# Field and maximal ideals

Given a field $F$ of 5 elements, show that $F[x]/\langle x^{3}+x+1 \rangle$ is not a field?

I went as follows: We can consider $F$ to be $\mathbb Z_{5}$, so $f(x)=x^{3}+x+1$ is irreducible in $F$, then $\langle f(x) \rangle$ is maximal, so $F[x]/\langle f(x)\rangle$ is a field!! Where is my mistake??

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Probably it should be $\ x^3 - x + 1\ =\ (x+2)\ (x^2 -2\ x - 2) \in \mathbb Z/5\:[x]$ – Bill Dubuque Jul 5 '11 at 23:13
Or, alternatively, the "not" could be a braino. – Bill Dubuque Jul 5 '11 at 23:31
nope!.......... – Hellen Jul 6 '11 at 0:08

Note $\rm\:F[x]/(p(x))\:$ is a field $\iff$ $\rm\:(p(x))\:$ is maximal $\iff$ $\rm\:0\ne p(x)\:$ is prime $\iff$ $\rm\:p(x)\:$ is irreducible. Recall that PID's are one-dimensional, i.e. primes $\ne 0$ are maximal, and, further, since a PID is a UFD,$\$ irreducible $\iff$ prime. In fact, PIDs are precisely the UFDs of dimension $\le 1$.
But a cubic $\rm\:f(x)\in F[x]\:$ is reducible iff it has root $\rm\in F\:,\:$ so for $\rm\: F = \mathbb Z/5\:$ you need only check if $\rm\ 0,\:\pm1,\:\pm2\$ are roots of $\rm\:f(x)\:.$
As it turns out, $\rm\:x^3 + x + 1\:$ has no roots, but $\rm\:x^3-x+1\:$ has the root $-2$. Hence once you determine the correct polynomial, it is clear whether or not the quotient ring is a field.