Let $F$ be a field. How do we show that maximal ideals of $F[x]$ are the principal ideals generated by the monic irreducible polynomials?

In Algebra by Artin, he says this proposition is proven analogously to:

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Here, he shows that if $n$ is prime, then $\mathbb Z/(n)$ is a field. Then we use the fact that $R/I$ is a field iff $I$ is maximal, and he concludes that $(n)$ is maximal.

The analogous proof would be that if $f(x)$ is monic irreducible, then $F[x]/(f)$ is a field. The only problem is that he has not proven that $F[x]$ modulo a monic irreducible polynomial is a field.

  • $\begingroup$ Related : math.stackexchange.com/q/350054 $\endgroup$ – Jean Marie Jan 31 at 23:55
  • $\begingroup$ $F[x]/(f)$ is a field because $(f)$ is a maximal ideal and $(f)$ is a maximal ideal because $(f)\subseteq (g)\iff g\mid f\iff g=f$ or $g=1 \iff (f)=(g)$ or $(g)=F[x]$. In the step $g\mid f \iff g=f$ or $g=1$ I am using that $F[x]$ is a UFD (and hence irreducible elements are prime elements). $\endgroup$ – yamete kudasai Feb 1 at 0:14

In general we have $F(x)/(f)$ is a field iff f(x) is irreducible.

If reducible, then we have a zero divisor, so it can't be a field. If irreducible, then all polynomial  can be subjected to Euclidean algorithm which gives you a multiplicative inverse.

The remaining field axioms follow from the fact that we have a ring quotient.


There''s a classic result in commutative algebra that you can apply:

Let $B$ be an integral domain, $A$ be a subring such that $B$ is integral over $A$. Then $B$ is a field if and only if $A$ is a field.


For any field $F$, $F[x]$ is a principal ideal domain; this is a very well-known and oft-quoted result, which I will accept here.

Now let

$M \subset F[x] \tag 1$

be a maximal ideal; since $F[x]$ is a principal ideal domain, we have

$M = (m(x)) \tag 2$

for some

$m(x) \in F[x]; \tag 3$

we may clearly take $m(x)$ to be monic, since the leading coefficient $\mu$ of $m(x)$, satisfying as it does $\mu \ne 0$, is a unit; thus $\mu^{-1} m(x)$ is monic and

$(\mu^{-1} m(x)) = (m(x)); \tag 4$

now if $m(x)$ were reducible in $F[x]$, we would have

$m(x) = p(x)q(x), \; p(x), q(x) \in F[x], \; \deg p(x), \deg q(x) \ge 1; \tag 5$

consider the ideal

$(p(x)) \subsetneq F[x]; \tag 6$

it is clearly proper: since $\deg p(x) \ge 1$, $(p(x))$ contains no polynomials of degree $0$, that is, contains no elements of $F$. Also,

$(m(x)) = (p(x)q(x)) \subset (p(x)), \tag 7$

which shows that $(m(x))$ is not a maximal ideal in $F[x]$; this contradiction implies that $m(x)$ is irreducible in $F[x]$. Finis.

  • 1
    $\begingroup$ In (4), did you want $(m(x))$ on the right-hand side? $\endgroup$ – J. W. Tanner Feb 1 at 1:13
  • $\begingroup$ @J.W.Tanner: 'deed I did! Thanks, will fix! $\endgroup$ – Robert Lewis Feb 1 at 1:15

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