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I want to show that the ideal $(x^3+x+1)$ is prime in $\mathbb{Z}/(2)[x]$. I know that $\mathbb{Z}/(2)[x]$ is isomorphic with $\mathbb{Z}[x]/(2)$ and also know that since $\mathbb{Z}/(2)[x]$ is commutative it would be enough to show that for every $p(x),q(x)\in\mathbb{Z}/(2)[x]$, if $p(x)q(x)\in(x^3+x+1)$ then either $p(x)\in(x^3+x+1)$ or $q(x)\in(x^3+x+1)$, or equivalently if $p(x)q(x)=(x^3+x+1)r(s)$ then either $x^3+x+1$ divides $p(x)$ or divides $q(x)$. I have also proved (using a simple proof by cases) that $x^3+x+1$ cannot be factorized in $\mathbb{Z}/(2)[x]$, but I don't know how I should proceed further.

Navid
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  • As you noted, being a cubic with no roots over $\Bbb F = Z_2$ shows that it is irreducible, and by the dupe, irreducibles are prime by EL = Euclid's Lemma (EL is true in any domain where gcds exists - here they exist by the Euclidean algorithm in $F[x])\ \ $ – Bill Dubuque Jan 09 '22 at 11:30
  • Alternatively, by the last linked dupe, in a PID irreducibles generate maximal (so prime) ideals (the Euclidean algorithm easily shows every ideal is principal in $F[x]).\ $ If you need further details then ask in comments below. – Bill Dubuque Jan 09 '22 at 11:39

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