Is $F[x]/\langle x^3-1\rangle$ a field?

Question

Let F be a field.

Is $F[x]/\langle x^3-1\rangle$ a field?

I know $x^3-1 = (x-1)(x^2+x+1)$. How can I use this to show the above is a field or not?

I am using the above to deduce $\langle x^3-1\rangle$ is maximal or not in $F[x]$.

Answer 1

I am using the above to deduce $\langle x^3-1\rangle$ is maximal in $F[x]$.

This is impossible. Just note that the factorization you mention gives you ideals properly containing this one. Generally, a reducible polynomial cannot yield a prime ideal.

Answer 2

You're declaring $x\ne x^2 \ne x^3\ne x$ and $x$ is a cube root of $1$. The underlying set of this "field" is $\{a+bx+cx^2 : a,b,c\in F\}$. In this "field" we have $$ (x-1)(x^2+x+1) = x^3 - 1 = 1-1 = 0, $$ so we're multiplying two nonzero things and getting zero. If you can show that that cannot happen in a field, then you have to conclude that this is not a field.

Answer 3

Your factorization $x^3-1=(x-1)(x^2+x+1)$ tells you that $\langle x^3-1\rangle\subseteq \langle x-1\rangle$ (which is maximal, but it's irrelevant).

The two ideals are not equal, because $x^3-1$ does not divide $x-1$.

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Consider the evaluation map at $a\in F$, defined over polynomials: $$ v_a\colon F[x]\to F, \qquad v_a(p)=p(a) $$ This map is a ring homomorphism and is obviously surjective, so its kernel is a maximal ideal. Since $a$ is a root of $p\in F[x]$ if and only if $x-a$ divides $p$, we have that $\ker v_a=\langle x-a\rangle$.