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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. – Mariano Suárez-Alvarez Oct 27 '10 at 3:46
@Mariano: Why i am interested in this is easy! Because, this particular case, is true for a commutative integral domain. – anonymous Oct 27 '10 at 4:05
@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! – anonymous Oct 27 '10 at 4:06
@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! – Mariano Suárez-Alvarez Oct 27 '10 at 4:07
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.) – Matt E Oct 27 '10 at 4:45
up vote 2 down vote accepted

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.

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