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I don't understand the three claims that some $F_a[X]/(p(x))$ has some $n$ elements in the following text (from Adkins' Algebra):

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For example Why $F_2[X]/(X^2+X+1)$ has $4$ elements and what are those? I think that since $(X^2+X+1)$ is an ideal so $F_2[X]={\{X^2+X+1}\}F_2[X]=(X^2+X+1)$ so $F_2[X]/(X^2+X+1)$ is singleton?

Same as the other two: Why $F_3[X]/(X^2+1)$ (or $F_2[X]/(X^3+X+1)$) has $9$ (or $8$) elements and what are those?

Text is elementry itself but here it doesn't explain them well. Simple detailed explanation would be much apprecaited.

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  • $\mathbb{F}_2[X]/(X^2+X+1)$ has the four elements ${0,1,X,X+1}$. All other elements can be reduced to one of these. – Michael Burr Nov 03 '16 at 02:24
  • @MichaelBurr, As $(X^2+X+1)$ is an ideal so $X^2+X+1 \times F_2[X] = F_2[X]$ is subset of $(X^2+X+1)$. So $\mathbb{F}_2[X]/(X^2+X+1)= \mathbb{F}_2[X]/\mathbb{F}_2[X]$? –  Nov 03 '16 at 02:34
  • You can't do that because it matters how $(X^2+X+1)$ sits inside $\mathbb{F}_2$, its isomorphism structure as a ring is not enough. – Michael Burr Nov 03 '16 at 02:37

2 Answers2

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Hint: What $R=\mathbb{F}_2/(X^2+X+1)$ means that $X^2+X+1=0$ in $R$. Therefore, any time that you see a power of $X$, you can reduce it to a lower power using $X^2=-X-1=X+1$ (we can drop the negatives since $-1=1$ in characteristic $2$).

For example, $$ X^3\equiv X(X+1)=X^2+X\equiv (X+1)+X=2X+1=1. $$

Michael Burr
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  • Thanks a lot. I could deduce ${0,1,X,X+1}$. I will work for the other two, too. –  Nov 03 '16 at 02:47
  • Can someone tell what's the logic behind that 1+1=0 in the characteristic 2 polynomial field? How do we decide that it is 0? - From Aziz Kurtoglu – For the love of maths Apr 29 '19 at 18:00
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    @MohammadZuhairKhan The definition of characteristic is the minimum number of $1$'s needed to add together to get $0$. – Michael Burr Apr 29 '19 at 18:56
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Since $x^2+x+1$ is irreducible (it is quadratic and has no root), $F_2/\langle x^2+x+1\rangle$ is a field. We can "mod out" any polynomials of degree $2$ or higher, so the only possibilities are $0,1,x,$ and $x+1$.

For example, $x(x+1)=x^2+x$, but since $x^2+x+1=0$, $x^2+x=-1=1$, so $x(x+1)=1$.

pancini
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