I am trying to prove that $ F_2[x]/(x^3 + x + 1)$ is a field, but that $F_3[x]/(x ^3 + x + 1)$ is not a field.

A solution here uses a lemma:

A degree 2 or 3 polynomial $f(x) ∈ F[x]$ is irreducible if and only if it has no linear factor $x − a ∈ F[x]$, i.e., no root in $F$.

But could someone tell me how to prove it? And how does it imply the fact that there is no root in $F$?

I am stuck at proving that it must have a root in F.

Thank in advance!


Suppose a polynomial $f$ of degree $2$ or $3$ were reducible; i.e. suppose we have $f(x) = g(x)h(x)$ where $g$ and $h$ are nonconstant polynomials. This means $\deg(g)$ and $\deg(h)$ are necessarily $\geq 1$. How is $\deg(f)$ related to $\deg(g)$ and $\deg(h)$ given that we are working over a field? What can we conclude?

  • $\begingroup$ I do know that if $deg(f)=3$, one of the deg(g) or deg(h) should be 1, but how does it implies the fact that it must be a root in F? $\endgroup$ – Y.X. Oct 10 '16 at 3:48
  • $\begingroup$ And here is I actually stuck on... $\endgroup$ – Y.X. Oct 10 '16 at 3:52
  • $\begingroup$ It is zero! Thus there exist an c in F such that f(x)=0, such f(x) has a root in F, isn't it? $\endgroup$ – Y.X. Oct 10 '16 at 3:57
  • 2
    $\begingroup$ Indeed! The only thing to worry about is if $f$ wasn't monic and we somehow factored as $f(x) = (bx-c)h(x)$ for some $b \in F$. But this isn't a problem since we are in a field: the leading constant term can be pulled out and incorporated into $h(x)$ instead, giving $f(x) = (x - c/b)(b(h(x))$, and $c/b \in F$ would be our root. $\endgroup$ – Kaj Hansen Oct 10 '16 at 3:59

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