Exploration 8: Galois theory

• If $F(\alpha)$ is degree $1$, its minimal polynomial is $x-\alpha$.
• We can show that $F[x]/(x-\alpha)\iso F$ using the first ring isoomorphism theorem.
  • The evaluation map at $\alpha$, $\varphi_\alpha:F[x]\to F$, has a kernel consisting of all polynomials with $\alpha$ as a root. In other words, all polynomials that have the factor $x-\alpha$, which is just the ideal $(x-\alpha)$.
  • Since the evaluation map is surjective, $\im\varphi_\alpha=F$.
  • The theorem states that $F[x]/\ker\varphi_\alpha\iso\im\varphi_\alpha$. In other words, $F[x]/(x-\alpha)\iso F$.

• By definition, quadratic extensions are extensions of degree $2$. Since finite field extensions are quotients by a suitable irreducible polynomial of the same degree, the quotient $F[x]/(x^2+bx+c)$ can be used to represent any degree $2$ extension.
• You can always find the roots of $x^2+bx+c$ using the quadratic formula $\frac12(-b\pm\sqrt{b^2-4c})$.
  • Since $2$ is in the denominator, this requires the field to not be of characteristic $2$ (otherwise $2=0$ and you can’t divide by zero).
• This means every root adjoined by quotienting by $(x^2+bx+c)$ can be expressed in terms of elements of $F$ together with $\sqrt{b^2-4c}$. Since $b^2-4c\in F$, this means we need only adjoin a square root of an element in $F$ to add every root of $x^2+bx+c$, implying that $F[x]/(x^2+bx+c)\iso F(\sqrt{b^2-4c})$.
• The minimal polynomial of $\sqrt{b^2-4c}$ is $x^2-(b^2-4c)$.
• Therefore, every quadratic extension $F[x]/(x^2+bx+c)$ is isomorphic to the quadratic extension $F[x]/(x^2-(b^2-4c))$. If we let $\delta=b^2-4c$, then this becomes $F[x]/(x^2-\delta)$, which is isomorphic to $F(\sqrt{\delta})$.

• First of all it is clear that all primitive $n$th roots of unity $\zeta_n$ share the same minimal polynomial ($\Phi_n$), and are therefore roots of $\Phi_n$.
• Now we need to show these are the only roots of $\Phi_n$. To do this, we make a degree argument.
• First, note that the polynomial $x^n-1$ contains all $n$th roots of unity, not just the primitive ones. Since every $n$th root of unity is a primitive $d$th root of unity for some divisor $d$ of $n$, and $x^n-1$ (by virtue of containing all $n$th roots of unity) must contain all these primitive $d$th roots of unity for each $d$, we can conclude that each $\Phi_d$ is a factor of $x^n-1$. This includes $\Phi_n$.
• Second, note that these $\Phi_d$ are disjoint. If $\prod_{d_1}$ and $\prod_{d_2}$ share a root $\zeta$, then by definition both of those polynomials must be the minimal polynomial of $\zeta$. But minimal polynomials are unique, which implies $\prod_{d_1}=\prod_{d_2}$.
• This proves the identity $x^n-1=\prod_{d\mid n}\Phi_d$.
• Finally, note that if any root $\zeta$ of $\Phi_n$ is not a primitive $n$th root of unity, then it must be a primitive $d$th root of unity for some $d\mid n$. But then $\Phi_d$ and $\Phi_n$ would share a root $\zeta$, which contradicts the fact that each $\Phi_d$ is disjoint. Thus the only roots of $\Phi_n$ are all the primitive $n$th roots of unity $\zeta_n$.

• If $K/F$ is normal, then by definition $K$ is the splitting field for some polynomial $f$ over $F$.
• Any polynomial $g$ irreducible over $F$ with a root in $K$ is a factor of $f$, since $K$ (being the splitting field of $f$) adds only roots of $f$ and nothing more.
• But since $f$ splits in its splitting field $K$, any factor of $f$ splits in $K$, therefore $g$ splits in $K$.

By the previous theorem, if $K$ contains a root $\alpha$, then it splits its minimal polynomial $f$. Therefore it contains all the roots of $f$, which are the conjugate roots of $\alpha$.

• All automorphisms fix the identities $0$ and $1$. So to prove the fixed elements of $\sigma$ form a subfield of $K$, we just check that they contain all additive inverses, all nonzero multiplicative inverses, and are closed under addition and multiplication.
  • Inverses: $\sigma(-a)=-\sigma(a)=-a$ and for nonzero $a$, $\sigma(a^{-1})=\sigma(a)^{-1}=a^{-1}$
  • Closure: $\sigma(a+b)=\sigma(a)+\sigma(b)=a+b$ and $\sigma(ab)=\sigma(a)\sigma(b)=ab$
• Since all $F$-automorphisms fix $F$ by definition, $F$ is included in the fixed elements, which we just proved is a subfield of $K$. That means the fixed elements must form an intermediate field of $K/F$.

We have already proved this for the case that $K$ is a simple extension $F(\alpha)$ of $F$. Since the Primitive Element Theorem guarantees that all finite separable extensions are simple, the proof extends to all finite separable extensions.

• First of all, since each of $\sigma,\tau,\sigma^{-1}$ fix $F$, the composition fixes $F$ as well.
• Second, since each of $\sigma,\tau,\sigma^{-1}$ are automorphisms in a normal extension, they represent permutations on the roots of polynomials.
• Permutations compose, so the composition also permutes the roots of polynomials, and is thus a $F$-automorphism of $K/F$.

• The order of $G(K/F)$ is precisely the number of $F$-automorphisms of $K$.
• Recall that a Galois extension is finite separable, and therefore has a primitive element $\alpha$ (which is a root). Since every root adjoined to $K$ can be expressed in terms of the primitive element $\alpha$, each $F$-automorphism $\sigma$ is characterized by where it sends $\alpha$.
• Since $K/F$ is normal, every root of $f$ is in $K$ and can be mapped to (i.e. are candidates for $\sigma(\alpha)$). Since $K/F$ is separable, these roots are distinct. Thus there are $\deg f$ possible $F$-automorphisms, one for each of the $\deg f$ distinct roots of $f$.
• In other words, the order of the Galois group $|G(K/F)|$ is equal to the degree of $f$, which is the degree of $K/F$ by definition.

If $\alpha$ and $\beta$ are roots of $f$, then $F(\alpha)$ and $F(\beta)$ are isomorphic. But this isomorphism is an $F$-automorphism in the splitting field $K$, since it fixes $F$ and swaps $\alpha$ and $\beta$. Therefore there is indeed a $F$-automorphism in $G(K/F)$ that maps arbitrary $\alpha$ to arbitrary $\beta$.

If $K/F$ is Galois of degree $p$, its Galois group $G(K/F)$ is also of order $p$, and therefore cyclic.

• Lemma 1: If $K/F$ is normal, then for any intermediate field $L$, $K/L$ is normal.

  $K/F$, being normal, is the splitting field for some polynomial $f$ over $F$. Since it is a splitting field, $K$ is already the minimal field containing $F$ together with the roots of $f$. Then $K$ is also the minimal field containing any intermediate field $L$ of $K/F$ together with the roots of $f$, meaning it is also the splitting field for $f$ over $L$, and thus $K/L$ is normal.

• Lemma 2: If $K/F$ is separable, then for any intermediate field $L$, $K/L$ is separable.

  • Recall that an extension $K/F$ is separable if every element of $K$ is the root of a separable polynomial over $F$. Recall that a separable polynomial has no repeated roots.
  • Every minimal polynomial $g$ over $L$ is potentially of lesser degree than the corresponding minimal polynomial $f$ over $F$, due to $L$ containing more roots than $F$. Thus, $g$ divides $f$ over $L$.
  • That means every root of $g$ is a root of $f$. That implies that $f\in F[x]$ is has distinct roots, $g\in L[x]$ has distinct roots too. So separable polynomials in $F[x]$ are also separable polynomials in $L[x]$.
  • $K/F$ is separable, so every element of $K$ is a root of some separable polynomial over $F$, which we just showed are also separable polynomials over $L$, implying that $K/L$ is separable.
• A Galois extension is one that is normal and separable. If $K/F$ is normal and separable, then $K/L$ is normal by Lemma 1, and separable by Lemma 2, thus Galois.

• Quadratic extensions are always expressible by adjoining a square root. Thus they are simple extensions, and simple extensions are finite and separable.
• Degree $2$ is the minimum degree required to split an irreducible quadratic polynomial. Since every irreducible with a root in the quadratic extension must split in the extension, quadratic extensions are normal.
• So quadratic extensions are finite separable and normal, meaning they’re Galois extensions.

• ($\to$) If $K/F$ is Galois of degree $p$, then $K=F(\alpha)$ for some $p$th root $\alpha$ of an element in $F$.
  • Since $K/F$ is Galois of degree $p$, its Galois group $G(K/F)$ is cyclic with a generator $\sigma$.
  • Since $p\ne 1$, $\sigma$ is not the identity. Therefore there must be some element $\alpha\in K$ not fixed by $\sigma$. Let $f$ be the minimal polynomial of $\alpha$.
  • The orbit of $\sigma$ under $G(K/F)$ contains all the roots of $f$, since $G(K/F)$ must act transitively on the roots of $f$. Since $K/F$ is separable, these roots are distinct, and therefore there are $p$ roots.
  • As $f$ is a degree $p$ irreducible polynomial in a field $F$ containing the $p$th roots of unity, it must be the polynomial $x^p-a$ for some $a\in F$. This implies $\alpha$ is a $p$th root.
  • Since every root is expressible in terms of $\alpha$, $F(\alpha)$ is a splitting field of $f$. But since $K/F$ is normal, $K$ is also a splitting field of $f$ containing the exact same roots. Thus they must be the same field: $K=F(\alpha)$ (not merely an isomorphism)
  • Thus $K$ is obtained by adjoining a $p$th root.
• ($\from$) If $K=F(\alpha)$ for some $p$th root $\alpha$ of an element $a$ in $F$, then $K/F$ is Galois of degree $p$.
  • $\alpha$ is a root of its minimal polynomial $x^p-a$, and therefore is a $p$th root of $a$. We can write the $p$th roots of $a$ as $\zeta_p^k\alpha$, and they are all in $K$ since we assume $\zeta_p\in F$. Thus $x^p-a$ splits in $K$.
  • The derivative of $x^p-a$ is $px^{p-1}$. Assuming we’re not in characteristic $p$, $x^p-a$ shares no roots with its derivative and is therefore separable.
  • Any polynomial with a root in $K$ can express that root in terms of $\alpha$, since $K=F(\alpha)$ is simple. Since the minimal polynomial of $\alpha$ splits in $K$ into distinct roots, any polynomial with a root in $K$ must split in $K$ as well into distinct roots. Therefore, $K/F$ is normal and separable and therefore Galois. Since $x^p-a$ is the minimal polynomial of $\alpha$ and is of degree $p$, $K/F$ is of degree $p$.

• Since $F_1$ and $F_2$ are fields and therefore integral domains, their subring $F_1\cap F_2$ is also an integral domain.
• All that is left is to show every nonzero element of $F_1\cap F_2$ has a multiplicative inverse. But since $F_1$ and $F_2$ both have the same multiplicative inverses for all their nonzero elements, so does $F_1\cap F_2$.

We already know that a single $F$-automorphism defines an intermediate field of $K/F$. A subgroup containing multiple $F$-automorphisms will fix elements $K^H$ corresponding to the intersection of the fixed elements of each individual $F$-automorphism. Being an intersection of subfields of $K$, $K^H$ must be a subfield of each of those subfields, and is therefore also a subfield of $K$ containing $F$ (i.e. an intermediate field of $K/F$.)

• First, for every intermediate field $L$, we can identify the permutations of $K$ that fix $L$, which is the Galois group $G(K/L)$. Call this map $\sigma=L\mapsto G(K/L)$.

• Second, for every subgroup $H\le G(K/F)$, we can identify the elements of $K$ fixed by every permutation in $H$, which is its fixed field $K^H$. Call this map $\tau=H\mapsto K^H$.

• The theorem states that $\sigma$ and $\tau$ are inverses. In other words, the subset relations below are actually equalities:

• This essentially only requires us to prove two things:

• For $\sigma=L\mapsto G(K/L)$: we must have $[K:L]=|G(K/L)|$.

  • Let $G=G(K/L)$. Recall that $G$ is exactly all the automorphisms of $K$ that fix $L$. This is a subgroup of $G(K/F)$.
  • Since $K/L$ is finite and separable, it has a primitive element $\alpha$. Let $f$ be its minimal polynomial.
  • Apply the orbit-stabilizer theorem on the action of $G$ on the roots of $f$. The orbit-stabilizer theorem states that the size of the orbit of $\alpha$ is equal to the size of the group divided by the size of the stabilizer of $\alpha$: $|O_\alpha|=|G|/|G_\alpha|$. Choosing $\alpha$ as a primitive element makes it so its stabilizer $G_\alpha$ is trivial, since if an automorphism fixes $\alpha$, it fixes the whole extension.
  • We know that $|G_\alpha|=1$, therefore the order of $G$ matches the order of the orbit of $\alpha$. We just need to show that there are $[K:L]$ elements in the orbit of $\alpha$.
  • Since the minimal polynomial of $\alpha$ is irreducible over $L$, its Galois group $G$ acts transitively on its roots. Therefore all of its roots are in the orbit of $\alpha$. Since $K/L$ is separable, the roots $\alpha$ are distinct, meaning the number of roots is equal to the degree $[K:L]$ of the extension.

• For $\tau=H\mapsto K^H$: we must have $|H|=[K:K^H]$.

  • Since $K^H$ is an intermediate field of $K/F$, we know $K/K^H$ is Galois, and therefore $|G(K/K^H)|=[K:K^H]$.
  • The automorphisms of $K$ that fix the fixed group of $H$ are exactly $H$, so $G(K/K^H)=H$ by definition. Therefore $|H|=[K:K^H]$.