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Abstract Algebra - Part II

Basics of field theory

Things to remember from before.

We already know quite a bit about fields.

Characteristic

If $F$ is a field, then there is a ring homomorphism $\Z\to F$ sending $1\to 1$. If this map is injective, then:

  • we say $F$ has characteristic zero
  • $F$ contains a copy of the rational numbers
  • The field $\Q$ is the prime subfield of $F$.

Otherwise the kernel of this map must be a prime ideal $p\Z$ of $\Z$. In this case:

  • we say that $F$ has characteristic $p$
  • $F$ contains a copy of $\Z/p\Z$.
  • $\Z/p\Z$ is the prime subfield of $F$.

Maps

If $f:F\to E$ is a homomorphism of fields, it is automatically injective (or zero).

The only field maps $f:\Q\to \Q$ and $f:\Z/p\Z\to \Z/p\Z$ are the identity.

Extensions

If $F$ is a field, and $F\subset E$ where $E$ is another field, then we call $E$ an extension field of $F$.

$E$ is automatically a vector space over $F$. The degree of $E/F$, written $[E:F]$, is the dimension of $E$ as an $F$-vector space.

Polynomials, quotient rings, and fields

We have the division algorithm for polynomials. $F[x]$ is a PID. An ideal is prime iff it is generated by an irreducible polynomial.

Let $p(x)$ be an irreducible polynomial of degree $d$ over $F$. Then:

  • $K=F[x]/(p(x))$ is a field
  • It is of degree $d$ over $F$.
  • $p(x)$ has a root in $K$ (namely the residue class of $x$)
  • The elements $1,x,\ldots, x^{d-1}$ are a basis for $K/F$.

Adjoining roots of polynomials

If $F\subset K$ is a field extension, and $\alpha\in K$, then $F(\alpha)$ is the smallest subfield of $K$ containing $F$ and $\alpha$. Similarly for $F(\alpha_1,\alpha_2,\ldots,\alpha_n)$.

If $p(x)$ is irreducible over $F$, and has a root $\alpha$ in $K$, then $F(\alpha)$ is isomorphic to $F[x]/p(x)$ via the map $x\mapsto \alpha$.

Key Theorem

Let $K$ be a field extension of $F$ and let $p(x)$ be an irreducible polynomial over $F$. Suppose $K$ contains two roots $\alpha$ and $\beta$ of $p(x)$. Then $F(\alpha)$ and $F(\beta)$ are isomorphic via an isomorphism that is the identity on $F$.

More generally:

Theorem: (See Theorem 8, DF, page 519) Let $\phi:F\to F’$ be an isomorphism of fields. Let $p(x)$ be an irreducible polynomial in $F[x]$ and let $p’(x)$ be the polynomial in $F’[x]$ obtained by applying $\phi$ to the coefficients of $p(x)$. Let $K$ be an extension of $F$ containing a root $\alpha$ of $p(x)$, and let $K’$ be an extension of $F’$ containing a root $\beta$ of $p’(x)$. Then there is an isomorphism $\sigma:F(\alpha)\to F’(\beta)$ such that the restriction of $\sigma$ to $F$ is $\phi$.

Algebraic Extensions

Definition

Definition: Let $F\subset K$ be a field extension. An element $\alpha\in K$ is algebraic over $F$ if it is the root of a nonzero polynomial in $F[x]$. Elements that aren’t algebraic are called transcendental.

An extension $K/F$ is algebraic if every element of $K$ is algebraic over $F$.

Basics

  • If $\alpha$ is algebraic over $F$, there is unique monic polynomial $m_{\alpha,F}(x)$ of minimal degree with coefficients in $F$ such that $m_{\alpha}(\alpha)=0$. (This follows from the division algorithm). This polynomial is called the minimal polynomial of $\alpha$ over $F$. Its degree is the degree of $\alpha$.
  • If $F\subset L$, then the minimal polynomial $m_{\alpha,L}(x)\in L[x]$ of $\alpha$ over $L$ divides the minimal polynomial $m_{\alpha,F}(x)$. Again, this follows from the division algorithm for $L[x]$.
  • $F(\alpha)$ is isomorphic to $F[x]/m_{\alpha,F}(x)$; and the degree $[F(\alpha):F]$ is the degree of $\alpha$.

Examples

If $n>1$ and $p$ is a prime, then the polynomial $x^{n}-p$ is irreducible over $\Q$, so $\alpha=\sqrt[n]{p}$ has degree $n$ over $\Q$.

The polynomial $x^3-x-1$ is irreducible over $\Q$ and has one real root $\alpha$. So $\alpha$ has degree $3$ over $\Q$ but degree $1$ over $\R$.

Finite extensions are algebraic

Suppose $K/F$ is finite and let $\alpha$ be an element of $K$. Then there is an $n$ so that the set $1,\alpha,\alpha^2,\ldots, \alpha^{n}$ are linearly dependent over $F$; so $\alpha$ satisfies a polynomial with $F$ coefficients, and is therefore algebraic.

As a partial converse, if $F(\alpha)/F$ is finite if and only if $\alpha$ is algebraic. If $\alpha$ is algebraic of degree $d$ over $F$, $F(\alpha)=F[x]/(m_{\alpha}(x))$ which is finite dimensional (with basis $1,x,x^2,\ldots, x^{d-1}$.)

Algebraic over algebraic is algebraic

Proposition: If $K/F$ is algebraic and $L/K$ is algebraic then $L/F$ is algebraic.

Proof: Let $\alpha$ be any element of $L$. It has a minimal polynomial $f(x)=x^{d}+a_{d-1}x^{d-1}+\cdots+a_{0}$ with the $a_{i}\in K$. Therefore $\alpha$ is algebraic over $F(a_0,\ldots, a_{d-1})$. Since the $a_i$ are in $K$, they are algebraic over $F$, and therefore $F(a_0,\ldots, a_{d-1})$ is finite over $F$ and so is $F(\alpha,a_0,\ldots, a_{d-1})$. Thus $F(\alpha)$ is contained in a finite extension of $F$ and so $\alpha$ is algebraic over $F$.

Field Degrees

Multiplicativity of degrees

Proposition: Suppose that $L/F$ and $K/L$ are extensions. Then $[K:F]=[K:L][L:F]$.

Proof: If $\alpha_1,\ldots, \alpha_n$ are a basis for $L/F$, and $\beta_1,\ldots, \beta_k$ are a basis for $K/L$, then the products $\alpha_{i}\beta_{j}$ are a basis for $K/F$.

Corollary: If $L/F$ is a subfield of $K/F$, then $[L:F]$ divides $[K:F]$.

Finitely generated extensions

A field $K/F$ is finitely generated if $K=F(\alpha_1,\ldots, \alpha_n)$ for a finite set of $\alpha_{i}$ in $K$.

Proposition: $F(\alpha,\beta)=F(\alpha)(\beta)$.

Proof: $F(\alpha,\beta)$ contains $F(\alpha)$ and also $\beta$. Therefore $F(\alpha)(\beta)\subset F(\alpha,\beta)$. On the other hand, since $\alpha$ and $\beta$ are in $F(\alpha)(\beta)$, we know that $F(\alpha,\beta)\subset F(\alpha)(\beta)$.

Finite is finitely generated

Proposition: A field $K/F$ is finite if and only if it is finitely generated. If it is generated by $\alpha_1,\ldots, \alpha_k$ then it is of degree at most $n_1 n_2\ldots n_k$ where $n_i$ is the degree of $\alpha_i$ over $F$.

Proof: If it’s finitely generated, then it’s a sequence of extensions $F(\alpha_1,\ldots, \alpha_{s-1})(\alpha_{s})$ each of degree at most $n_i$. So $K/F$ is finite. Conversely, if $K/F$ is finite (and of degree greater than 1), choose $\alpha_1\in K$ of degree greater than 1. Then $F(\alpha)\subset K$ and $[K:F(\alpha)]$ is smaller than $[K:F]$. Now choose $\alpha_2$ in $K$ but not $F(\alpha_1)$, and so on. This process must terminate.

Corollary: If $\alpha$ and $\beta$ are algebraic over $F$, so are $\alpha+\beta$, $\alpha\beta$, and (if $\beta\not=0$) $\alpha/\beta$.

Proof: All these elements lie in $F(\alpha,\beta)$ which is finite over $F$.

Corollary: If $K/F$ is a field extension, the subset of $K$ consisting of algebraic elements over $F$ is a field (called the algebraic closure of $F$ in $K$).

Towers of algebraic extensions are algebraic

Propositoin: If $L/K$ is algebraic and $K/F$ is algebraic so is $L/F$.

Proof: Choose $\alpha\in L$. Then $\alpha$ satisfies a polynomial $f(x)=x^{d}+a_{d-1}x^{d-1}+\cdots+a_{0}$ where the $a_i$ are in $K$. Therefore $\alpha$ is algebraic over $E=F(a_0,a_1,\ldots, a_{d-1})$. But $E/F$ is finitely generated hence finite. Therefore $[E(\alpha):F]=[E(\alpha):E][E:F]$ is finite. Thus every element of $L$ is algebraic over $F$.

Composites

If $K_1$ and $K_2$ are subfields of a field $K$, then $K_1 K_2$ is the smallest subfield of $K$ containing these two fields. Then $[K_1 K_2:F]$ is divisible by both $[K_1:F]$ and $[K_2:F]$ and in addition

\[[K_1 K_2:F]\le [K_1:F][K_2:F].\]

In particular, if $[K_1:F]$ and $[K_2:F]$ are relatively prime, then $[K_1 K_2:F]=[K_1:F][K_2:F]$.

Classical Constructions (Ruler and Compass)

Classical ruler and compass constructions allow one to:

  • find the point of intersection of two lines.
  • find the point of intersection of a line and a circle.
  • find the points of intersection of two circles.

Constructions

If we begin with a line segment of length 1, we can:

  • construct a perpendicular, and then construct all integer lengths along that line
  • construct all points with integer coordinates in the plane
  • using similar triangles, construct all points in the plane with rational coordinates

Extensions

Now suppose we can construct all points with coordinates in a field $F$. Then:

  • intersections of lines joining points of over $F$ meet in points with coordinates in $F$
  • intersections of a line joining two points with coordinates in $F$ with a circle of radius in $F$ yields points in a quadratic extension of $F$.
  • intersections of two circles with radii in $F$ yields points with coordinates in a quadratic extension of $F$.

Gauss’s Theorem on constructibility

Theorem: If a line segment of length $\alpha$ is constructible by ruler and compass, then $\alpha$ lies in a field obtained from $\Q$ by a sequence of quadratic extensions, and $[F(\alpha):F]=2^k$ for some integer $k\ge 0$.

Corollary: One cannot “double the cube” , trisect an angle, or square the circle.

Here doubling the cube means given a length $\alpha$ construct a length $\beta$ so that the cube with side length $\beta$ has double the volume of the cube with side length $\alpha$. This is impossible because $\sqrt[3]{2}$ does not meet Gauss’s criterion.

Squaring the circle means, given $\alpha$, constructing a length $\beta$ so that a square of side $\beta$ has the same area as a circle of radius $\alpha$. This is impossible because $\pi$ is not algebraic (we won’t prove this).

Trisecting the angle means constructing an angle with one-third the measure of a given angle $\theta$. If we can trisect $\theta$, we can construct a length of $\cos(\theta/3).$ If $\theta=\pi/3$, then $\theta/3=\pi/9$ or $\beta=\cos 20^{\circ}$. One can show that, if $u=2\beta$, then

\[u^3-3u-1=0.\]

This polynomial has no rational roots (it is irreducible mod $2$ for example).

A pentagon is constructible because the $\cos(2\pi/5)$ is the root of a quadratic polynomial.

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