Let $F$ be a field, $f(x)$ is a polynomial in $F[x]$. $E = F[x]/(f)$ is a field if and only if $f(x)$ is irreducible.

Question

Can anyone help me with a proof for this theorem:

Let $F$ be a field, $f(x)$ is a polynomial in $F[x]$. $E = F[x]/(f)$ is a field if and only if $f(x)$ is irreducible.

Answer 1

You know that $F[X]/(f)$ is a field iff $(f)$ is a maximal ideal, but this is true iff $f$ is irreducible since $F[X]$ is a PID.

Answer 2

Hints:

(1) If $\,R\,$ is a commutative unitary ring, then an ideal $\,M\leq R\,$ is maximal iff the quotient ring $\,R/M\,$ is a field

(2) In the polynomial ring $\,\Bbb F[x]\,$ over a field $\,\Bbb F\,$ , a non-constant polynomial $\,f(x)\,$ is irreducible iff it is a prime element in the ring iff the principal ideal $\,(f(x))\in\Bbb F[x]\,$ is prime iff this ideal is MAXIMAL.

(3) You may want to use the fact that the ring $\,\Bbb F[x]\,$ is a UFD, an Euclidean domain, and/or a PID.

Answer 3

If $f$ is reducible, say $f=gh$ with $\deg g>0$ and $\deg h>0$, then the images of $g$ and $h$ in $F[x]/(f)$ are nonzero while their product is $0$, so $F[x]/(f)$ fails to be a domain. Also if $f$ is a unit or zero then $F[x]/(f)$ fails to be a field (it is the null ring or isomorphic to $F[x]$, respectively). So $f$ being irreducible is a necessary condition for $F[x]/(f)$ to be a field.

Now let $f$ be irreducible and $a$ a nonzero element of $F[x]/(f)$, so it is represented by a polynomial $g\notin(f)$. Since $F[x]$ is a PID one has $(f,g)=(d)$ for some monic polynomial $d$, which by assumption can be written $d=sf+tg$. Now $(d)\supsetneq(f)$ and since $f$ is irreducible this can only be if $d=1$. Now the image of $t$ is a multipliciative inverse of $a$, showing that $F[x]/(f)$ is a field.

Answer 4

Since $F[x]$ is an Euclidean Domain, then $F[x]$ is a principal ideal domain (PID) and recall that in a PID, an (nonzero) ideal is maximal iff it is prime. Hence, in this case $$ F[x]/(f(x)) \text{ is a field } \iff (f(x)) \text{ is maximal} \iff (f(x)) \text{ is prime}. $$ However you can notice that $(f(x))$ is prime if and only if $f(x)$ is irreducible:

If $(f(x))$ is prime, then $f(x)$ cannot be a unit as $(f(x))$ is a proper ideal. If $f(x)=a(x)b(x)$, then $a(x)$ or $b(x)$ are in $(f(x))$ because $(f(x))$ is prime, say $b(x)=r(x)f(x)$. Then $f(x)=a(x)r(x)f(x)$ and since $F[x]$ is an integral domain, $1=a(x)r(x)$, so $a(x)$ is a unit. Hence $f(x)$ is irreducible. Conversely assume $f(x)$ is irreducible. As noted above it is enough to prove that $(f(x))$ is a maximal ideal. Let $(m(x))$ be an ideal containing $(f(x))$ (recall that $F[x]$ is a PID). Then $f(x)=q(x)m(x)$ for some $q(x)\in F[x]$. Since $f(x)$ is irreducible, at least one of $q(x)$ or $m(x)$ is a unit. If $m(x)$ is a unit, then $(m(x))=F[x]$. Otherwise, if $q(x)$ is a unit, then $m(x)\in (f(x))$ and $(m(x))=(f(x))$, proving that $(m(x))$ is maximal.

In general: In a Principal Ideal Domain (like $F[x]$), a nonzero element is prime if and only if it is irreducible. Also, in an integral domain, a prime element is always irreducible (i.e. implication $\Longrightarrow$ holds).