Use Kronecker's theorem to construct a field with four elements by adjoining a suitable root of $x^4-x$ to $\mathbb Z/2\mathbb Z$.

Definition: A polynomial $f(x)\in F[x]$ splits over $F$ if it is a product of linear factors in $F[x]$.

Kronecker's theorem: Let $f(x)\in F[x]$ where $F$ is a field. There exists a field $E$ containing $F$ over which $f(x)$ split.

What I did is: $$x^4-x=x(x-1)(x^2+x+1).$$

I know that $x^2+x+1$ is irreducible, then $$(\mathbb Z/2\mathbb Z)[x]/(x^2+x+1)=\{[0],[1],[1+x],[x]\}$$ is a field.

But I don't see I can conclude. Moreover, What do they want to say by "adjoining a suitable root of $x^4-x$" ?

  • 1
    $\begingroup$ You started by saying $x^2-x$, was this a typo? $\endgroup$ – Alan Oct 12 '14 at 14:46
  • $\begingroup$ I corrected it. Tks. $\endgroup$ – idm Oct 12 '14 at 15:15

Typically, when we adjoin a root to a field, it looks something like $\mathbb C=\mathbb R (i)$, which is the closure of $\mathbb R$ when we add a root of $x^2+1=0$. So here, call your root of $x^2+x+1$ something, so say $j$ satisfies $j^2+j+1=0$ mod 2. Now, look at the field $\mathbb Z/2\mathbb Z (j)$, the elements of which are going to be $0,1,j,j+1$. We get this is a field automatically since j is a root of an irreducible polynomial. They may want you to actually work out the multiplication table/etc., but this shouldnt be too hard


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