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How do we find the roots of ${x^3 + x + 1}$ in $ {Z_2[x]} $

The elements of the congruence class are: $$0, 1, x, x + 1, x^2, x^2 + 1, x^2 + x, x^2 + x + 1$$ as they have to be of the form $ax^2 + bx + c \in Z_2$.

However, I am unable to find the roots.

Specifically, I am unsure why $x^2$ and $x^2 + x + 1$ are roots in $x^3 + x + 1$

Any help would be appreciated.

abstract
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2 Answers2

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Do you mean finding the other roots of $x^3+x+1$ in $\mathbf F_8=\mathbf Z_2[x]/(x^3+x+1)$, in addition to the class $\omega$ of $x$?

$\omega^2$ is another root, because of the Frobenius morphism: $$(\omega^2)^3+\omega^2+1=(\omega^3+\omega+1)^2=0.$$ The sum of the roots is $0$,so the third root is the sum of the first two: $\,\omega^2+\omega$.

If you mean finding the roots of $x^3+x+1$ in $\mathbf Z_2$, there are none, as is easily checked, hence the polynomial is irreducible in $\mathbf Z_2[x]$ (a reducible polynomial of degree $3$ in a field has a root in that field).

Bernard
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  • It might be more clear if you did not use the same symbol for the variable of the ring and for the polynomial. – Tobias Kildetoft Jun 16 '15 at 08:34
  • The OP says " I actually have to show that $\mathbb{Z}_2[x]/(x^3+x+1)$ is a field, hence I am trying to find the roots". So I think, he just wants to find the roots of $x^3+x+1$ over $\mathbb{F}_2$. – Dietrich Burde Jun 16 '15 at 08:35
  • @Dietrich Burde: this was not clear to me, but it's a possible interpretation indeed. – Bernard Jun 16 '15 at 08:49
  • How come my calculations tell me $x^2+x+1$ is a root and I read $x^2+x$ is? Substituting $x^2+x$ in the polynomial, I get (of Wolfram gets) $x^6+3x^5+3x^4+x^3+x^2+x+1$, and $x^6=x^2(-x^2-x)=x^2+x+x+1=x^2+1$, $x^5=x(-x^2-x)=x+1-x=1$, $x^4=-x^2-x$ and $x^3=-x-1$, which, brought into Wolfram, yields $x^2-x+2$, which modulo 2 is $x^2-x≠0$. Am I missing something? – MickG Jun 16 '15 at 08:59
  • Your value for $x^5$ is false: I find $x^2+x+1$. Your way of calculating is very heavy: don't forget in $\mathbf Z_2$, $-1=1=3. – Bernard Jun 16 '15 at 09:32
  • You're right, I mis-multiplied and had a $-x$ instead of a $-x^2$. With that, I do get your value. Will edit my own answer. – MickG Jun 17 '15 at 06:58
  • Why do you state the sum of the roots must be zero? – MickG Jun 17 '15 at 07:02
  • Because the coefficient of $x^2$ in the equation is $0$(see relations between elementary symmetric functions of the roots and coefficients). – Bernard Jun 17 '15 at 08:48
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Older Answer left as extra

If I take you correctly, what you are after are the roots of the polynomial $t^3+t+1$ viewed as an element of $\frac{\mathbb{Z}_2[x]}{(x^3+x+1)}[t]$. Let me name that ideal $I$. You are asking why $x^2+I$ and $x^2+x+1+I$ are roots. Let us just substitute those roots into the polynomial $f(t)=t^3+t+1$, shall we? \begin{align*} f(x^2)={}&x^6+x^2+1=x^2(-x^2-x)-x^2+1=-x^4-x^3-x^2+1={} \\ {}={}&x^2+x+x+1-x^2+1=0, \end{align*} definitely a root then, and: \begin{align*} f(x^2+x+1)={}&(x^2+x+1)^3+x^2+x+1+1=(x^2+x+1)(x^4+x^2+1)+x^2+x={} \\ {}={}&x^6+x^4+x^2+x^5+x^3+x+x^4+x^2+1+x^2+x={} \\ {}={}&x^2(-x^2-x)-x^2-x+x^2+x(-x^2-x)-x-1+x-x^2-x+x^2+1+x^2+x={} \\ {}={}&x^2+x+x+1-x^2-x+x^2+x(-x^2-x)-x-1+x-x^2-x+x^2+1+x^2+x={} \\ {}={}&1-x+x^2-x^2+x+1-x-1+x-x^2-x+x^2+1+x^2+x={} \\ {}={}&-x-1+x-x^2-x+x^2+1+x^2+x=-1-x+1+x^2=-1, \end{align*} definitely not a root then. $x$ is the another root, as always in this kind of quotients. As noted by Bernard, the last root is close to the one tried above: it's $x^2+x$: \begin{align*} f(x^2+x)={}&(x^2+x)^3+x^2+x+1=(x^2+x)(-x^2-x+x^2)+x^2+x+1={} \\ {}={}&-x^3-x^2+x^2+x+1=-x^3+x+1=x+1+x+1=0. \end{align*} Let us now factor the polynomial since we have its three roots: $$f(t)=(t-x)(t-x^2-x)(t-x^2).$$ In carrying out my calculations, I used the fact that: \begin{align*} x^4=x\cdot x^3=x(-x-1)=-x^2-x, && x^3=-x-1, \\ \begin{aligned} x^6=x^2\cdot x^4={}&x^2(-x^2-x)=-x^4-x^3={} \\ {}={}&x^2+x+x+1=x^2+1 \end{aligned} && \begin{aligned} x^5={}&x\cdot x^4=x(-x^2-x)={} \\ {}={}&x+1-x^2=x^2+x+1 \end{aligned} \end{align*} which is because $x^3+x+1=0$ in your quotient field.

End of old answer which answers actual question, with a little addition

Yes indeed, this is a field, since the polynomial generating $I$ is clearly irreducible over $\mathbb{Z}_2$, as it has no roots. We could have seen that from the start, but now that we have the 3 roots in the quotient we see none of them is in $\mathbb{Z}_2$. Also, this quotient is the splitting field of that polynomial.

$f(x)$ has no roots in $\mathbb{Z}_2$ since $\mathbb{Z}_2=\{0,1\}$, and $f(0)=0^2+0+1=1$ and $f(1)=1^2+1+1=1+1+1=0+1=1$. It is a third-degree polynomial, so having no roots implies being irreducible, as if it were reducible it would have to factor as a first-degree polynomial times a second-degree one, and hence have a root. Being $f$ irreducible, the quotient of $\mathbb{Z}_2[x]$ with respect to $I=(f(x))$ is a field.

MickG
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