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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.

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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.

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(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.

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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.

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