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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    $\begingroup$ Take polynomial $f(x) = x$ and check that it can't have inverse. $\endgroup$
    – falagar
    Aug 15, 2010 at 13:50
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    $\begingroup$ The favourite question of my students at this point: "Why isn't $1/x$ a polynomial?" :-) $\endgroup$ Aug 15, 2010 at 14:01
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    $\begingroup$ @SB: Well, a simpler way of looking at it is that a polynomial has to be continuous everywhere; 1/x clearly isn't. $\endgroup$ Aug 15, 2010 at 14:55
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    $\begingroup$ 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. :) $\endgroup$
    – BBischof
    Aug 15, 2010 at 22:17
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    $\begingroup$ @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. $\endgroup$
    – vonbrand
    Mar 12, 2014 at 0:05

6 Answers 6


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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    $\begingroup$ 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!!! $\endgroup$
    – BBischof
    Aug 15, 2010 at 22:15
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    $\begingroup$ I agree--this is extremely elegant, exactly what you want in a proof. I upvoted it! $\endgroup$
    – user452
    Aug 19, 2010 at 12:49
  • $\begingroup$ @BillDubuque Read this wishing to understand it, but isn’t this peculiar? “it suffices to exhibit _”. Please fill me in on what you mean by “exhibit”? Do you mean that it suffices to show that a non-identity does X? $\endgroup$ Oct 14, 2020 at 5:28
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    $\begingroup$ @Jon To "exhibit" a counterexample to: every $\rm R$-algebra element is a unit, it suffices to show there exists a nonunit in some $\rm R$-algebra $\endgroup$ Oct 15, 2020 at 2:57
  • $\begingroup$ @BillDubuque I just read the following question and answers. I got curious on the following: Why can we construct a field in $F_p$ but we can't in $R$? $\endgroup$
    – Red Banana
    Jun 8, 2022 at 3:57

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.

  • $\begingroup$ So what if I consider $ \Bbb{Z}_n[x]$? $\endgroup$
    – Manjoy Das
    May 3, 2020 at 17:02
  • $\begingroup$ @ManjoyDas, again, what is the inverse of $x$? Show a polynomial $p(x)$ over $\mathbb{Z}_n$ such that $x p(x) = 1$. $\endgroup$
    – vonbrand
    May 4, 2020 at 3:12
  • $\begingroup$ I wanted a general proof that such $ p(x)$ does not exist in case of $ \mathbb{Z}_n[x]$ rather than finding an arbitrary example. $\endgroup$
    – Manjoy Das
    May 4, 2020 at 5:01
  • $\begingroup$ @ManjoyDas, "a general proof" can't exist, as there ate polynomials (the constant ones) that do have inverses. Any polynomial of degree 1 or more can't have an inverse. $\endgroup$
    – vonbrand
    Jul 11, 2021 at 14:29

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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    $\begingroup$ The degree equation only works for polynomial rings over domains. $\endgroup$
    – azimut
    May 8, 2015 at 8:59

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 multiplicative inverse.


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

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    $\begingroup$ While this is true when $D$ is an integral domain, it fails for general commutative rings (but still $D[X]$ is never a field). $\endgroup$ Aug 15, 2010 at 15:55

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.

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    $\begingroup$ Well. If $F=\Bbb{Z}_4$, then we have $$(1+2x)(1+2x)=1+4x+4x^2=1.$$ $\endgroup$ Mar 18, 2019 at 14:59
  • $\begingroup$ may be just restrict R to be a field. my answer was not for any ring .. :) im wrong $\endgroup$
    – kjce
    Mar 18, 2019 at 15:13

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