Theorem 3.10.2 (E

Dec. 15, 2006

3.10. solving cubic polynomials. In this section, we are going to review classical result on solving polynomials by using non-classical language. I think this experience also serve a good start for Galois theory in general.

Definition 3.10.1. A character from a group G to a field K is group homomorphism χ : G → K^∗. The set of characters is denoted Hom_gp(G, K^∗).

Let Hom(G, K) be the set of functions from G to K. It’s clear that Hom(G, K) is a K-vector space.

Theorem 3.10.2 (E. Artin). Homgp(G, K^∗) is linearly independent in Hom(G, K).

Proof. Suppose on the contrary that Homgp(G, K^∗) is not linearly in- dependent. Pick a linearly dependent subset {χ1, ..., χn} of minimal n.

There are ai ∈ K such thatP

aiχi = 0, i.e.

Xa_iχ_i(g) = 0, (∗)

for all g ∈ G. We can rewrite it as

Xaiχi(gh) = 0, (∗∗)

for all g, h ∈ G. Multiply (∗) by χ₁(h), we get

Xa_iχ_i(g)χ₁(h) = 0. (∗ ∗ ∗)

Compare (∗) with (∗ ∗ ∗), we get

Xa_i(χ_i(h) − χ₁(h))χ_i(g) = 0 for all g ∈ G.

Thus P_n

i=2ai(χi(h) − χ1(h))χi = 0 ∈ Hom(G, K). It follows that the n − 1 elements {χ2, ..., χn} is linearly dependent, which is a contradic-

tion to the minimality. ¤

Corollary 3.10.3. Let F/K be an extension. The set of K-homomorphisms from F to K is linearly independent in the K-vector space of linear maps from F to K.

Sketch. Take G = F^∗. ¤

Let K be a field containing n-th root of unity ζ. Let F/K be a Galois extension with Galois group ∼= Zn generated by σ. We consider

ψ_ζ := 1 + ζσ + ζ²σ²+ ... + ζⁿ⁻¹σⁿ⁻¹ ∈ Hom(F, K).

Any element of the form ψ(x) is called a Lagrange resolvent.

By direct computation, we have the following properties.

Proposition 3.10.4. Keep the notation as above, we have:

1. σ(ψζ(x)) = ζ⁻¹ψζ(x).

2. ψ₁(x) ∈ K.

3. (ψ_ζ(x))ⁿ ∈ K.

4. (ψ_ζ(x))(ψ_ζ⁻¹(x)) ∈ K.

5. P

ζ∈µnζ^−rψ_ζ(x) = nσ^r(x).

Now we can use this technique to solve cubic equations. Let f (x) = x³ + px + q ∈ K[x] be an irreducible polynomial with discriminant D = −4p³− 27q² ∈ K. We assume that K contains a primitive 3-root of unity ζ. We have extension K ⊂ L := K[√

D] ⊂ F := K[u₁, u₂, u₃].

Note that F/L is Galois with Galois group ∼= Z3. Step 1. ψ_ζ 6= 0 ∈ Hom(F, K), in fact ψ_ζ(u₁) 6= 0.

Step 2. ψ_ζ(u₁) 6= L and (ψ_ζ(u₁))³ ∈ L, thus F = L[ψ_ζ(u₁)].

And similarly, ψ_ζ²(u1) ∈ L, (ψ_ζ²(u1))³ ∈ L. Moreover, ψζ(u1)ψ_ζ²(u1) ∈ L.

Step 3. Solve ψζ(u1),ψ_ζ²(u1) . Recall that

∆ := (u1−u2)(u2−u3)(u3−u1) = u²₁u2+u³₂u3+u²₃u1−u1u²₂−u2u²₃−u3u²₁. ψζ(u1)³ = u³₁+u³₂+u³₃+3ζ(u²₁u2+u²₂u3+u²₃u1)+ζ²(u1u²₂+u2u²₃+u3u²₁)+6u1u2u3. Let v₁ = u²₁u₂+ u²₂u₃+ u²₃u₁, v₂ = u₁u²₂+ u₂u²₃+ u₃u²₁, then

v₁+ v₂ = (u₁+ u₂+ u₃)(u₁u₂+ u₂u₃+ u₃u₁) − 3u₁u₂u₃ = 3q, v₁− v₂ = ∆.

Thus ψ_ζ(u₁)³ can be expressed in terms of p, q, ∆.

Step 4. solve u₁, u₂, u₃ in terms of ψ_ζ(u₁),ψ_ζ²(u₁).

By the property 5 above, we have

3u₁ = ψ₁(u₁) + ψ_ζ(u₁) + ψ_ζ²(u₁),

3u₂ = 3σ(u₁) = ψ₁(u₁) + ζ⁻¹ψ_ζ(u₁) + ζ⁻²ψ_ζ²(u₁), 3u₃ = 3σ²(u₁) = ψ₁(u₁) + ζ⁻²ψ_ζ(u₁) + ζ⁻¹ψ_ζ²(u₁).

And note that ψ1(u1) = 0. So one can solve cubic polynomial explicitly.

3.11. cyclic extension. The discussion in the previous section can be extended to a more general setting.

Definition 3.11.1. We say that an extension is cyclic (resp. abelian) if it’s algebraic Galois and Gal_F/K is cyclic (resp. abelian). An cyclic extension of order n is an cyclic extension whose Galois group is iso- morphic to Z_n.

The following theorem characterize cyclic extension except some ex- ceptional case.

Theorem 3.11.2. Suppose that char(K) = 0 or char(K) = p - n.

Suppose furthermore that there is a primitive n-th root of unity in K, say ζ. Then F/K is a cyclic of order n if and only if F = K(u) where u is a root of irreducible polynomial xⁿ− a ∈ K[x].

Before we get into the proof. Let’s consider the ”difference” be- tween u and σ(u) for σ ∈ Gal_F/K. Let F/K be a finite Galois exten- sion. Then in this circumstance, norm and trace (which we will define more generally later) are nothing but N_F/K(u) :=Q

σ∈GalF/Kσ(u) and T_F/K := P

σ∈GalF/Kσ(u). It’s easy to see that T (u − σ(u)) = 0 and N(u/σ(u)) = 1. The follows lemma says that the converse is also true for cyclic extension, which will play the central role in the study of cyclic extension.

Lemma 3.11.3. Let F/K be an cyclic extension with σ the generator of the Galois group.

(1) If T_F/K(u) = 0, then there exists an v ∈ F such that u = v − σ(v).

(2) (Hilbert’s Theorem 90) If N_F/K(u) = 1, then there exists an v ∈ F such that u = v/σ(v).

Proof of the Theorem 3.11.2. Let u be a root of xⁿ − a, then all the roots are uζⁱ for i = 0, ..., n − 1. Since ζ ∈ K. We can produce an element in Galois group by considering σ_i : u 7→ uζⁱ. Thus we have {σ_i}i=0,...,n−1 ⊂ Gal_KF . It’s clear that Gal_KF = {σ_i}i=0,...,n−1 =<

σ₁ >. Thus F = K(u) is a cyclic extension over K.

Conversely, suppose that F/K is a cyclic extension of order n. Since there is a primitive n-th root ζ ∈ K, one has N(ζ) = ζⁿ = 1. By the Lemma, there exist an v such that ζ = v/σ(v). Let u = v⁻¹, then σ(u) = ζu. Hence σ(uⁿ) = uⁿ∈ K. Therefore u satisfies xⁿ− a ∈ K[x]

for some a ∈ K.

Moreover, for uζⁱ and uζ^j, there is an automorphism sending uζⁱ to uζ^j. So they have the same minimal polynomial p(x) dividing xⁿ− a.

One the other hand, p(x) has n distinct roots uζⁱ for i = 0, ..., n − 1.

It follows that p(x) = xⁿ− a is irreducible. One has [K(u) : K] = n

and thus F = K(u). ¤

Theorem 3.11.4. Suppose that char(K) = p 6= 0. Then F/K is a cyclic extension of order n if and only if F = K(u), where u is a root of an irreducible polynomial x^p− x − a ∈ K[x].

Proof. The proof is parallel to the previous one.

Let u be a root of x^p − x − a, then all the roots are u + i for i = 0, ..., p − 1. It’s clear that F = K(ζ) is a cyclic extension over K with Galois group generated by σ such that σ(u) = u + 1.

Conversely, suppose that F/K is a cyclic extension of order n. One has T (1) = p = 0. By the Lemma, there exist an v such that 1 =

v − σ(v). Let u = −v, then σ(u) = u + 1. Hence σ(u^p) = u^p+ 1 and σ(u^p− u) = u^p− u. Therefore u satisfies x^p − x − a ∈ K[x] for some a ∈ K.

Moreover, for u + i and u + j, there is an automorphism sending uζⁱ to uζ^j. So they have the same minimal polynomial p(x) dividing x^p − x − a. One the other hand, p(x) has p distinct roots u + i for i = 0, ..., p − 1. It follows that p(x) = x^p− x − a is irreducible. One

has [K(u) : K] = n and thus F = K(u). ¤

It remains to define norm and trace, and prove the main lemma 3.11.3.

Definition 3.11.5. Let [F : K] be a finite separable extension. Let Σ be the set of K-embeddings of F into K. For any u ∈ F , we define the norm, denoted

N_F/K(u) := (Y

σ∈Σ

σ(u)).

Similarly, we define the trace as

T_F/K(u) := (Σσ∈Σσ(u)).

Example 3.11.6. If F/K is finite Galois extension, then the set of all K-embeddings of F is nothing but the Galois group of F (since F is normal). Therefore, N_F/K(u) = Q

σ∈GalF/Kσ(u) and T_F/K(u) = P

σ∈GalF/Kσ(u)

Proof of Lemma 3.11.3. We only prove that T (u) = 0 implies u = v − σ(v). The other implication is easy.

Step 1. Find an element z ∈ F with T (z) 6= 0. This is an immediate consequence of independency of automorphism.

Step 2. We normalize it to get w ∈ F with T (w) = 1. In fact, we take w := _{T (z)}^z .

Step 3. Let

v = uw + (u + σ(u))σ(w) + ... + (u + σ(u) + ... + σⁿ⁻²(u))σⁿ⁻²(w).

Then by direct computation and T (u) =P

σ(u) = 0, we are done.

For the norm, if N(u) = 1, then u 6= 0. Take

v = uy + uσ(u)σ(y) + ... + uσ(u)...σⁿ⁻¹(u)σⁿ⁻¹(y).

By independency of automorphism, there exist a y such that v is non- zero. One checks that u⁻¹v = σ(v). We are done. ¤ 3.12. radical extension.

Definition 3.12.1. F/K is said to be an radical extension if F = K(u₁, ..., u_n) such that for 1 ≤ i ≤ n, uⁿ_iⁱ ∈ K(u₁, ..., u_n−1).

For a polynomial f (x) ∈ K[x]. We say f (x) = 0 is solvable by radical if its splitting field E is contained in some radical extension.

Remark 3.12.2. In the definition, it’s not necessary that the splitting field itself is a radical extension over K.

The main observation is the following:

Proposition 3.12.3. Let F/K be a radical and Galois extension over K. Write F = K(u1, ..., un) such that for 1 ≤ i ≤ n, uⁿ_iⁱ ∈ K(u1, ..., un−1).

Let m = Q

ni and assume that char(K) - m. Suppose furthermore that K contains a primitive m-th root of unity. Then Gal_F/K is solvable.

Proof. Let K_i := K(u₁, ..., u_i). And let G_i = K_i⁰. One sees that K₁ is cyclic over K, hence Galois over K. Hence G₁ C G₀ = Gal_F/K. Consider next F/K₁ which is radical and Galois. Then K₂ is cyclic over K₁ and hence similarly, G₂C G₁. Therefore, we have a solvable series {e} = G_nC G_n−1C ... C G₀ = Gal_F/K with G_i−1/G_i cyclic. We

are done. ¤

One can actually generalize it to the following general setting:

Theorem 3.12.4. Let F/K be a radical extension, and K ⊂ E ⊂ F . Then Gal_E/K is solvable. As a consequence, if f (x) = 0 is solvable by radical, then Gf is solvable.

Proof. We first reduce to simpler situation.

Step 1. Let G = Gal_E/K and K₀ = G⁰. It’s clear that F/K₀ is radical, and E/K₀ is Galois for Gal_E/K0 = G⁰⁰ = G and G⁰⁰⁰ = G⁰. Thus F/K₀ is radical and E/K₀ is Galois with Galois group Gal_E/K.

We thus replacing K by K₀ and assume that E/K is Galois.

Step 2. Reduce to the case that E = F/K is Galois. To see this, let σ : F → K be an K-embedding. One can show that σ(F ) is again a radical extension. One can also prove that if F₁, F₂ ⊂ K are radical extension over K, then F₁F₂ is a radical extension over K. Hence let N be the compositum of σ(F ) for all σ. It follows that N is radical over K. Moreover, N is normal over K.

Since E/K is Galois, in particular, E is normal over K and E is a stable intermediate subfield of N/K. Then one has a homomorphism Gal_N/K → Gal_E/K. This is surjective because N is normal. Thus it suffices to prove that Gal_N/K is solvable.

Step 3. By the same trick an in Step 1. We may assume that N/K is Galois. Therefore, it suffices to show that if F/K is Galois and radical, then Gal_F/K is solvable.

Step 4. Since F/K is separable, we may assume that (char(K), ni) = 1. Let m =Q

ni.

Let ζ be a primitive m-th root of unity. We claim that F (ζ) is Galois over K. Grant this for the time being, then F (ζ) is Galois over K(ζ) and K(ζ)⁰ C Gal_{F (ζ)/K}. Moreover, Gal_{F (ζ)/K}/K(ζ)⁰ ∼= GalK(ζ)/K. By Proposition 3.12.3, K(ζ)⁰ is solvable. K(ζ)/K is cyclotomic, hence Gal_K(ζ)/K is solvable. Thus, Gal_{F (ζ)/K} is solvable.

Now F/K is Galois, GalF/K ∼= GalF (ζ)/K/F⁰ which is solvable.

Step 5. To prove the claim, suppose that F is a splitting field of separable polynomial f₁, .., f_n ∈ K[x]. Then F (ζ) is nothing but a splitting field of separable polynomials f₁, ..., f_n, x^m − 1. Thus we are

done. ¤

Theorem 3.12.5. Let E be a finite dimensional Galois extension over K with solvable Galois group. Assume that char(K) - [E : K], then there is a radical extension F/K containing E.

Proof. We prover by induction on [E : K]. Let n = [E : K] and assume the theorem is true for all Galois extension of degree < n.

Let ζ be a primitive n-th root of unity. Then E(ζ)/K(ζ) is Galois.

If [E(ζ) : K(ζ)] < n then we are done by induction hypothesis and the fact that K(ζ)/K is radical.

By replacing E, K by E(ζ), K(ζ) respectively, we my assume that K has m-th root of unity.

Gal_E/K is solvable, let H be a subgroup of index q, for some prime q. Then H⁰/K is a cyclic extension, hence a radical extension. By induction hypothesis, E/H⁰ is radical. We are done. ¤ Corollary 3.12.6. Let f (x) ∈ K[x] be a polynomial of degree n > 0.

Suppose that char(K) - n!, then f (x) = 0 is solvable by radical if and only if G_f is solvable.