0

I've come across this here: for a proof that x+1 is a generator.

\begin{align*} &x+1, \quad (x+1)^2=x^2+2x+1=2x, \\ &(x+1)^3=2x(x+1)=2x^2+2x=2x-2=2x+1\\ &(x+1)^4=(2x+1)(x+1)=2x^2+3x+1=2. \end{align*}

But I'm not sure I understand why in \begin{align*} &(x+1)^3=2x(x+1)=2x^2+2x=2x-2=2x+1\\ \end{align*} \begin{align*} 2x^2+2x=2x-2\\ \end{align*}

I tried dividing, subtracting, I multiplied all factors and made a table for Z3[x] x^2.

oliver
  • 675
  • 4
    Remember that in $F/(x^2+1)$ for any field $F$, $x^2\equiv -1$, so $2x^2 \equiv 2\cdot -1 = -2$. In general, since you're 'modding out' by a second-degree polynomial, you know that any reduced term must be linear; that is, all the elements of $F/(x^2+1)$ are of the form $ax+b$ for some $a,b\in F$. – Steven Stadnicki Dec 02 '19 at 16:55
  • Hi Steven, thank you. I thought this is only for fields in GF(2)? I understand that the terms are liniear. I'm just really confused to 2x^2 +2x\equiv 2\cdot 2x-2 – oliver Dec 02 '19 at 16:56
  • 2
    no; that's the definition of $/(x^2+1)$ here — you're considering $x^2+1$ to be effectively equal to zero, so $x^2$ is equivalent to $-1$ (that is, they're in the same residue class). – Steven Stadnicki Dec 02 '19 at 16:57
  • I'm sorry for the slow pick up. Wouldn't x squared be congruent to 1? I probably have to hit the books again on this one. Would you happen to recommend a good ressource for this particular topic? I had thought I understood, but clearly not. – oliver Dec 02 '19 at 16:59
  • 1
    If $x^2+1$ is congruent to $0$, then by subtracting $1$ from each you have $x^2+1-1$ is congruent to $0-1$ or in other words $x^2$ is congruent to $-1$, not $1$. – JMoravitz Dec 02 '19 at 17:01
  • @JMoravitz thank you! it clicked! – oliver Dec 02 '19 at 17:02
  • 1
    By definition if $K/b$ we are declaring $b \equiv 0$. so $x^2 + 1\equiv 0$ by definition. $(x+1)^2 = (x^2+1) + 2x \equiv 2x$ and $(x+1)^3 = (x+1)^2 (x+1)=2x(x+1) = 2x^2 + 2x = 2(x^2+1)-2 + 2x\equiv 2*0 -2 +2 x \equiv 2x-2$. Now as this is $F_3[x]$ we have $3\equiv 0$ so $-2 \equiv 1$ so $(x+1)^2 \equiv 2x+1$ – fleablood Dec 02 '19 at 17:09
  • 1
    Aternative $a \equiv a \pm (x^2 + 1)$ so $2x^2 \equiv 2(x^2 +1 -1)\equiv 2(-1)$. – fleablood Dec 02 '19 at 17:11
  • 1
    In the quotient ring $,x^2+1 = 0,\Rightarrow, x^2 = -1,$ so $,x,$ is like $,i.,$ In fact some write $,\Bbb F_3[x]/(x^2+1) \cong \Bbb F_3[i],$ where $,i,$ denotes the class $,x + (x^2+1)\Bbb F_3[x],$ in the quotient ring, like $\Bbb R[x]/(x^2+1) \cong \Bbb R[i]\cong \Bbb C\ \ $ – Bill Dubuque Dec 02 '19 at 17:13
  • 1
    So we can use complex arithmetic: evaluate at $,x = i,$ then work in $,\Bbb Z[i]\subset \Bbb C,$ then, finally, reduce $!\bmod 3.,$ This works because quotient ring reciprocity yields $$\Bbb F_3[x]/(x^2+1) \cong (\Bbb Z/3)[x]/(x^2+1) \cong \Bbb Z[x]/(3,x^2+1)\cong (\Bbb Z[x]/(x^2+1))/3\cong \Bbb Z[i]/3 $$ – Bill Dubuque Dec 02 '19 at 17:34
  • @BillDubuque Thank you, I'm super grateful for the excellent explanations you gave. I understand well now after pondering on it. – oliver Dec 03 '19 at 19:00
  • @fleablood thank you I understand now after following your factoring. – oliver Dec 03 '19 at 19:00
  • 1
    @oliver Glad it helped. I expanded the linked answer to explicitly show the isomorphism - which might clarify it a bit. – Bill Dubuque Dec 03 '19 at 19:20

0 Answers0