We already know quite a bit about fields.
If \(F\) is a field, then there is a ring homomorphism \(\Z\to F\) sending \(1\to 1\). If this map is injective, then:
Otherwise the kernel of this map must be a prime ideal \(p\Z\) of \(\Z\). In this case:
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.
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.
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:
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 \(x\mapsto \alpha\).
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 \(\phi\).
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 \(F\).
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 \(\R\).
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 \(1,x,x^2,\ldots, x^{d-1}\).)
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 \(F\).
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 \([K:F]\).
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 \(F(\alpha,\beta)\subset F(\alpha)(\beta)\).
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 \(K\)).
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 \(F\).
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 ruler and compass constructions allow one to:
If we begin with a line segment of length 1, we can:
Now suppose we can construct all points with coordinates in a field \(F\). Then:
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.