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Let $R$ be a ring and $f(X) \in R[x]$ be a non-constant polynomial. We know that the number of roots, of $f(X)$ in $R$ has no relation, to its degree if $R$ is not commutative, or commutative but not a domain. But,

  • The number of roots, of a non-zero polynomial over commutative integral domain, is at most its degree.

How does one prove above result?

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    Again, no indication what so ever of what you tried, nor even of what you *know* in this context, of why you are interested in this, or anything, really. Again, this is undistinguishable from you copying problems from random sources.2010-10-27
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    @Mariano: Why i am interested in this is easy! Because, this particular case, is true for a *commutative integral domain*.2010-10-27
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    @Mariano: Well, as far as my tries, goes, i only know that cancellation laws are valid in an integral domain, but couldn't figure out to use it here!2010-10-27
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    @Chandry1: One can always pick one of the hypotheses of a problem to be one's motivation. I can write a little python script to do it for me!2010-10-27
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    Any domain embeds into its field of fractions, and so the problem reduces to the corresonding question over a field, where it is solved by the division algorithm (if you like; there are probably lots of ways to prove it). (Passing to the field of fractions is one standard approach to problems that are initially posed over an integral domain.)2010-10-27

2 Answers 2

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This is a consequence of the fact that over a commutative ring with identity $A$, an element $a \in A$ is a zero of a polynomial $f \in A[x]$ if and only if $f(x) = (x - a)q(x)$ for some polynomial $q(x) \in A[x]$.

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It follows from the division algorithm, the fact that the evaluation maps gives a homomorphism from $R[x]$ to $R^R$ (functions from $R$ to $R$ with pointwise operations), and that $R$ is a domain. It is the exact same argument as for fields, with the division algorithm suitably restricted to certain kinds of polynomials over $R$ as divisors.