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If $K$ is a field. Then in $K[X]$ there is an Euclidean algorithm and if $K$ is replaced by any arbitrary commutative ring $R$, then almost we have an Euclidean algorithm, by the following result:

Theorem: Let $f,g\in R[X]$ be polynomials, and assume that the leading coefficient of $g$ is a unit in $R$. Then there exist unique polynomials $q,r\in R[X] $ such that $$f=gq+r$$ and $\deg(r)< \deg (g)$

If $g$ is any arbitrary polynomial, is the conclusion of the theorem false?

Thank you all.

quid
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user126033
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1 Answers1

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The conclusion of the theorem for arbitrary $g$ is false for example for $R=\mathbb{Z}$, indeed for any domain that is not a field. Consider $f=X$ and $g=2X$, or generally $f=X$ and $g = aX$ where $a$ is not invertible.

quid
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