# Separable extensions

There are two ways to do great mathematics. The first is to be smarter than everybody else. The second way is to be stupider than everybody else — but persistent, Raoul Bott.

Let f(x) be a polynomial, f(x) = anxn + an-1xn-1 + ··· + a1x + a0 ∈ F[x], with degree(f(x)) = n. We have already demonstrated that there exists an extension K/F that contains all the roots of f.

Definition. If F is a field with f(x)∈ F[x] and the factorization of f(x) = c(x -α1)m1(x -α2)m2···(x -αk)mk, mk ≥ 1, we say that mi is the multiplicity of the root αi. If mi = 1, αi is said to be a simple root. Otherwise, if mi ≥ 2, αi is said to be a multiple or repeated root.

If all of the roots of f are simple, then we say that f is separable, e.g., x2-5x +6 (3 and 2 are simple roots), x2 -3x +2 (1 and 2 are simple roots). On the contrary, if f has at least one multiple root, f is said to be inseparable, e.g., x4 -4x3 + 4x2 (0 and 2 are multiple roots), (x-1)2(x - 2)2(x2 +1) (1 and 2 are multiple roots, i and -i are simple roots).

The derivative of a polynomial f(x) = anxn + an-1xn-1 + ··· + a0 = $\sum_{i=0}^n a_ix^i$, written by f’(x) is well defined for any field F in F[x]. It is defined as the polynomial nanxn-1 + (n-1)an-1xn-2 + ··· + a1 = $\sum_{i=1}^n ia_ix^{i-1}$ ∈ F[x] and degree(f’(x)) ≤ n-1.

Properties of the Derivate. Let f(x), g(x) ∈ F[x], a ∈ F. Then, we have the basic properties as the derivate of calculus,

1. (f(x) + g(x))’ = f’(x) + g’(x)
2. (af(x) + bg(x))’ = af’(x) + bg’(x)
3. (f(x)g(x))’ = f(x)g’(x) + g(x)f’(x)
4. ((f(g(x)))’ = f’(g(x))g’(x)

Bezout’s identity in F[x]. Let F be a field, and let f(x) and g(x) be polynomials in F[x]. If f and g are relatively prime, then there exist polynomials h(x) and k(x) in F[x] such that f(x)h(x) + g(x)k(x) = 1

Proof.

f and g are relatively prime ⇒ there is no polynomial of positive degree in F[x] that divides both f(x) and g(x).

F is a field ⇒ F[x] is a PID ⇒ [A PID is an integral domain in which every ideal is generated by a single element] So the ideal in F[x] generated by f and g, ⟨f(x), g(x)⟩ is generated by a single element, ⟨f(x), g(x)⟩ = a(x) for some a(x) ∈ F[x]. Thus a(x) divides both f(x) and g(x). This means that a(x) is a constant multiple of a unit, i.e. a constant polynomial, say a ∈ F, since these polynomials are coprime ⇒ it contains 1 (c·a ∈ ⟨f(x), g(x)⟩ for any element c ∈ F, in particular a^(-1)*a = 1), 1 ∈ ⟨f(x), g(x)⟩ and therefore there exist polynomials h(x) and k(x) in F[x] such that f(x)h(x) + g(x)k(x) = 1∎

Alternative definition. Let K be a splitting field of a polynomial f(x) ∈ F[x]. Let α be a root of this polynomial f(x), then (x -α) would be a factor of f(x), that is, (x - α) divides f(x) in K[x]. If (x - α)s divides f(x) in K[x], s > 1, α is a multiple root of f(x).

Theorem. Criterion for multiple zeros. Let f(x) be a polynomial over a field F (f(x) ∈ F[x]), f(x) has a multiple zero a (root with multiplicity greater than 1) in some extension E of F iff f(x) and f'(x) have a common factor of positive degree in F[x] ↭ f(a) = f’(a) = 0.

Example. f(x) = x2 -4x + 4. f(2) = 0, 2 is a zero of f(x) with multiplicity 2. f’(x) = 2x -4, f’(2) = 0. Notice that (x -2) is a common factor of f(x) and f’(x) because f(x) = (x -2)2 and f’(x) = 2(x -2).

Proof.

⇒) Let f(x) has a multiple zero “a” in some extension E/F ⇒ ∃g(x)∈E[x]: f(x) = (x -a)2g(x).

f’(x) = [Properties of the Derivate] (x -a)2g’(x) + 2(x -a)g(x) = (x -a)·[(x -a)g’(x) + 2g(x)] ⇒ (x -a) is a factor of both f(x) and f’(x) in some extension E of F ⇒ [Factor Theorem. A polynomial f(x) has a factor (x -α) if and only if α is a root, i.e., f(α) = 0.] f(a) = f’(a) = 0.

⇐) Suppose that f(x) and f’(x) have a common factor of positive degree (x -a) ↭ [Factor Theorem] Let a be a zero of the common factor, f(a) = f’(a) = 0.

a is a zero of f(x) ↭ (x - a) divides f(x) ⇒ ∃q(x): f(x) = (x -a)q(x). Then, f’(x) = (x -a)q’(x) + q(x) ⇒ 0 = f’(a) = q(a) ⇒ q(a) = 0 and x -a is a factor of q(x) and [f(x) = (x -a)q(x)] f(x) is divisible by (x -a)2 ⇒ “a” is a multiple zero of f(x).

Definition. A field F is called perfect if either has characteristic zero or when F has characteristic p > 0, every element of F is a pth power, i.e., Fp = {ap | a ∈ F} = F. ↭ Every irreducible polynomial over F is separable (has distinct roots), e.g., ℚ, ℂ, and $\mathbb{F_q}$. In particular, all fields of characteristic zero and all finite fields are perfect.

In a field K with characteristic p, the Frobenius map x → xp is a field homomorphism. Indeed, $(x + y)=\sum_{k=0}^p{p \choose k}x^py^{p-k} = x^p+y^p$ because all the intermediate terms are zero ($p | {p \choose k}$). A field homomorphism is necessarily injective, because ker(Φ) is a proper ideal of K, hence it is {0} or K (it would be the identity⊥).

If K = $\mathbb{F_p}$, i.e., a finite field, then the Frobenius map is surjective, because an injective function from a finite set to itself must be surjective. However, in a field such as K = $\mathbb{F_p(t)}$={$\frac{f}{g}| f, g ∈ \mathbb{F_p(t)},g ≠ 0$}, the field of rational functions with coefficients in the field with p elements, the Frobenius map is not surjective, e.g., t is not in the image of Φ, suppose $Φ(\frac{f}{g})=\frac{f^p}{g^p}=t ⇒$ p·(deg(f) - deg(g)) = deg(t) = 1 ⊥ def(f) - deg(g) is an integer and 1 is not divisible by p.

Recall. Freshman’s Dream or Exponentiation. Let R be a commutative ring with unity of characteristic p. Then, (a + b)p=ap + bp

Theorem. Every finite field is perfect.

Proof.

Suppose that F has characteristic p, then we claim that Φ: F → F defined by Φ(x) = xp (the Frobenius endomorphism) is a field isomorphism or automorphism.

1. Φ(a +b) = (a +b)p = [Freshman’s Dream or Exponentiation. For completeness’ sake, we repeat the argument] $a^p+(\begin{smallmatrix}p\\1\end{smallmatrix})a^{p-1}b+(\begin{smallmatrix}p\\2\end{smallmatrix})a^{p-2}b^2+···+(\begin{smallmatrix}p\\p-1\end{smallmatrix})ab^{p-1}+b^p = a^p+b^p$ = Φ(a)+Φ(b) because all other terms cancel. Notice that $(\begin{smallmatrix}p\\i\end{smallmatrix})=\frac{p!}{i!(p-i)!}$ are integers divisible by p (there is a p in the numerator, but not in the denominator and p is prime).
2. Φ(ab) = (ab)p=apbp = Φ(a)Φ(b).
3. Φ(a) = Φ(b) ⇒ ap = bp ⇒ [Φ distributes over addition and (-1)p = -1 in $\mathbb{F_p}$ for any prime p] (a -b)p = 0 ⇒ [F is a field] a - b = 0 ⇒ a = b, Φ is injective.
4. Φ is one-to-one homomorphism, and F is finite ⇒ Φ is onto ⇒ Fp = F.

Proposition. Let F be any field.

1. An irreducible polynomial f ∈ F[x] is separable iff the greatest common factor of f and f' is 1, it is expressed or written as (f, f') = 1, f is relatively prime to its derivate.
2. If F has characteristic zero, F is a finite field or perfect, then every polynomial in F[x] is separable.
3. If F has characteristic zero, F is a finite field or perfect, then any finite extension K/F is separable.

Proof.

Let K be a finite extension of F where both f and f’ split completely (e.g., let K be the splitting field of the polynomial f·f’), then (f, f’)= 1 ↭ [Criterion for multiple zeros.] f and f’ have no common roots in K.

(1) Suppose a ∈ K is a root of f ⇒ f(x) = (x -a)g(x) in K[x] ⇒ f’(x) = (x-a)g’(x) + g(x)

a ∈ K is a multiple root of f ↭ (x -a)2 divides f ↭ [f(x) = (x -a)g(x)] (x -a) divides g(x) ↭ (x -a) divides f’ ↭ a is a root of f'

f is separable ↭ [By definition] f does not have multiple roots in K ↭ f and f’ have no common roots in K ↭ [Criterion for multiple zeros] f and f’ are coprime in F[x]

(2) Let f be irreducible.

f is not separable ⇒ [An irreducible polynomial f ∈ F[x] is separable iff the greatest common factor of f and f' is 1, it is expressed or written as (f, f') = 1. ] (f, f’) ≠ 1 ⇒ ∃h: h = (f, f’), deg(h) > 0, h divides f and f’. Since f is irreducible, only 1 and f are possible ⇒ [deg(h)>0] h = f

Futhermore, since h divides f’, we are left with two options, f’ = 0 or deg(f’) ≥ deg(f) = deg(h) ⇒ [Differentiating a polynomial reduces the degree, deg(f’)< deg(f)] f’ = 0. In summary, for any field F, f ∈ F[x], f irreducible and not separable ⇒ f' = 0.

Example: F = $\mathbb{F_2}(t)$, f(x) = x2 -t ∈ F[x] is not separable ⇒ f’(x) = 2x = [char(F)=2] 0.

There are two options:

1. If char(F) = 0, f = anxn + an-1xn-1 + ··· + a0 ⇒ [n > 0 (f is irreducible so f cannot be constant -constants are units, so they cannot be irreducibles-), and an ≠ 0 ] nan ≠ 0 ⇒ f’ = annxn-1 +an-1(n-1)xn-2 + ··· + a1x = $\sum_{i=1}^n ia_ix^{i-1}$ ≠ 0 ⇒ every irreducible polynomial over a characteristic zero field is separable.

Alternatively, f’(x) = 0 only when iai = 0, i = 1, 2, ···, n ⇒ [char(F) = 0] ai = 0, i = 1, 2, ···, n ⇒ f(x) = a0, f is a constant, so it is not an irreducible polynomial over a field ⊥. Therefore, f(x) has no multiples zeros ↭ f and f’ are relatively prime ↭ f is separable.

2. Let’s suppose F is finite, char(F) = 0, or char(F) = p and every element of F is a pth power of some other element ⇒ The Frobenius map F → F, α → αp is surjective (∀β ∈ F, ∃α ∈ F: αp = β, i.e., α is a pth root of β) or, in other words, every element of F has a pth root.

f = anxn + an-1xn-1 + ··· + a0, n > 0, an ≠ 0

Suppose f’ = annxn-1 +an-1(n-1)xn-2 + ··· + a1x = 0 ⇒ nan = (n-1)an-1 = ··· = a1 = 0

[(aixi)’=(iaixi-1)] If ai≠0, then i = 0 in F ⇒ [char(F) = p] p divides i, e.g., char(F)=3, a15x15 + a12x12 + a9x9 + ··· + a3x3 + a0

Therefore, f must be of the form f = anxpbn + an-1xpbn-1 + ··· + a1xpb1 + a0 ⇒ [The Frobenius map is surjective, ∀β ∈ F, ∃α ∈ F: αp = β, let’s take a’i: (ai’)p = ai] f = $(a_n^‘x^{b_n})^{p}+(a_{n-1}^‘x^{b_{n-1}})^{p}+···+(a_{1}^‘x^{b_{1}})^{p}+(a’_0)^p$ = [···]

Consider Char(F) = p & the Binomial Theorem, $(a+b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k}b^k$ where ${n \choose k}$ denotes the binomial coefficient, which is given by: ${n \choose k} = \frac{n!}{k!(n-k)!}$

[···] = $(a_n^‘x^{b_n}+a_{n-1}^‘x^{b_{n-1}}+···+a_{1}^‘x^{b_{1}}+a’_0)^p = (g(x))^p$

Therefore, f(x) = (g(x))p where $g(x)=a_n^‘x^{b_n}+a_{n-1}^‘x^{b_{n-1}}+···+a_{1}^‘x^{b_{1}}+a’_0$, i.e., f cannot be irreducible ⊥.

(3) If F has characteristic zero, F is a finite field or perfect, then any finite extension K/F is separable.

∀ α ∈ K, let p(x) be its irreducible polynomial ⇒ [Char(F) = 0, F is a finite field or perfect, by (2)] p(x) is separable ⇒ K/F is separable. ∎

Theorem. If f(x) ∈ F[x] is irreducible over F, then all roots of f(x) have the same multiplicity.

Proof.

Let $\bar F$ be the algebraic closure of F. Let α and β be the roots of f(x) in $\bar F$ with multiplicity k and k’ respectively. We already know that $F(α)≈\frac{F[x]}{⟨f(x)⟩}≈F(β)$

Let’s define the isomorphism σ: F(α) → F(β), it is better understood from g(α) → g(x) + ⟨f(x)⟩ → g(β). More specifically, let a0 + a1α + ··· + anαn ∈ F(α) → a0 + a1x + ··· + anxn → a0 + a1β + ··· + anβn, Figure 2.b.

σ: F(α) → F(β), σ(α) = β is a mapping that is homomorphism, one-to-one, and onto. σ can be extended to an isomorphism σ* from $\bar F$ → $\bar F$ which induces a ring homomorphism μ: $\overline {F[x]} → \overline {F[x]}$ given by μ(a0 + a1x + ··· + arxr) = σ*(a0) + σ*(a1)x + ··· + σ*(ar)xr

μ acts as the identity on f(x), i.e., μ(f(x)) = f(x) ⇒ μ(x -α)k = [σ(α) = β] (x- β)k, then f(x) = (x- β)k··· ⇒ k ≤ k’ (k’ is β multiplicity). Mutatis mutandis, the same reasoning applies to beta respect to α ⇒ k’ ≤ k ⇒ k = k’.

Corollary. If f(x) ∈ F[x] is irreducible over F, then f(x) = a$\prod_{i=1}^r (x-α_i)^k$ where αi are the roots of f(x) in the splitting field over F, and k is the multiplicity of each and every root.

Proof. Let f(x) be irreducible polynomial over F. Let α1, α2, ···, αr be the roots of f(x). If α1 has multiplicity k ⇒ αi have multiplicity k ∀i, 1 ≤ i ≤ r ⇒ f(x) = a (x - α1)k(x - α2)k ··· (x - αr)k = a$\prod_{i=1}^r (x-α_i)^k$∎

Bitcoin donation

JustToThePoint Copyright © 2011 - 2024 Anawim. ALL RIGHTS RESERVED. Bilingual e-books, articles, and videos to help your child and your entire family succeed, develop a healthy lifestyle, and have a lot of fun. Social Issues, Join us.

This website uses cookies to improve your navigation experience.
By continuing, you are consenting to our use of cookies, in accordance with our Cookies Policy and Website Terms and Conditions of use.