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I'm currently studying Polynomial Rings, but I can't figure out why they are Rings, not Fields. In the definition of a Field, a Set builds a Commutative Group with Addition and Multiplication. This implies an inverse multiple for every Element in the Set.

The book doesn't elaborate on this, however. I don't understand why a Polynomial Ring couldn't have an inverse multiplicative for every element (at least in the Whole numbers, and it's already given that it has a neutral element). Could somebody please explain why this can't be so?

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Take polynomial $f(x) = x$ and check that it can't have inverse. –  falagar Aug 15 '10 at 13:50
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The favourite question of my students at this point: "Why isn't $1/x$ a polynomial?" :-) –  a.r. Aug 15 '10 at 14:01
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@SB: Well, a simpler way of looking at it is that a polynomial has to be continuous everywhere; 1/x clearly isn't. –  J. M. Aug 15 '10 at 14:55
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@SB. Greater or equal than zero. –  a.r. Aug 15 '10 at 15:18
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You might find it useful to look at the field of fractions of a polynomial ring. By looking at this and comparing the differences, I think it will be illuminating for you. Plus it will get you thinking in the direction of localization, which is a good direction to be thinking. :) –  BBischof Aug 15 '10 at 22:17

5 Answers 5

up vote 88 down vote accepted

HINT $\rm\qquad\rm x \; f(x) = 1 \;$ in $\rm R[x]\:\ \Rightarrow\: \ 0 = 1 \ $ in $\rm R \ \ $ by evaluating at $\rm\ x = 0\:. $

NOTE $\:\ $ This has a very instructive universal interpretation: if $\rm\: x\:$ is a unit in $\rm\: R[x]\:$ then so too is every $\rm\: R$-algebra element $\rm\: r\:,\:$ as follows by evaluating $\ \rm x \ f(x) = 1 \ $ at $\rm\ x = r\:.\:$ Therefore to present a counterexample it suffices to exhibit any nonunit in any $\rm R$-algebra. $\:$ A natural choice is the nonunit $\;\rm 0\in R\:,\:$ which yields the above proof.

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This is a good answer! The type of answer I really want to remember when my students ask this question! So much better than the degree argument! In short, I love it!!! –  BBischof Aug 15 '10 at 22:15
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I agree--this is extremely elegant, exactly what you want in a proof. I upvoted it! –  trb456 Aug 19 '10 at 12:49

Because by definition, the only polynomial that can have a negative degree is $0$, which is defined to have a degree of $-\infty$. Non-zero constants have degree $0$. You then have the degree equation: $\deg (fg) = \deg (f) + \deg (g)$ for any polynomials $f,g$. By inspection, any polynomial of degree $n \geq 1$ would need as an inverse a polynomial of degree $-n$, which does not exist (i.e. what Agusti Roig said!) The set you want does exist, however: it is called the field of rational functions, and is precisely the set of ratios of polynomials. It is constructed the same way that the field of rational numbers is from the ring of integers.

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Consider $\mathbb{C}[x]$ the ring of polynomials with coefficients from $\mathbb{C}$. This is an example of polynomial ring which is not a field, because $x$ has no multiplicativ inverse.

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The units of $D[x]$ are exactly the units of $D$, when $D$ is a domain.

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While this is true when $D$ is an integral domain, it fails for general commutative rings (but still $D[X]$ is never a field). –  Robin Chapman Aug 15 '10 at 15:55
    
@Robin: edited, thanks. –  lhf Aug 15 '10 at 20:28

For $F[x]$ to be a field, you need to show there is an inverse for each element that isn't 0. Now $x \in F[x]$, and clearly $x \ne 0$ (considered as a polynomial). But if you multiply $x$ by any non-zero polynomial, the result will always contain $x$ or higher powers, so it has no inverse.

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