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 September 22nd, 2012, 10:53 AM #1 Newbie   Joined: Sep 2012 Posts: 11 Thanks: 0 Finite field Hello guys. Can i ask you for some help about two problems i am fighting with? The first one is: Given is a polynom $p(x)=x^{6} + x + 1$ in $\mathbb{Z}_{2}[x]$. I have to show that $\mathbb{Z}_{2}[x]/p(x)$ is a finite field. Is it enough if i show that the polynom is irreducible over $\mathbb{Z}_{2}[x]$ ? The second problem: To prove that $\mathbb{F}[x]/(x^{2} + x^{2} + x+ 1)$ cannot be a field, no matter what the field $\mathbb{F}$ is. I would be very thankful if you could give me some hints how to solve these problems. Thank you in advance!
 September 22nd, 2012, 01:47 PM #2 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: Finite field for your first question: yes. if p(x) is irreducible over F, then (p(x)), the ideal generated by p(x) is maximal. for if I is an ideal with (p(x)) < I < F[x], then since F[x] is principal, we have I = (f(x)) with f(x)|p(x), contradicting the irreducibility of p(x), or f(x) is a unit, in which case I = F[x] (the units of F[x] are the units of F), because I contains 1. now, for any domain R, with I maximal in R, R/I is a field...because it has no non-trivial proper ideals. for your second question: is there a typo?
 September 22nd, 2012, 02:26 PM #3 Newbie   Joined: Sep 2012 Posts: 11 Thanks: 0 Re: Finite field Deveno, thank you very much! Yes, there is a mistake in the second problem. It should be $\mathbb{F}[x]/(x^{3}+x^{^2}+x+1)$ I'm sorry about that.
 September 22nd, 2012, 02:32 PM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Finite field That's important because $x^3+ x^2+ x+ 1$ now has an even number of terms and x= -1 is a root. Of course, since every field has a multiplicative identity, 1, every field contains -1.
 September 23rd, 2012, 12:54 PM #5 Newbie   Joined: Sep 2012 Posts: 11 Thanks: 0 Re: Finite field HallsofIvy, thank you very much!

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