1

The field $K$ is constructed in the following way: $K=F_2[X]/(X^4+X^3+1)$, where $F_2$ is short for $\mathbb{Z}$/$2$$\mathbb{Z}$.

Let $a$ be the class of $X$ in $K$ (so $a=X+(X^4+X^3+1))$.

The field $K$ contains a unique subfield $L$ with $4$ elements.

I need to show that $a^3+a+1$ is in that subfield $L$.

I know that a subfield of field $K$ is a subset containing $0,1$ and that it is closed with respect to addition, multiplication and inverses. But I don't know how to show that the element $a^3+a+1$ is in $L$. How can I do this?

user272027
  • 125

3 Answers3

2

We just need to perform a little of linear algebra to check that $a^3+a+1$ is an element of order $3$. In the given field, $a^4=a^3+1$, so: $$ (a^3+a+1)^2 = a^6+a^2+1 = a^2(a^3+1)+a^2+1 = a+a^3,$$ $$ (a^3+a+1)^3 = (a^3+a)^2+(a^3+a) = a^6+a^2+a^3+a = 1 $$ and the set $\{0,a^3+a+1,a^3+a,1\}$ is a subfield with four elements.

Jack D'Aurizio
  • 353,855
2

The field $K$ that you have has $2^4=16$ elements.

Let $b=a^3+a+1$, then $$b^2=(a^3+a+1)^2=a^6+a^2+1=a^2(a^3+1)+a^2+1=a^5+1=a(a^3+1)+1=a^3+a.$$ This gives, $$b^2=b+1$$ Furthermore, we get $$b^4=b \implies b^3=1.$$ Thus $b$ has order $3$. Now we know that $K$ has a unique subfield $L$ of size $4$, so $|L^{\times}|=3$ (cyclic group of order $3$) and by uniqueness it should be generated by $b$ (element of order $3$).

Anurag A
  • 41,067
1

Nothing wrong with the other methods (+1 to y'all). Just proffering a different track.

The multiplicative group of $\Bbb{F}_{16}$ is cyclic of order fifteen (your $a$ is one of the generators, but we won't need that). The multiplicative group of the subfield $\Bbb{F}_4$ is cyclic of order three. Because $15/3=5$ the elementary properties of the cyclic groups imply that the fifth power of any element of $z\in\Bbb{F}_{16}$ is in that subfield.

Let's calculate. Applying this to $z=a$ shows that $$ a^5=a\cdot a^4=a(1+a^3)=a+a^4=a+(1+a^3)=a^3+a+1\in\Bbb{F}_4. $$ If we had not been lucky on the first attempt, we could have done the same with $z=a^2$ or, more simply, used the fact that $\Bbb{F}_4$ is closed under addition as well.

I am not saying that this way would be better than the alternatives. Because there is only a single irreducible quadratic polynomial in $\Bbb{F}_2[x]$, namely $p(x)=x^2+x+1$, the other answerers knew that $p(x)$ has to be the minimal polynomial of $a^3+a+1$. So they took advantage. In a slightly bigger example we would have more choices for the minimal polynomial. Linear algebra could still be used to find the minimal polynomial. The method I used will always produce elements of the prescribed subfield, but the calculations become longer (even if you exploit the trusted square-and-multiply to the fullest).

There are many ways to skin this cat.

Jyrki Lahtonen
  • 133,153