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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.2010-08-15
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    The favourite question of my students at this point: "Why isn't $1/x$ a polynomial?" :-)2010-08-15
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    @Augusti Roig: The definition of a polynomial is that the deg(x) has to be greater than one, right? So 1/x is clearly not a polynomial. Feel free to correct that if it's wrong xD2010-08-15
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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.2010-08-15
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    @SB. Greater or equal than zero.2010-08-15
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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. :)2010-08-15
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    @Soul: Minor nit: It's not true that a field is a commutative group with multiplication, because $0$ has no inverse. Rather, a field is a ring in which the **nonzero** elements form a group under multiplication.2011-08-09
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    @J.M., polynomials over arbitrary fields aren't "continuous" anywhere. In fact, you can use $x$ just as a mark, and not ever consider the polynomial as a function. Yes, for $\mathbb{R}[x]$ this makes sense, but don't get boxed in.2014-03-12
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    @vonbrand I remember when my abstract algebra teacher taught me formal sums, I was like "okay what are you saying? Is this a sum or not a sum?"2015-04-21

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Hint $\rm\quad\rm x \, f(x) = 1 \,$ in $\,\rm R[x]\ \Rightarrow \ 0 = 1 \, $ in $\,\rm R, \, $ by evaluating at $\rm\ x = 0 $

Remark $\ $ 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!!!2010-08-15
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    I agree--this is extremely elegant, exactly what you want in a proof. I upvoted it!2010-08-19
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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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    The degree equation only works for polynomial rings over domains.2015-05-08
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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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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).2010-08-15
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    @Robin: edited, thanks.2010-08-15
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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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Let $R$ be a field. If $f$ is a polynomial whose degree is at least 1 in $R[x]$, then $f$ cannot have an inverse.

For if $f(x)g(x) = 1$, where $f$ and $g$ are polynomials whose degree is at least 1 in $R[x]$, then the leading coefficient of $g$ would have to be $0$, which is impossible.