Algebra, Artin

Created: October 30, 2024
Updated: November 08, 2024

A textbook written by Michael Artin (son of Emil Artin!) on algebra.

Chapter 15, “Fields”

  • Examples of Fields
    • Definition: $F \subset K$ fields, $K/F$ is a field extension
    • Definition: A number field is a subfield of $\mathbb C$
    • Definition: Algebraic number field, a number field where all elements are algebraic over $\mathbb Q$
    • Definition: Finite fields contain a finite amount of elements
    • Remark: Any finite field contains contains a copy of $\mathbb F _ p$ and so is a field extension of $\mathbb F _ p$
    • Definition: Extensions of $F = \mathbb C(t)$ (rational functions over $\mathbb C$) are called function fields
    • Remark: Function fields can be specified implicitly in terms of an equation $f(t, x) = 0$ where $f$ is an irreducible complex polynomial in the variables of $t$ and $x$
    • Example: $f(t, x) = x^2 - t^3 + t$ leads to the function field $p + q \sqrt{t^3 - t}$ where $p$ and $q$ are rational functions in $t$
  • Algebraic and transcendental elements
    • Definition: If $\alpha \in K$ and $K/F$, then $\alpha$ is algebraic over $F$ if $\alpha$ is the root of a monic polynomial with coefficients in $F$.
    • Definition: If $\alpha$ is not algebraic over $F$, then it is transcendental
    • Remark: Consider the substitution homomorphism $\phi : F[x] \to K$ defined by $x \mapsto \alpha$. If $\phi$ is injective, then $\alpha$ is transcendental. Otherwise, the kernel is nonzero and $\alpha$ is algebraic.
    • Remark: Since $F[x]$ is a principal ideal domain, the kernel of $\phi$ is generated by a single element $f$.
    • Proposition: The generator for the kernel is the minimal polynomial, and the minimal polynomial has all the properties you would expect
    • Definition: $F(\alpha)$ is the smallest subfield of $K$ that contains $F$ and $\alpha$
    • Definition: $F[\alpha]$ is the ring generated by $\alpha$ over $F$
    • Remark: $F(\alpha)$ is isomorphic to the field of fractions of $F[\alpha]$.
    • Remark: If $\alpha$ is transcendental over $F$, then $F[x] \to F[\alpha]$ is an isomorphism and $F(\alpha)$ is isomorphic to the field $F(x)$ of rational functions. In this case, every field extension is isomorphic.
    • Proposition: $F[x]/\langle f \rangle \cong F[\alpha]$.
    • Proposition: If $\alpha _ 1, \ldots, \alpha _ k$ are algebraic, $F[\alpha _ 1, \ldots, \alpha _ k]$ is equal to the field $F(\alpha _ 1, \ldots, \alpha _ k)$.
    • Proposition: Degree of minimal polynomial is equal to degree of field extension and powers of algebraic number give a basis
    • Proposition: If $\alpha$ and $\beta$ are algebraic over $F$, then $F(\alpha)$ and $F(\beta)$ are isomorphic by the map that sends $\alpha$ to $\beta$ and is the identity on $F$ iff the irreducible polynomials $\alpha$ and $\beta$ over $F$ are equal
    • F: Definition: Suppose $K$ and $K’$ are extensions of the same field $F$, an isomorphism $\phi : K \to K’$ that restricts to the identity on the subfield $F$ is called an $F$-isomorphism, and $K$ and $K’$ are isomorphic extension fields
    • Proposition: Isomorphisms of field extensions map roots to roots
  • The degree of a field extension:
    • Definition: Degree of a field extension
    • Lemma: $F/K$ has degree 1 iff $F = K$
    • Lemma: Element $\alpha$ of a field extension $K$ has degree $1$ over $F$ iff $\alpha$ is an element of $F$
    • Proposition: If $F$ is a field where $\text{char }F \ne 2$, you can obtain any extension $K$ by adjoining a square root, and if the field is missing a square root, adding it gives a degree 2 field extension
    • Remark: Not all cubic extensions can be obtained by adjoining a cube root
    • Proposition: $F(\alpha)$ degree of $\alpha$ over $F$ is the degree of the field extension
    • Proposition: $\alpha$ is algebraic iff $F(\alpha) : F$ is finite
    • Proposition: Tower law
    • Proposition: If $K/F$ is a field extension and $\alpha \in K$, if $\alpha$ is algebraic over $F$ then its degree over $F$ divides $n$.
    • Proposition: Degree of intermediate field can’t be more than the degree of the outside field
    • Proposition: Every field extension generated by finitely many algebraic elements is a finite extension. A finite extension is generated by finitely many elements
    • Proposition: If $K/F$ is a field extension, the set of algebraic elements is a subfield
    • Corollary: If $K/F$ is a field extension of prime degree, then if $\alpha$ is not in $F$, the degree of the field extension is $p$
    • Corollary: Let $\mathcal K$ be an extension field of a field $F$, let $K$ and $F’$ be subfields of $\mathcal K$ that are finite extensions of $F$, and let $K’$ denote the subfield of $\mathcal K$ generated by the two fields $K$ and $F’$ together. Then if $[K’ : F] = N$, $[K : F] = m$, and $[F’ : F] = n$. Then $m$ and $n$ divide $N$, and $N \le mn$.
    • Example: $[\mathbb Q (\alpha _ 1, \alpha _ 2, \alpha _ 3) : \mathbb Q] = 6$ where $\alpha _ i$ are the roots of $x^3 - 2$.
    • Example: If $\alpha = \sqrt[3]{2}$ and $\beta$ is a root of $x^4 + x + 1$, since $3$ and $4$ are relatively prime, $[\mathbb Q(\alpha, \beta) : \mathbb Q] = 12$
    • Example: Computing powers of $\gamma = \sqrt 2 + \sqrt 3$ and spotting a linear dependence gives you the minimal polynomial

Chapter 16, “Galois Theory”

  • Symmetric functions
    • Definition: symmetric polynomials are the polynomials that are invariant under permutations of the variables $f(u _ 1, \ldots, u _ n) \mapsto f(u _ {1\sigma}, \ldots, u _ {n\sigma})$.
    • Remark: The symmetric polynomials form a subring of the polynomial ring $R[u]$.
    • Definition: $\sigma$ defines an automorphism of $R[u]$, and since it is the identity on constant polynomials, it’s called an $R$-automorphism.
    • Remark: $S _ n$ then acts on $R[u]$ by $R$-automorphisms.
    • Remark: A polynomial is symmetric if the coefficients of monomials in the same orbit have the same coefficient.
    • Definition: Orbit sum: the sum of all polynomials in the same orbit, these form a basis of symmetric polynomials: $1$, $u _ 1 + u _ 2 + u _ 3$, $u^2 _ 1 + u^2 _ 2 + u^2 _ 3$, $u _ 1 u _ 2 + u _ 1 u _ 3 + u _ 2 u _ 3$, and so on.
    • Definition: Elementary symmetric functions are special symmetric polynomials defined by
      • $s _ 1 = \sum _ i u _ i = u _ 1$
      • $s _ 2 = \sum _ {i < j} u _ i u _ j = u _ 1 u _ 2 + u _ 1 u _ 3 + \cdots$
      • $s _ 3 = \sum _ {i < j < k} u _ i u _ j u _ k = u _ 1 u _ 2 u _ 3 + \cdots$
      • …and so on.
    • Remark: Elementary symmetric functions are the coefficients of the polynomial with roots $u _ 1, \ldots, u _ n$ up to sign, e.g. $P(x) = (x - u _ 1)(x - u _ 2) = x^2 - (u _ 1 + u _ 2)x + (u _ 1 u _ 2)$.
    • Lemma: Can obtain the coefficients of a polynomial by evaluating the roots on the elementary symmetric functions, e.g. Vieta’s formulas
    • Theorem: Symmetric functions theorem: Every symmetric polynomial $g(u _ 1, \ldots, u _ n)$ with coefficients in a ring $R$ can be written in a unique way as a polynomial in the elementary functions (not the same as being a basis)
      • This explains some of the A-level “roots of polynomials” topic, where you’d be asked to calculate e.g. $\alpha^2 + \beta^2$ just from the coefficients $(x-\alpha)(x-\beta) = x^2 - Bx + C$. $\alpha^2 + \beta^2$ is a symmetric polynomial, but it is not an elementary symmetric function.
      • This result says that it’s always possible to find an expression where any symmetric polynomial in the roots can be written uniquely as a polynomial in the elementary symmetric functions, so if you know the roots, you also know how to calculate any symmetric function of the roots
    • Examples:
      • $u^2 _ 1 + \cdots + u^2 _ n = s _ 1^2 - 2s _ 2$
      • $u _ 1 u _ 2^2 + u _ 2 u _ 1^2 + u _ 1 u _ 3^2 + u _ 3 u _ 1^2 + u _ 2 u _ 3^2 + u _ 3 u _ 2^2 = s _ 1 s _ 2 - 3 s _ 3$
    • Corollary: Suppose $K/F$ is a splitting field for some polynomial $f \in F[x]$ with roots $\alpha _ 1, \ldots, \alpha _ n \in K$. Then if $g$ is a symmetric polynomial, $g(\alpha _ 1, \ldots, \alpha _ n) \in K$.
    • Proposition: Can build symmetric polynomials out of symmetric polynomials where we substitute the orbit of any polynomial under $S _ n$ for the variables
  • The discriminant
    • Definition: The discriminant is the product of the squares of the differences of the roots, i.e. if $u _ 1, \ldots, u _ n$ are the roots, then $D(u) = \prod _ {i < j} (u _ i - u _ j)^2$.
    • Remark: $D(u)$ is a symmetric polynomial with integer coefficients
    • Remark: If $\alpha _ 1, \ldots, \alpha _ n$ are elements of a field, then $D(\alpha) = 0$ iff two of the elements are equal
    • Remark: The symmetric function theorem says that the discriminant can always be as a polynomial of the elementary symmetric functions
    • Remark: For roots $u _ 1, u _ 2$, we have $D = (u _ 1 - u _ 2)^2 = s _ 1^2 - 4s _ 2$, which is the normal formula for the discriminant of a quadratic, and it’s the difference between the two roots!
    • Remark: $D$ gets very complicated for large $n$
  • Splitting fields
    • Definition: A splitting field for $f$ is extension field $K/F$ such that:
      • $f$ splits completely in $K$
      • $K$ is generated by the roots
    • Lemma: If $F \subseteq L \subseteq K$ are fields and $K$ is a splitting field for some polynomial over $F$, it is also a splitting field for the same polynomial over $L$
    • Lemma: Every polynomial $f(x)$ in $F[x]$ has a splitting field.
    • Lemma: A splitting field is a finite extension of $F$, every finite extension is contained in a splitting field
    • Theorem: Splitting theorem: Let $K$ be an extension of a field $F$ that is a splitting field of a polynomial $f(x)$ with coefficients in $F$. If an irreducible polynomial $g(x)$ with coefficients in $F$ has one root in $K$, then it splits completely in $K$.
  • Isomorphisms of field extensions:
    • Definition: $K$ and $K’$ extension fields of $F$. An $F$-isomorphism $\sigma : K \to K’$ is an isomorphism whose restriction to the subfield $F$ is the identity map.
    • Definition: The $F$-automorphisms of a finite extension $K$ form a group called the Galois group of $K$ over $F$, denotes $G(K/F)$.
    • (this is slightly different terminology to [[Course - Galois Theory HT25]]U, where these are referred to as the $F$-linear automorphisms of $K$, rather than the $F$-automorphisms of $K$).
    • Definition: A finite extension $K/F$ is a Galois extension if $ \vert G(K/F) \vert = [K : F]$ (again, slightly different but equivalent definition is given in [[Course - Galois Theory HT25]]U, namely that there exists a non-constant polynomial $f \in F[x]$ such that $K$ is the splitting field of $f$).
    • Example: $G(\mathbb C/\mathbb R) \cong \mathbb Z _ 2$, and any other quadratic extension is analogous, e.g. $G(\mathbb Q(\sqrt 2) / \mathbb Q)$.
    • Lemma: If $K$ and $K’$ are extensions of a field $F$ and $\sigma$ is an $F$-isomorphism between them. Then if $\alpha$ is a root of $f$ in $K$, then $\sigma(\alpha)$ is a root of $f$ in $K’$.
    • Lemma: Suppose $K$ is generated over $F$ by elements $\alpha _ 1, \ldots, \alpha _ n$. If $\sigma$ fixes all the generators, it’s the identity.
    • Lemma: Let $f$ be irreducible with coefficients in $F$ and $\alpha$, $\alpha’$ are roots of $f$ in $K$ and $K’$ respectively. Then there’s a unique $F$-isomorphism $\sigma : F(\alpha) \to F(\alpha’)$ that sends $\alpha$ to $\alpha’$. If $F(\alpha) = F(\alpha’)$, then $\sigma$ is an $F$-automorphism.
    • Proposition: Suppose $f$ is a polynomial with coefficients in $F$, any extension field $L/F$ contains at most one splitting field of $f$ over $F$ (and possibly zero)
    • Proposition: Any two splitting fields for $f$ over $F$ are isomorphic.
  • Fixed fields
    • Definition: Let $H$ be a group of automorphisms of a field $K$. The fixed field of $H$, denoted $K^H$ is $\{\alpha \in K \mid \sigma(\alpha) = \alpha \text{ for all } \sigma \in H\}$.
    • Proposition: $K^H$ is a subfield of $K$.
    • Proposition: $H$ is a subgroup of the Galois group $G(K/K^H)$.
    • Theorem: Let $H$ be a finite group of automorphisms of a field $K$ and let $F$ denote the fixed field $K^H$. Let $\beta _ 1$ be an element of $K$, and let $\{\beta _ 1, \ldots, \beta _ r\}$ be the $H$-orbit of $B _ 1$. Then:
      • The irreducible polynomial for $\beta _ 1$ over $F$ is $g(x) = (x - \beta _ 1) \ldots (x - \beta _ r)$.
      • $\beta _ 1$ is algebraic over $F$ and its degree over $F$ is equal to the order of its orbit, so the degree of $\beta _ 1$ over $F$ divides the order of $H$.
    • Definition: An extension field $K/F$ is called algebraic if every element of $K$ is algebraic over $F$.
    • Lemma: Let $K / F$ be algebraic that is not a finite extension of $F$. There exist elements in $K$ whose degrees over $F$ are arbitrarily large.
    • Theorem: Fixed Field Theorem: Let $H$ be a finite group of automorphisms of a field $K$, and let $F = K^H$ be its fixed field. Then $K$ is a finite extension of $F$, and its degree $[K : F]$ is equal to the order $ \vert H \vert $ of the group.
    • Example: Let $K = \mathbb C(t)$ and let $\sigma, \tau$ be the automorphisms of $K$ that are the identity on $\mathbb C$ and $t \mapsto it$ and $\tau \mapsto t^{-1}$. Then $\sigma$ and $\tau$ generate the group of automorphisms $H$ that are isomorphic to the dihedral group.
    • Lemma: $u = t^4 + t^{-4}$ is transcendental over $\mathbb C$
    • Theorem: Lüroth’s theorem: Let $F$ be a subfield of $\mathbb C(t)$ of rational functions that contains $\mathbb C$ and is not $\mathbb C$ itself. Then $F$ is isomorphic to a field $\mathbb C(u)$ of rational functions.
  • Galois extensions
    • Definition: Intermediate field
    • Definition: When is an intermediate field proper


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