# 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?

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: 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.

• Related : math.stackexchange.com/q/350054 Jan 31, 2019 at 23:55
• $F[x]/(f)$ 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). Feb 1, 2019 at 0:14

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 $$F$$ other than $$0$$. Also,

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

for

$$p(x) \notin (p(x)q(x)) \tag 8$$

lest for some

$$r(x) \in F[x] \tag 9$$

we have

$$p(x) = r(x)p(x)q(x), \tag{10}$$

or

$$p(x)(r(x)q(x) - 1) = 0, \tag{11}$$

whence

$$r(x)q(x) = 1, \tag{12}$$

which yields

$$\deg r(x) + \deg q(x) = \deg 1 = 0, \tag{13}$$

impossible in light of the assumption $$\deg q(x) \ge 1$$; thus we have shown that

$$(m(x)) = (p(x)q(x)) \subsetneq (p(x)) \subsetneq F[x] \tag{14}$$

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

• In (4), did you want $(m(x))$ on the right-hand side? Feb 1, 2019 at 1:13
• @J.W.Tanner: 'deed I did! Thanks, will fix! Feb 1, 2019 at 1:15

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.

It is really easy to show that $$\mathbb{Z}/p\mathbb{Z}$$ is a field when $$p$$ is prime, because it easily stems from Bézout's identity.

However, Bézout's identity also holds over any Euclidean domain. Let $$g(x)\in F[x]$$, with $$g(x)$$ not divisible by $$f(x)$$. We want to show that $$g(x)+(f)$$ is invertible in $$F[x]/(f)$$.

Since $$f$$ is irreducible, $$1$$ is a greatest common divisor of $$f$$ and $$g$$, hence there are polynomials $$u,v\in F[x]$$ such that $$1=f(x)u(x)+g(x)v(x)$$ Then, clearly, $$v(x)+(f)$$ is the inverse of $$g(x)+(f)$$ in $$F[x]/(f)$$.

Note that we actually proved that, for every irreducible element $$n$$ in a Euclidean domain $$R$$, the quotient ring $$R/(n)$$ is a field.

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.