Field Extensions

Study Sheet

Field Extensions

New fields from irreducible polynomials, their degrees, and the tower law

Building Fields from Polynomials

Concept
Extensions and the quotient F[x]/(p(x))F[x]/(p(x))

A field EE containing FF is an extension of FF, written E/FE/F. If p(x)p(x) is irreducible over FF, then F[x]/(p(x))F[x]/(p(x)) — polynomials over FF, computed modulo p(x)p(x) — is a field, and the class α\alpha of xx is a root of pp in it. So every irreducible polynomial has a root in some extension field (Kronecker's theorem).

Working "modulo p(x)p(x)" is like clock arithmetic with polynomials: every polynomial is replaced by its remainder on division by pp, so the elements are the polynomials of degree less than deg⁡p\deg p. Addition is ordinary, and products are reduced using the rule p(α)=0p(\alpha)=0. This is how C\mathbb C arises from R\mathbb R: computing with a+bxa+bx modulo x2+1x^2+1 is exactly computing with a+bia+bi, since the rule becomes i2=−1i^2=-1.

Example
A field with four elements

Over Z2\mathbb Z_2, x2+x+1x^2+x+1 is irreducible. In Z2[x]/(x2+x+1)\mathbb Z_2[x]/(x^2+x+1) the elements are 0,1,α,α+10,1,\alpha,\alpha+1, with the rule α2=α+1\alpha^2=\alpha+1 (since −1=1-1=1). The nonzero elements multiply by

⋅1αα+111αα+1ααα+11α+1α+11α\begin{array}{c|ccc}\cdot&1&\alpha&\alpha+1\\\hline1&1&\alpha&\alpha+1\\\alpha&\alpha&\alpha+1&1\\\alpha+1&\alpha+1&1&\alpha\end{array}

Every nonzero element has an inverse (the 11 appears in each row), so this is a field with 44 elements — and it is not Z4\mathbb Z_4, where 2⋅2=02\cdot2=0.

Example
The complex numbers as a quotient

In R[x]/(x2+1)\mathbb R[x]/(x^2+1) write ii for the class of xx. Then (2+3i)(1−i)=2−2i+3i−3i2=2+i+3=5+i(2+3i)(1-i)=2-2i+3i-3i^2=2+i+3=5+i, using i2=−1i^2=-1 — the familiar complex multiplication. So R[x]/(x2+1)≅C\mathbb R[x]/(x^2+1)\cong\mathbb C.

Side note
Where it goes wrong

The polynomial must be irreducible. In Z2[x]/(x2+1)\mathbb Z_2[x]/(x^2+1), the class β\beta of x+1x+1 satisfies β2=x2+1=0\beta^2=x^2+1=0 although β≠0\beta\ne0: a zero divisor, so this quotient is not a field. That happens because x2+1=(x+1)2x^2+1=(x+1)^2 over Z2\mathbb Z_2.

Try it
Try it: a power of α\alpha

In Z2[x]/(x2+x+1)\mathbb Z_2[x]/(x^2+x+1), compute α3\alpha^3.

Show the answer

α3=α⋅α2=α(α+1)=α2+α=(α+1)+α=2α+1=1\alpha^3=\alpha\cdot\alpha^2=\alpha(\alpha+1)=\alpha^2+\alpha=(\alpha+1)+\alpha=2\alpha+1=\mathbf1. So α\alpha has order 33 in the group of nonzero elements.

Degree and Bases

Concept
The degree of an extension

An extension EE of FF is a vector space over FF; its dimension is the degree [E:F][E:F]. If pp is irreducible of degree nn, then [F[x]/(p):F]=n[F[x]/(p):F]=n, with basis 1,α,…,αn−11,\alpha,\ldots,\alpha^{n-1}. In particular, over Zp\mathbb Z_p this gives a field with pnp^n elements.

The degree measures how much bigger EE is than FF. C\mathbb C has degree 22 over R\mathbb R (basis 1,i1,i); Q(2)\mathbb Q(\sqrt2) has degree 22 over Q\mathbb Q (basis 1,21,\sqrt2); R\mathbb R has infinite degree over Q\mathbb Q. For a finite field the degree pins down the size: a degree-nn extension of Zp\mathbb Z_p has exactly pnp^n elements, one for each choice of nn coefficients.

Example
Three degrees over Q\mathbb Q

[Q(2):Q]=2[\mathbb Q(\sqrt2):\mathbb Q]=2 with basis 1,21,\sqrt2. [Q(23):Q]=3[\mathbb Q(\sqrt[3]2):\mathbb Q]=3 with basis 1,23,431,\sqrt[3]2,\sqrt[3]4, because x3−2x^3-2 is irreducible. [Q(54):Q]=4[\mathbb Q(\sqrt[4]5):\mathbb Q]=4, since x4−5x^4-5 is irreducible (Eisenstein, p=5p=5).

Example
Counting elements

Over Z3\mathbb Z_3, x2+1x^2+1 has no root (0,1,4≡10,1,4\equiv1 are the values x2x^2 takes, so x2+1∈{1,2,2}x^2+1\in\{1,2,2\}), so it is irreducible, and Z3[x]/(x2+1)\mathbb Z_3[x]/(x^2+1) is a field with 32=93^2=9 elements: all a+bαa+b\alpha with a,b∈Z3a,b\in\mathbb Z_3.

Side note
Where it goes wrong

Adjoining two square roots does not give degree 22: Q(2+3)\mathbb Q(\sqrt2+\sqrt3) has degree 44, since the minimal polynomial of 2+3\sqrt2+\sqrt3 is x4−10x2+1x^4-10x^2+1. And the degree of F[x]/(p)F[x]/(p) equals deg⁡p\deg p only when pp is irreducible.

Try it
Try it: size of a finite field

How many elements does Z5[x]/(x3+x+1)\mathbb Z_5[x]/(x^3+x+1) have, given that x3+x+1x^3+x+1 has no root in Z5\mathbb Z_5?

Show the answer

A cubic with no root is irreducible, so this is a field of degree 33 over Z5\mathbb Z_5, with 53=1255^3=\mathbf{125} elements.

Algebraic Elements and Minimal Polynomials

Concept
Algebraic elements and the minimal polynomial

An element α\alpha of an extension of FF is algebraic over FF if it is a root of some nonzero polynomial in F[x]F[x]. Its minimal polynomial is the monic polynomial of least degree with α\alpha as a root; it is irreducible, divides every polynomial in F[x]F[x] vanishing at α\alpha, and [F(α):F][F(\alpha):F] equals its degree.

The minimal polynomial is the most efficient description of α\alpha over FF. To find it: produce some polynomial that α\alpha satisfies, typically by isolating radicals and squaring, then check irreducibility. Numbers like π\pi and ee are not algebraic over Q\mathbb Q (they are transcendental), and Q(π)\mathbb Q(\pi) has infinite degree.

Example
The minimal polynomial of 2+3\sqrt2+\sqrt3

Let α=2+3\alpha=\sqrt2+\sqrt3. Then α2=5+26\alpha^2=5+2\sqrt6, so (α2−5)2=24(\alpha^2-5)^2=24, i.e. α4−10α2+1=0\alpha^4-10\alpha^2+1=0. This quartic is irreducible over Q\mathbb Q, so it is the minimal polynomial and [Q(2+3):Q]=4[\mathbb Q(\sqrt2+\sqrt3):\mathbb Q]=4.

Example
A cube root of unity

ω=−1+3 i2\omega=\frac{-1+\sqrt3\,i}2 satisfies ω3=1\omega^3=1 but ω≠1\omega\ne1. Since x3−1=(x−1)(x2+x+1)x^3-1=(x-1)(x^2+x+1), ω\omega is a root of x2+x+1x^2+x+1, which has no rational root and so is irreducible: the minimal polynomial of ω\omega over Q\mathbb Q is x2+x+1x^2+x+1.

Side note
Where it goes wrong

A polynomial that α\alpha satisfies need not be minimal. 2\sqrt2 is a root of x4−4=(x2−2)(x2+2)x^4-4=(x^2-2)(x^2+2), but its minimal polynomial is x2−2x^2-2. Also the minimal polynomial depends on the base field: ii has minimal polynomial x2+1x^2+1 over Q\mathbb Q but x−ix-i over C\mathbb C.

Try it
Try it: a minimal polynomial

Find the minimal polynomial of 1+21+\sqrt2 over Q\mathbb Q.

Show the answer

If α=1+2\alpha=1+\sqrt2 then (α−1)2=2(\alpha-1)^2=2, so α2−2α−1=0\alpha^2-2\alpha-1=0. This quadratic has no rational root (α\alpha is irrational), so the minimal polynomial is x2−2x−1\mathbf{x^2-2x-1}.

The Tower Law

Concept
Degrees multiply in towers

If F⊆E⊆KF\subseteq E\subseteq K are fields with finite degrees, then [K:F]=[K:E] [E:F][K:F]=[K:E]\,[E:F]. Consequently [E:F][E:F] divides [K:F][K:F], and the degree of any element of KK over FF divides [K:F][K:F].

The tower law is the field-theory twin of Lagrange's theorem. It is the main tool for computing degrees: adjoin one element at a time, and multiply. It also proves impossibility results — if [K:Q]=3[K:\mathbb Q]=3, then KK cannot contain 2\sqrt2, whose degree 22 does not divide 33.

Example
Q(2,3)\mathbb Q(\sqrt2,\sqrt3) has degree 44

[Q(2):Q]=2[\mathbb Q(\sqrt2):\mathbb Q]=2. Is 3∈Q(2)\sqrt3\in\mathbb Q(\sqrt2)? If 3=a+b2\sqrt3=a+b\sqrt2 then 3=a2+2b2+2ab23=a^2+2b^2+2ab\sqrt2, forcing ab=0ab=0, and then 3=a23=a^2 or 3=2b23=2b^2, impossible in Q\mathbb Q. So x2−3x^2-3 stays irreducible over Q(2)\mathbb Q(\sqrt2), giving [Q(2,3):Q]=2⋅2=4[\mathbb Q(\sqrt2,\sqrt3):\mathbb Q]=2\cdot2=4, with basis 1,2,3,61,\sqrt2,\sqrt3,\sqrt6.

Example
2∉Q(23)\sqrt2\notin\mathbb Q(\sqrt[3]2)

[Q(23):Q]=3[\mathbb Q(\sqrt[3]2):\mathbb Q]=3. If 2\sqrt2 were in it, then Q⊆Q(2)⊆Q(23)\mathbb Q\subseteq\mathbb Q(\sqrt2)\subseteq\mathbb Q(\sqrt[3]2) would give 2∣32\mid3. So 2∉Q(23)\sqrt2\notin\mathbb Q(\sqrt[3]2) — proved without any computation.

Example
Q(23,i)\mathbb Q(\sqrt[3]2,i) has degree 66

[Q(23):Q]=3[\mathbb Q(\sqrt[3]2):\mathbb Q]=3, and Q(23)⊂R\mathbb Q(\sqrt[3]2)\subset\mathbb R does not contain ii, so x2+1x^2+1 is irreducible over it: [Q(23,i):Q(23)]=2[\mathbb Q(\sqrt[3]2,i):\mathbb Q(\sqrt[3]2)]=2. Total degree 3⋅2=63\cdot2=6.

Side note
Where it goes wrong

Do not multiply degrees blindly. Q(2,8)=Q(2)\mathbb Q(\sqrt2,\sqrt8)=\mathbb Q(\sqrt2) has degree 22, not 44, because 8=22\sqrt8=2\sqrt2 is already in Q(2)\mathbb Q(\sqrt2). Always check whether the new element is already in the field built so far.

Try it
Try it: a tower

What is [Q(2,5):Q][\mathbb Q(\sqrt2,\sqrt5):\mathbb Q]?

Show the answer

[Q(2):Q]=2[\mathbb Q(\sqrt2):\mathbb Q]=2, and 5∉Q(2)\sqrt5\notin\mathbb Q(\sqrt2) (the same argument as for 3\sqrt3), so the second step also has degree 22: total 4\mathbf4.

Why It Works

Let ff be a polynomial not divisible by pp. Since pp is irreducible, gcd⁡(f,p)=1\gcd(f,p)=1, so the Euclidean algorithm gives a,ba,b with af+bp=1af+bp=1. Modulo pp this says af≡1af\equiv1: the class of ff has inverse the class of aa. Every nonzero class is invertible.

If the minimal polynomial mm of α\alpha factored as ghgh with smaller degrees, then 0=m(α)=g(α)h(α)0=m(\alpha)=g(\alpha)h(\alpha) in a field, so g(α)=0g(\alpha)=0 or h(α)=0h(\alpha)=0 — a polynomial of smaller degree with root α\alpha, contradicting minimality.

Let e1,…,eme_1,\ldots,e_m be a basis of EE over FF and k1,…,knk_1,\ldots,k_n a basis of KK over EE. Every element of KK is ∑jcjkj\sum_j c_jk_j with cj∈Ec_j\in E, and each cj=∑iaijeic_j=\sum_i a_{ij}e_i with aij∈Fa_{ij}\in F. So the mnmn products eikje_ik_j span KK over FF; reversing the argument shows they are independent. Hence [K:F]=mn[K:F]=mn.

If mm has degree nn, divide any polynomial in α\alpha by mm: only the remainder, of degree less than nn, matters. So 1,α,…,αn−11,\alpha,\ldots,\alpha^{n-1} span F(α)F(\alpha), and they are independent because a dependence would be a nonzero polynomial of degree less than nn vanishing at α\alpha.