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well, I know similar question has been asked in mse but I want to ask is the intention of the proof by aluffi in his text,which seems strange for me.. that is,why does he introduce the field of fraction? may just want to use that $K[x]$ is UFD?

I think this statement could be proved without doing it.use the fact that if $(x-a)^m|(x-b)^ng(x) （a≠b）$ then $(x-a)^m|g(x)$ we could prove it ,right?R is integral domain

here is the proof of aluffi's algebra chapter0：

The number of roots of $f$ in $R$ is less than or equal to the number of roots of f viewed as a polynomial over the field of fractions $K$ of $R$;so we may replace $R$ by $K$. Now, $K[x]$ is a UFD, and the roots of f correspond to the irreducible factors of f of degree 1. Since the product of all irreducible factors of $f$ has degree $n$,the number of factors of degree $1$ can at most be $n$, as claimed.

Mugenen
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1 Answers1

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You're right. There is no need to go to $K$ nor to appeal to $K[x]$ begin a UFD.

Polynomial division by monic polynomials works over every commutative ring and gives: $$ f(x) = (x-a)q(x) + f(a) $$ In particular, if $a$ is a root of $f$, then $f(x) = (x-a)q(x)$. Since $\deg q < \deg f$,we have by induction that $q$ has at most $n-1$ roots and so $f$ has at most $n$ roots.

Wait. Where have we used that $R$ is a domain? We must have because $x^2-1$ has four roots in $\mathbb Z/8\mathbb Z$.

Let's see. If $b$ is a root of $q$, then $q(b)=0$ and so $f(b)=(b-a)q(b)=0$. So, every root of $q$ is a root of $f$. But that does not help the induction argument.

Conversely, if $b\ne a$ is a root of $f$, then $0=f(b)=(b-a)q(b)$ and that implies $q(b)=0$ only when $R$ is a domain. In this case, yes, every root of $f$ is either $a$ or a root of $q$, which is what is needed in the induction argument.

lhf
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  • Sir, In your answer you had written that "Polynomial division by monic polynomials works over every ring"? What if the ring is Non-commutative? Does it still works?(For, example in case of ring of polynomials over Hamilton's quaternions?) How? Please reply. – Akash Patalwanshi Sep 10 '21 at 03:39
  • I think the factor theorem fails if $R$ is non commutative ring. For example: In case of ring of polynomials over Hamilton's quaternions. Please reply – Akash Patalwanshi Sep 10 '21 at 04:02
  • Sir, As shown in this answer ( https://math.stackexchange.com/a/1730015/168676 ). The polynomial $f(x)=x^2-(j+i)x-k$ has a factor $x-j$ however, $f(j)≠0$. so how could we say that, polynomial division by monic polynomial works over every ring? – Akash Patalwanshi Sep 10 '21 at 04:57
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    @AkashPatalwanshi, I assumed the ring to be commutative. I've updated my answer. – lhf Sep 10 '21 at 10:32