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Let $R$ be commutative ring and $R[x]$ be ring of polynomial in one variable.

True/false

"If $R$ is field then $R[x]$ is an ED."

I think the above statement is true from result, "If $R$ is commutative ring such that polynomial ring $R[x]$ is ED then $R$ is necessarily a field."

user173049
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    The result you quote is related, but is actually the converse of what you want to show. You're asked to determine if "$R$ a field $\implies$ $R[x]$ and ED," and the result instead says that "$R[x]$ an ED $\implies$ $R$ a field". – vociferous_rutabaga Sep 03 '14 at 05:20
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    Dear user, you shouldn't use abbreviations like ED: electrons are cheap! Why should anyone bother to answer your question if writing out two words in full is too tiring for you? – Georges Elencwajg Sep 03 '14 at 17:50
  • @Georges Do you think the same for UFD, PID, gcd, lcm, etc? And why not be more polite when expressing such opinions? (esp. since the OP is new). – Bill Dubuque Sep 06 '14 at 14:05
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    @Bill: yes to all. – Georges Elencwajg Sep 06 '14 at 17:01
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    @Bill: there was definitely nothing impolite in my comment and I would be grateful to you for refraining to think you are entitled in any way to give me politeness lessons. – Georges Elencwajg Sep 06 '14 at 17:16

2 Answers2

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Outline: Look up the definition of ED. Then show that if $a(x)$ and $b(x)$ are polynomials in $R[x]$, with $b(x)$ not the zero polynomial, there exist polynomials $q(x)$ and $r(x)$ in $R[x]$, such that $r(x)=0$ or $\operatorname{deg}(r(x))\lt \operatorname{deg}(b(x))$, and $$a(x)=q(x)b(x)+r(x).$$

André Nicolas
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The Euclidean algorithm is simply the high-school polynomial long division algorithm, which has a simple proof by induction on the degree $\,d(f),\,$ namely to divide $\,f\,$ by $\,g\,$

$\ d(f) < d(g)\,\Rightarrow\, f\, =\, 0\cdot q + f\ $ (Base case). $ $ For the inductive step, note

$\ d(f)\ge d(g)\,\Rightarrow\, \exists k\!:\ f'=f-x^kg\ $ has $\,d(f')<d(f),\,$ so, by induction, $f'\! = q\,g+r,\,\ d(r)<d(g)$ hence we conclude: $\ f\, =\, f' + x^kg\, =\, (x^k\!+q)g + r,\,\ d(r)<d(g)$

Bill Dubuque
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