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I have to calculate a residue field of $\mathbb{F}_2[X]$ with exactly 16 Elements, but have real clue, how to do that.

My Idea would be $\mathbb{F}_2[X]\backslash(X^4+X^3+X^2+X+1)$.

But I don't know how to show that this is irreducible in $\mathbb{F}_2$

Any Ideas

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    This doesn't quite work. Most notably because the product of the cosets of $X$ and $X^3+X^2+X+1=(X+1)^3$ becomes zero, so the residue ring will not be a field this time (a field cannot have zero-divisors). You need to use an irreducible quartic. Do you know how to check whether a quartic in $\Bbb{F}_2[X]$ is irreducible? – 2017-01-03
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    @JyrkiLahtonen Thanks,I edited my post. With $X^4+X^3+X^2+X+1$ this should work?! I don't know how to proove that a quartic is irreducible in $\mathbb{F}_2. That's exactly my problem at the moment. – 2017-01-03
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    Yes, $X^4+X^3+X^2+X+1$ works. Well done! Irreducibility verification involves ruling out lower degree factors. The presence of linear factors can be overruled by checking that the polynomial has no zeros. Ruling quadratic factors takes a bit more, because you first need to build up a list of irreducible quadratics. In this case that list will be very short, because there is only a single irreducible quadratic. – 2017-01-03
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    @JyrkiLahtonen So I have to show that there is nothing that makes $$X^4+X^3+X^2+X+1 = (X^2+X+1)(?)$$ Because deg(?) = 2 and it $?\neq X^2+X+1$ its irreducible? – 2017-01-03
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    Correct again! Because $?$ has to be of degree two either it is equal to $X^2+X+1$ or it has a linear factor (a possibility hopefully ruled out earlier). – 2017-01-03
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    If you feel like it you can then post an answer. That way you get more feedback, we can identify any possible remaining weak links in your reasoning and we also get this question *answered* :-) If you do, @-ping me. I will give it a look and feedback and/or an upvote. – 2017-01-03

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