Exercise 13.3.7

Proof. (a) $u$ is a prime of the PID ${𝔽}_2[u]$, but the reduction modulo $u$ is $\bar{f}=x^4+x^2+1$, which is not irreducible, since $x^4+x^2+1={(x^2+x+1)}^2$. Nor the Schönemann-Eisenstein not the reduction modulo $u$ applies here.

We give a direct proof of the irreducibility of $f$, by proving that $f$ has no factor with degree 1 ou 2. By Gauss’s Lemma, if $f$ has no factorisation in ${𝔽}_2[[u][x]$, then $f$ is irreducible in ${𝔽}_2[u](x)$.

$\text{∙}$ If $f$ has a factor with degree 1, then $f$ has a root $\frac{a(u)}{b(u)}\in {𝔽}_2(u)$, where $a(u),b(u)$ are relatively prime. Then

therefore $a\mid b^4$, and $a\wedge b=1$, hence $a\mid 1$, and similarly $b\mid 1$. Thus $a,b\in {𝔽}_2$, where $a\ne 0,b\ne 0$, so that $a(u)=b(u)=1$. But $f(1)=1\ne 0$, therefore $f$ has no factor with degree 1 in $F[x]$.

$\text{∙}$ If $f$ has a factor with degree 2, then

where $a,b,c,d,e,f$ are in ${𝔽}_2[u]$. Since $a(u)d(u)=1$, $a(u)$ and $d(u)$ are units of ${𝔽}_2[u]$, so are in $F_2$, therefore $a(u)=d(u)=1$, and similarly $c(u)=f(u)=1$, so that

The comparison of the terms in $x^3$ gives $b(u)+e(u)=0$, so that

This is a contradiction, since $f$ has a nonzero term $\mathit{ux},u\ne 0$. Therefore $f$ is irreducible over $F$.

Sage verification:

F = Frac(GF(2)[’u’]); u = F.gen()
f = x^4 + (u+1)*x^2 + u*x +1
f.is_irreducible()

gcd(f,f.derivative(x))

This is a verification that $f$ is irreducible, separable. (b) For the given polynomial $c_1={\sigma }_1=0,c_2={\sigma }_2=u+1,c_3={\sigma }_3=-u,c_4={\sigma }_4=1$. For $D_f(y)$ we have (cf. Ex.13.3.6,13.3.4 and (13.3),(13.32)):

Sage verification:

    R.<x1,x2,x3,x4,y1,y2,y3,y4> = PolynomialRing(QQ, order = ’degrevlex’)
    elt = SymmetricFunctions(QQ).e()
    e = [elt([i]).expand(4).subs(x0=x1, x1=x2, x2=x3, x3 = x4) for i in range(5)]
    J = R.ideal(e[1]-y1, e[2]-y2, e[3]-y3,e[4]-y4)
    G = J.groebner_basis()
    D = x1^3*x2^2*x3;
    D=D+D.subs(x1=x3,x2=x4,x3=x1)+D.subs(x1=x2,x2=x1,x3=x4)
      +D.subs(x1=x4,x2=x3,x3=x2)
    D=D+D.subs(x1=x2,x2=x3,x3=x1)
      +D.subs(x1=x2,x2=x3,x3=x1).subs(x1=x2,x2=x3,x3=x1)
    d1=x1*x2+x3*x4;d2=d1.subs(x2=x3,x3=x2);d3=d1.subs(x1=x3,x3=x1);
    S1=d1+d2+d3; S2=d1*d2+d1*d3+d2*d3; S3=d1*d2*d3

    S.<c1,c2,c3,c4,u> = PolynomialRing(ZZ, order = ’degrevlex’)
    A=(D+D.subs(x1=x2,x2=x1)).reduce(G).subs(y1=c1,y2=c2,y3=c3,y4=c4)
    A=A.subs(c1=0,c2=u+1,c3=-u,c4=1).change_ring(GF(2));
    B=(D*D.subs(x1=x2,x2=x1)).reduce(G).subs(y1=c1,y2=c2,y3=c3,y4=c4)
    B=B.subs(c1=0,c2=u+1,c3=-u,c4=1).change_ring(GF(2));
    a=S1.reduce(G).subs(y1=c1, y2 = c2,y3=c3,y4=c4)
    a=a.subs(c1=0,c2=u+1,c3=-u,c4=1).change_ring(GF(2));
    b=S2.reduce(G).subs(y1=c1, y2 = c2,y3=c3,y4=c4)
    b=b.subs(c1=0,c2=u+1,c3=-u,c4=1).change_ring(GF(2));
    c=S3.reduce(G).subs(y1=c1, y2 = c2,y3=c3,y4=c4)
    c=c.subs(c1=0,c2=u+1,c3=-u,c4=1).change_ring(GF(2));
    show("A= ",A,",  B= ",B)
    show("a=",a,", b=",b,", c=",c)
    show((y^3+a*y^2+b*y+c).factor())

(c) As per (b), $D_f(y)=y^2+u^{2}y+u^5+u^3+u^2={(y+u)}^2+u^2(y+u)+u^5=u^4(Y^2+Y+u)$, where $Y=(y+u)∕u^2$.

Then the splitting field of $D_f(y)$ is $F(D(f))={𝔽}_2(u,\alpha )$, where ${\alpha }^2+\alpha +u=0$. Since $u={\alpha }^2+\alpha$, ${𝔽}_2(u,\alpha )={𝔽}_2(\alpha )$ and $g(y)=y^2+\mathit{uy}+1=y^2+({\alpha }^2+\alpha )y+1$. Then $g(y+1)={(y+1)}^2+({\alpha }^2+\alpha )(y+1)+1=y^2+({\alpha }^2+\alpha )y+({\alpha }^2+\alpha )$ is irreducible in ${𝔽}_2[\alpha ][y]$ (and in ${𝔽}_2(\alpha )[y]$) by the Schönemann-Eisenstein criterion. Therefore $g(y)$ is irreducible over the splitting field ${𝔽}_2(\alpha )$ of $D(f)$

This criterion applies here because $[{𝔽}_2(\alpha ):{𝔽}_2]=[{𝔽}_2(u,\alpha ):{𝔽}_2]=[{𝔽}_2(u,\alpha ):{𝔽}_2(u)][{𝔽}_2(u):{𝔽}_2]=\infty$ ( $u$ is a variable), therefore $\alpha$ is transcendental over ${𝔽}_2$, so that we may consider $\alpha$ as a variable $x$: ${𝔽}_2[\alpha ]\simeq {𝔽}_2[x]$. Hence ${𝔽}_2[\alpha ]$ is a PID, and $\alpha$ is a prime in ${𝔽}_2[\alpha ]$. Gauss’s Lemma and the Schönemann-Eisenstein Criterion apply here.

As an alternative proof of the irreducibility of $g(y)=y^2+\mathit{uy}+1$ over $F(D(f))$, without this criterion, it is sufficient to prove that $g(y)$ has no root in $F(D(f))={𝔽}_2(\alpha )$, where ${\alpha }^2+\alpha +u=0$ as above. Such a root is under the form $a(\alpha )∕b(\alpha )$, where $a(\alpha )$ and $b(\alpha )$ are relatively prime in the PID ${𝔽}_2[\alpha ]$. Then

hence $a(\alpha )\mid b^2(\alpha )$, with $a(\alpha ),b(\alpha )$ relatively prime, thus $a(\alpha )\mid 1$, $a(\alpha )=1$, and similarly $b(\alpha )=1$. But $g(1)=u\ne 0$, so $g$ has no root in $F(D(f))$.

Thus $y^2+\mathit{uy}+1$ is irreducible over the splitting field of $D_f(y)$, therefore $G_f\simeq D_8$ by Proposition 13.3.6. □