-2

My professor said that $(\mathbb F_2 [x] / (x^3 + x +1))^×$ has order 7, I am wondering, why this is true?

And what is the structure of this group? Is it a cyclic group?

Could someone clarify this to me, please?

Intuition
  • 3,269
  • 2
    Do you know how many elements $\mathbb{F}_2[x]/(x^3 + x + 1)$ has? Can you think of an element here that cannot be a unit? As to whether it's cyclic, there aren't that many groups of order $7$. – ronno Oct 11 '23 at 08:28
  • @ronno for your first question, no I do not, can you clarify please how we can know the number of elements of this group? – Intuition Oct 13 '23 at 13:03
  • Cosets of the ideal $(x^3 + x + 1)$ are uniquely represented by polynomials of degree $\le 2$ (why?) – ronno Oct 13 '23 at 13:57

2 Answers2

3

Polynomial $q(x)=x^3+x+1$ is irreducible over $\Bbb F_2$. Because if it was reducible, then it could be factored like $(x+c)\cdot p_2(x)$ where $p_2$ is a polynomial of degree 2.

Now if $q$ had a factor of degree 1, then $q(0)=0$ or $q(1)=0$ because the only linear factors over $\Bbb F_2[x]$ are $x$ or $x+1$; but $q(0)=q(1)=1$.

Hence, $\Bbb F_2[x]/q(x)$ has order 8 and is a (finite) field, and therefore all elements except 0 are units:

$$(\Bbb F_2[x]/q(x))^\times = \Bbb F_2[x]/q(x)\setminus\{0\}$$

Because the order of the multiplicative group is $8-1=7$ which is a prime, it must be a cyclic group, and indeed one finds that $x$ generates $(\Bbb F_2[x]/q(x))^\times$:

$$\begin{align} x^1 &= x \\ x^2 &= x^2 \\ x^3 &= x+1 \\ x^4 &= x\cdot x^3 = x(x+1) = x^2 + x \\ x^5 &= x\cdot x^4 = x^3+x^2 = x^2+x+1 \\ x^6 &= (x^3)^2 = (x+1)^2 = x^2+1\\ x^7 &= x\cdot x^6 = x^3+x = 1 \end{align}$$

emacs drives me nuts
  • 10,390
  • 2
  • 12
  • 31
2

Note that $x^3+x+1$ is an irreducible polynomial in $\mathbb{F}_2[x]$. Indeed, it has degree $3$, so to check irreducibility, one needs to show that the polynomial has no zero in $\mathbb{F}_{2}=\{0, 1\}$ which is easy to verify. Now, there is a general result which says that a quotient ring of the form $$ F[x]/(p(x)) $$ is a field whenever $p(x)$ is an irreducible polynomial over a field $F$. Therefore, $$ E = \mathbb{F}_{2}[x]/(x^3+x+1) $$ is a field. By definition, every non-zero element in $E$ is (multiplicatively) invertible. This means that the group of nonzero elements $$ E^{\times} = E - \{0\} $$ has exactly $7$ elements. As @ronno points out in the comments, there are not too many groups of order $7$ (because $7$ is prime). Can you see which group is it?

Prism
  • 11,162