a simpler proof:

Take $a\in R$ with $a\ne0$. Write $X=aq+r$ with $r=0$ or $\deg(r)<\deg(a)$. This implies that $r=0$ because $\deg(a)=0$. Now compare the leading terms of $X=aq$ and get $1=a q_n$.

lhf
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  • Interesting. I actually intended my hint to be towards a much less direct way (using that non-zero prime ideals in PIDs are maximal). – Tobias Kildetoft May 25 '15 at 12:22
  • If $R[X] $ is ED, the proof same?? – Sara May 25 '15 at 13:13
  • @Sara, if you know that every ED is a PID, then yes. Otherwise, see my edited answer. – lhf May 25 '15 at 13:47
  • @lhf thank yuo very much :) – Sara May 25 '15 at 13:59
  • @lhf nice answer, can you explain to me $X=aq$ and get $1=aq_n$ , when you says $X$, are you saying for any polynomial in $R[x]$, and $q \in R[x]$ too, so, can have a different degree,I am not able to rearrange to get $1 = aq_n $ .. thank you for your time – Francisco Jul 10 '15 at 00:33
  • @Francisco, $X$ is the indeterminate, as in the question. – lhf Jul 10 '15 at 09:36
  • Hello, could you explain why can we write $1=a p+X q$? Is it because $(a)$ and $(X)$ are coprime? How do we know that? – NotAbelianGroup Jun 10 '19 at 07:49
  • The 2nd proof is a special case of: ideals in Euclidean domains are principal - generated by any element of minimal Eucldiean value (applied to the ideal $(1)),,$ e.g. see here. – Bill Dubuque Jul 09 '22 at 09:02