22. Where algebra goes¶

You started this book with a letter standing for a number. You end it with groups of symmetries deciding which equations can be solved. Where does algebra go from here? Into almost all of modern mathematics. This last chapter is a guided tour: the language of categories that algebraists now speak, algebraic geometry where equations become shapes, algebraic number theory where Kummer's ideals grew up, Lie groups where symmetry moves continuously, the Langlands programme that connects them, the computer algorithms that do algebra for you, and the open problems that nobody can yet solve. Each section gives the main idea with honest statements and at least one example you can check, and points to where the full story is told.

In this chapter

You will learn:

what categories, functors and natural transformations are, and why "universal properties" describe so many constructions
how polynomial equations define geometric shapes, and Hilbert's basis theorem and Nullstellensatz
algebraic integers, rings of integers, class groups and units, and why \(e^{\pi\sqrt{163}}\) is nearly an integer
Lie groups and Lie algebras: rotations, the matrix exponential and the bracket
an honest one-page picture of the Langlands programme, with a modular form you can check by counting points
Gröbner bases: how computers solve systems of polynomial equations
some open problems: the inverse Galois problem, the Jacobian conjecture, and others

You need: the whole of Part IV and V, especially Chapter 18, Chapter 20 and Chapter 21. This chapter states more than it proves; where a proof is given it is complete, and where it is not, the statement says where to find it.

Levels: all sections are expert in content, but written as a survey: 22.1, 22.2 and 22.6 can be read by anyone who has met rings; 22.3 and 22.5 build on Chapters 18 to 21; 22.4 needs some calculus and linear algebra.

22.1 Categories and functors¶

Level: expert. By now you have seen the same pattern many times. Groups have homomorphisms, rings have ring homomorphisms, vector spaces have linear maps, modules have module homomorphisms. Each time there is a kernel, an image, a first isomorphism theorem, products, quotients. Mathematicians noticed in the 1940s that one should study the pattern itself.

Definition. A category \(\mathcal{C}\) consists of:

a collection of objects \(A, B, C, \ldots\);
for each pair of objects a set \(\operatorname{Hom}(A, B)\) of morphisms (arrows) \(f: A \to B\);
a composition \(g \circ f: A \to C\) for \(f: A \to B\) and \(g: B \to C\), which is associative, \(h \circ (g \circ f) = (h \circ g) \circ f\);
an identity \(1_A: A \to A\) for each object, with \(f \circ 1_A = f = 1_B \circ f\).

Examples. Set (sets and functions), Grp (groups and homomorphisms), Ab (abelian groups), Ring, Vect\(_F\) (vector spaces over \(F\) and linear maps), \(R\)-Mod, Top (topological spaces and continuous maps). Less obviously: a single group \(G\) is a category with one object, whose morphisms are the elements of \(G\) and whose composition is the group law; and a partially ordered set is a category with one morphism \(a \to b\) whenever \(a \le b\).

An isomorphism in a category is a morphism with a two-sided inverse. Notice that this one definition gives bijections in Set, group isomorphisms in Grp and homeomorphisms in Top, with nothing about elements.

Functors. A functor \(F: \mathcal{C} \to \mathcal{D}\) sends objects to objects and morphisms to morphisms, preserving composition and identities: \(F(g \circ f) = F(g) \circ F(f)\), \(F(1_A) = 1_{F(A)}\). A contravariant functor reverses arrows.

The forgetful functor Grp \(\to\) Set forgets the multiplication.
Abelianization: \(G \mapsto G^{\mathrm{ab}} = G/[G, G]\), where \([G,G]\) is the subgroup generated by commutators \(ghg^{-1}h^{-1}\). A homomorphism \(G \to H\) sends commutators to commutators, so it induces \(G^{\mathrm{ab}} \to H^{\mathrm{ab}}\).
Units: \(R \mapsto R^\times\), from Ring to Grp; and \(R \mapsto GL_n(R)\) for commutative rings.
Duals: \(V \mapsto V^* = \operatorname{Hom}(V, F)\) is a contravariant functor on Vect\(_F\); a linear map \(T: V \to W\) gives the transpose \(T^*: W^* \to V^*\).
The fundamental group \(\pi_1\) sends a space with a base point to a group, and continuous maps to homomorphisms. This is how topology turns geometric questions into algebra: since \(\pi_1\) of the circle is \(\mathbb{Z}\) and that of the disc is trivial, there is no continuous retraction of a disc onto its boundary circle (it would give a homomorphism \(\{1\} \to \mathbb{Z}\) that is onto), which is the key step of Brouwer's fixed point theorem in the plane.
Galois theory is an order-reversing correspondence between two posets, subgroups and intermediate fields (Chapter 20), each viewed as a category: a pair of contravariant functors that are inverse to each other.

Natural transformations. A finite-dimensional vector space \(V\) is isomorphic to its dual \(V^*\) (same dimension), but to write down an isomorphism you must choose a basis. It is also isomorphic to its double dual \(V^{**}\), by \(v \mapsto (\varphi \mapsto \varphi(v))\), with no choice at all. Samuel Eilenberg and Saunders Mac Lane invented categories in 1945, in "General theory of natural equivalences", precisely to make this difference precise: the second isomorphism is natural, meaning that for every linear map \(T: V \to W\) the square

\[ \begin{array}{ccc} V & \xrightarrow{\ \ T\ \ } & W \\ \downarrow & & \downarrow \\ V^{**} & \xrightarrow{\ T^{**}\ } & W^{**} \end{array} \]

commutes; a natural transformation is such a family of maps between two functors.

Universal properties. The most useful idea in the subject: describe an object by how it maps to or from everything else. The product of \(A\) and \(B\) is an object \(P\) with maps \(p_A: P \to A\), \(p_B: P \to B\) such that for any object \(X\) with maps \(f: X \to A\), \(g: X \to B\) there is exactly one \(h: X \to P\) with \(p_A h = f\) and \(p_B h = g\). In Set this is the Cartesian product, in Grp the direct product, in Top the product space, and in a poset the greatest lower bound. A product, if it exists, is unique up to unique isomorphism, and the proof is one paragraph that works in every category at once. Free groups, polynomial rings ("a ring homomorphism out of \(R[x]\) is decided by where \(x\) goes"), quotients ("a homomorphism killing \(N\) factors through \(G/N\)"), tensor products and fields of fractions are all defined this way. Alexander Grothendieck rebuilt algebraic geometry in this language in the late 1950s and 1960s; today it is the common tongue of algebra, topology and geometry, and of parts of computer science (type theory and functional programming).

22.2 Algebraic geometry: equations as shapes¶

Level: expert. The circle \(x^2 + y^2 = 1\) is a shape and an equation. Algebraic geometry studies the dictionary between the two.

Let \(k\) be a field and \(k^n\) the space of \(n\)-tuples. For a set \(S\) of polynomials in \(k[x_1, \ldots, x_n]\), the affine variety

\[ V(S) = \{a \in k^n : f(a) = 0 \text{ for all } f \in S\} \]

is their common zero set. For a set of points \(X \subseteq k^n\), its ideal \(I(X)\) is the set of polynomials vanishing on all of \(X\). Examples in the plane: \(V(x^2 + y^2 - 1)\) is the circle; \(V(xy)\) is the union of the two axes; \(V(x - y^2)\) is a parabola; and \(V(y - x^2, z - x^3)\) in \(k^3\) is the twisted cubic \(\{(t, t^2, t^3)\}\).

Some first facts, each a one-line check: \(V(S) = V(\text{ideal generated by } S)\); bigger sets of equations give smaller varieties; \(V(I) \cup V(J) = V(IJ)\); \(V(I) \cap V(J) = V(I + J)\). The two maps \(V\) and \(I\) reverse inclusions, exactly like the Galois correspondence.

Intersecting a circle and a parabola. \(V(x^2 + y^2 - 1,\ x - y^2)\): substituting \(x = y^2\) gives \(y^4 + y^2 - 1 = 0\), so \(y^2 = \tfrac{-1 + \sqrt5}{2}\) (the other root is negative). So \(x = \tfrac{\sqrt5 - 1}{2} \approx 0.618\), the golden ratio's reciprocal, and \(y = \pm 0.786\). Two real points; over \(\mathbb{C}\) there are two more, with \(x = \tfrac{-1-\sqrt5}{2}\) and \(y\) imaginary. 22.6 does this by machine.

Bézout's theorem counts such intersections: two plane curves of degrees \(d\) and \(e\) with no common component meet in at most \(de\) points; over \(\mathbb{C}\), counting points at infinity (in the projective plane) and multiplicities, in exactly \(de\). Circle and parabola: \(2 \times 2 = 4\), as found. Two circles meet in at most 2 points of the real plane, but \(2 \times 2 = 4\): the missing two are the same for every pair of circles, the "circular points at infinity" \((1 : i : 0)\) and \((1 : -i : 0)\).

Theorem 22.1 (Hilbert's basis theorem, 1890). Every ideal of \(k[x_1, \ldots, x_n]\) is finitely generated. More generally, if \(R\) is Noetherian (every ideal finitely generated), so is \(R[x]\).

Proof. Let \(I \subseteq R[x]\) be an ideal. For each \(d\), the leading coefficients of the elements of \(I\) of degree \(d\), together with 0, form an ideal \(L_d\) of \(R\), and \(L_d \subseteq L_{d+1}\) (multiply by \(x\)). An increasing chain of ideals in a Noetherian ring stops (the union is an ideal, finitely generated, so all its generators appear at some stage), say at \(L_N\). Each of \(L_0, \ldots, L_N\) is finitely generated; for each generator pick a polynomial in \(I\) of that degree with that leading coefficient. These finitely many polynomials generate \(I\): given \(f \in I\) of degree \(d\), its leading coefficient lies in \(L_{\min(d, N)}\), so a combination of the chosen polynomials (times powers of \(x\)) has the same degree and leading coefficient, and subtracting it lowers the degree; by induction on degree, \(f\) is in the ideal they generate. Then induct on the number of variables, starting from the field \(k\), whose only ideals are \(0\) and \(k\). \(\blacksquare\)

So every variety is defined by finitely many equations, even if you start from infinitely many. Paul Gordan, whose problem in invariant theory Hilbert solved this way without constructing anything, is said to have called it "theology, not mathematics"; the quote may be apocryphal, but the debate about non-constructive proofs it represents was real, and 22.6 shows how it was eventually answered with algorithms.

Theorem 22.2 (Hilbert's Nullstellensatz, 1893). Let \(k\) be algebraically closed (for example \(\mathbb{C}\)).

(Weak form.) If \(f_1, \ldots, f_m \in k[x_1, \ldots, x_n]\) have no common zero in \(k^n\), then \(g_1 f_1 + \cdots + g_m f_m = 1\) for some polynomials \(g_i\).
(Strong form.) For every ideal \(J\), \(I(V(J)) = \sqrt{J}\), the radical of \(J\): the polynomials \(f\) with \(f^r \in J\) for some \(r \ge 1\).

("Nullstellensatz" is German for "theorem of zeros".) The weak form says the only reason equations can fail to have a common solution is the obvious one: some combination of them is \(1 = 0\). The strong form makes \(V\) and \(I\) an exact dictionary:

\[ \{\text{radical ideals of } k[x_1, \ldots, x_n]\} \;\longleftrightarrow\; \{\text{affine varieties in } k^n\}, \]

with maximal ideals corresponding to points, \((x_1 - a_1, \ldots, x_n - a_n) \leftrightarrow (a_1, \ldots, a_n)\). Example of the radical: \(V(x^2) = \{0\}\) on the line, and \(I(\{0\}) = (x) = \sqrt{(x^2)}\). The proof needs some commutative algebra (Zariski's lemma: a field finitely generated as a ring over \(k\) is finite over \(k\)); see Atiyah and Macdonald, chapters 5 and 7, or Cox, Little and O'Shea, chapter 4. Algebraically closed is necessary: over \(\mathbb{R}\), \(x^2 + y^2 + 1\) has no zeros, yet the ideal it generates does not contain \(1\).

Where it went. Grothendieck's schemes (from 1960) replaced varieties by rings themselves, every commutative ring becoming a "space", which let geometry work over \(\mathbb{Z}\) and finite fields. With it Pierre Deligne proved the Weil conjectures on counting points of varieties over finite fields (1974), and modern number theory, including the proof of Fermat's last theorem, speaks this language.

22.3 Algebraic number theory and class groups¶

Level: expert. Chapter 18 ended with Kummer and Dedekind repairing unique factorization by ideals. Here is the theory they started.

A number field is a finite extension \(K\) of \(\mathbb{Q}\), such as \(\mathbb{Q}(i)\), \(\mathbb{Q}(\sqrt{-5})\) or \(\mathbb{Q}(\sqrt[3]2)\). In it, the right analogue of the integers is the set of algebraic integers: elements that are roots of a monic polynomial with integer coefficients. (Chapter 10 proved the first theorem of the subject: a rational algebraic integer is an ordinary integer.)

Theorem 22.3. The algebraic integers in \(K\) form a ring, \(\mathcal{O}_K\), the ring of integers of \(K\).

Proof. The key criterion: \(\alpha\) is an algebraic integer if and only if \(\mathbb{Z}[\alpha]\) is contained in a finitely generated \(\mathbb{Z}\)-module \(M \subseteq \mathbb{C}\) with \(\alpha M \subseteq M\) and \(M \ne 0\). (If \(\alpha^n = -(c_{n-1}\alpha^{n-1} + \cdots + c_0)\) then \(M = \mathbb{Z} + \mathbb{Z}\alpha + \cdots + \mathbb{Z}\alpha^{n-1}\) works. Conversely, if \(m_1, \ldots, m_n\) generate \(M\), write \(\alpha m_i = \sum a_{ij} m_j\) with \(a_{ij} \in \mathbb{Z}\); then \((\alpha I - A)\mathbf m = 0\) with \(\mathbf m \ne 0\), so \(\det(\alpha I - A) = 0\), a monic integer equation for \(\alpha\).) If \(\alpha\) and \(\beta\) are algebraic integers of degrees \(n\) and \(m\), then \(M = \sum \mathbb{Z}\alpha^i\beta^j\) (\(i < n\), \(j < m\)) is finitely generated and stable under multiplication by \(\alpha + \beta\) and \(\alpha\beta\). So they are algebraic integers. \(\blacksquare\)

Examples. For a square-free integer \(d\), the ring of integers of \(\mathbb{Q}(\sqrt d)\) is \(\mathbb{Z}[\sqrt d]\) if \(d \equiv 2, 3 \pmod 4\), and \(\mathbb{Z}\big[\tfrac{1 + \sqrt d}{2}\big]\) if \(d \equiv 1 \pmod 4\). So \(\mathcal{O} = \mathbb{Z}[i]\) for \(\mathbb{Q}(i)\), \(\mathbb{Z}[\sqrt{-5}]\) for \(\mathbb{Q}(\sqrt{-5})\), but \(\mathbb{Z}[\omega]\), \(\omega = \tfrac{-1 + \sqrt{-3}}{2}\), for \(\mathbb{Q}(\sqrt{-3})\): the number \(\omega\) is a root of \(x^2 + x + 1\), monic, so it is an integer even though it has a denominator.

Dedekind domains. Every \(\mathcal{O}_K\) is a Dedekind domain: every nonzero ideal factors uniquely as a product of prime ideals (Dedekind, 1870s). Elements may not factor uniquely, and the ideal class group measures by how much: it is the group of nonzero (fractional) ideals modulo the principal ones. It is always finite (Minkowski's geometry of numbers gives a bound to check), and its order is the class number \(h_K\). \(\mathcal{O}_K\) is a UFD exactly when \(h_K = 1\) (for Dedekind domains, UFD and PID are the same). In Chapter 18, \(\mathbb{Z}[\sqrt{-5}]\) had \(h = 2\).

Class number one. Gauss conjectured, in the language of quadratic forms, that the imaginary quadratic fields with class number 1 are exactly \(\mathbb{Q}(\sqrt{-d})\) for

\[ d = 1, 2, 3, 7, 11, 19, 43, 67, 163 . \]

Kurt Heegner proved it in 1952, but his paper was thought to have a gap; Alan Baker (1966) and Harold Stark (1967) gave proofs by different methods, and Heegner's was then recognized as essentially correct. Two curiosities come from \(163\):

Euler's polynomial \(n^2 + n + 41\) is prime for \(n = 0, 1, \ldots, 39\) (at \(n = 40\) it is \(1681 = 41^2\)). By a theorem of Georg Rabinowitsch (1913), \(n^2 + n + A\) is prime for \(n = 0, \ldots, A - 2\) exactly when \(\mathbb{Q}(\sqrt{1 - 4A})\) has class number 1, and \(1 - 4 \cdot 41 = -163\).
\(e^{\pi\sqrt{163}} = 262537412640768743.99999999999925\ldots\), within \(10^{-12}\) of the integer \(640320^3 + 744\). This is explained by the theory of modular functions: the \(j\)-function takes an integer value at \(\tfrac{1 + \sqrt{-163}}{2}\) because the class number is 1.

Whether infinitely many real quadratic fields have class number 1 is still unknown (also a question of Gauss).

Units. The units of \(\mathcal{O}_K\) need not be just \(\pm 1\). In \(\mathbb{Z}[\sqrt2]\), \((1 + \sqrt2)(-1 + \sqrt2) = 1\), and all powers \((1 + \sqrt2)^n\) are units: \((1 + \sqrt2)^2 = 3 + 2\sqrt2\), \((1 + \sqrt2)^4 = 17 + 12\sqrt2\). A unit \(a + b\sqrt2\) has norm \(a^2 - 2b^2 = \pm 1\), so units are solutions of Pell's equation: \(3^2 - 2 \cdot 2^2 = 1\), \(17^2 - 2 \cdot 12^2 = 1\). Dirichlet's unit theorem (1846) says the unit group is (roots of unity) \(\times\ \mathbb{Z}^{r_1 + r_2 - 1}\), where \(r_1\) is the number of real embeddings of \(K\) and \(r_2\) the number of pairs of complex ones. For real quadratic fields that is \(\{\pm 1\} \times \mathbb{Z}\), generated by one fundamental unit.

How primes split. In \(\mathbb{Z}[i]\), a prime \(p\) splits, stays prime or ramifies according to \(p \bmod 4\) (Theorem 18.4). In every number field, how a prime \(p\) factors in \(\mathcal{O}_K\) is controlled by the Galois group, through the Frobenius element at \(p\). When the Galois group is abelian, the rule always depends only on congruences, as with \(p \bmod 4\); that is class field theory (Teiji Takagi and Emil Artin, 1920s), the far-reaching generalization of Gauss's quadratic reciprocity law: for distinct odd primes \(p, q\), \(\left(\tfrac{p}{q}\right)\left(\tfrac{q}{p}\right) = (-1)^{\frac{p-1}{2}\cdot\frac{q-1}{2}}\). When the Galois group is not abelian, the rule is no longer about congruences, and that is where 22.5 begins.

22.4 Lie groups and Lie algebras: symmetry that moves¶

Level: expert. The symmetry groups of this book have mostly been finite: the six symmetries of a triangle, the Galois group of a quintic. But a sphere can be rotated by any angle about any axis. Its rotation group \(SO(3)\) is a continuous group, and it is also a smooth 3-dimensional space (three numbers fix a rotation: an axis direction, two numbers, and an angle).

Definition. A Lie group is a group that is also a smooth manifold (a space that looks locally like \(\mathbb{R}^n\)), with multiplication and inversion smooth maps. Examples: \((\mathbb{R}, +)\); the circle \(SO(2) \cong U(1) = \{e^{i\theta}\}\); the rotation group \(SO(3)\); \(GL_n(\mathbb{R})\) and \(SL_n(\mathbb{R})\); the unitary groups \(U(n)\), \(SU(n)\).

Sophus Lie began the subject in the 1870s, trying to do for differential equations what Galois had done for polynomial equations: use their continuous symmetries to solve them.

Rotations in practice. The orientation of a spacecraft, or of a rover's camera mast, is an element of \(SO(3)\). Composing two rotations is multiplying in the group, and the group is not commutative (turn a book \(90^\circ\) about two different axes in the two orders and compare). Flight software commonly represents orientations by unit quaternions, which form the group \(SU(2)\); each rotation corresponds to exactly two unit quaternions, \(\pm q\), because \(SU(2) \to SO(3)\) is a 2-to-1 homomorphism. Quaternions avoid the "gimbal lock" of three-angle descriptions and are cheap to renormalize.

The Lie algebra. The key idea: a Lie group is curved, but its tangent space at the identity is a vector space, and most of the group's structure can be recovered from it. For a matrix group \(G\), the Lie algebra is

\[ \mathfrak{g} = \{X : e^{tX} \in G \text{ for all real } t\}, \qquad e^{X} = I + X + \frac{X^2}{2!} + \frac{X^3}{3!} + \cdots . \]

Example: \(SO(2)\). Let \(X = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\). Then \(X^2 = -I\), \(X^3 = -X\), \(X^4 = I\), and sorting the series into its \(I\) and \(X\) parts,

\[ e^{tX} = \Big(1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \cdots\Big)I + \Big(t - \frac{t^3}{3!} + \cdots\Big)X = \begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{pmatrix}, \]

the rotation by angle \(t\). This is Euler's formula \(e^{it} = \cos t + i\sin t\) (Chapter 9) in matrix form, with \(X\) playing \(i\). The Lie algebra \(\mathfrak{so}(2)\) is the line of multiples of \(X\).

Example: \(SO(3)\). Differentiating \(R^{\mathsf T} R = I\) along a path through \(I\) shows the Lie algebra \(\mathfrak{so}(3)\) consists of the skew-symmetric matrices \(X^{\mathsf T} = -X\), a 3-dimensional space with basis

\[ L_x = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}, \quad L_y = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}, \quad L_z = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \]

the infinitesimal rotations about the three axes (\(e^{tL_z}\) is rotation by \(t\) about the \(z\)-axis).

The bracket. The Lie algebra is not closed under matrix multiplication, but it is closed under the commutator \([X, Y] = XY - YX\), which measures how far the group is from commutative. For \(\mathfrak{so}(3)\), multiplying out gives

\[ [L_x, L_y] = L_z, \qquad [L_y, L_z] = L_x, \qquad [L_z, L_x] = L_y, \]

the same rule as the cross product \(\mathbf i \times \mathbf j = \mathbf k\). A Lie algebra is a vector space with a bilinear bracket that is antisymmetric and satisfies the Jacobi identity \([X,[Y,Z]] + [Y,[Z,X]] + [Z,[X,Y]] = 0\). \(SU(2)\) and \(SO(3)\) have isomorphic Lie algebras: a Lie algebra sees the group near the identity, not globally.

The classification. Wilhelm Killing (1888-1890) and Élie Cartan (1894) classified the simple Lie algebras over \(\mathbb{C}\) (those with no proper ideals, like simple groups): four infinite families, \(A_n = \mathfrak{sl}_{n+1}\), \(B_n = \mathfrak{so}_{2n+1}\), \(C_n = \mathfrak{sp}_{2n}\), \(D_n = \mathfrak{so}_{2n}\), and five exceptional algebras \(G_2, F_4, E_6, E_7, E_8\) of dimensions \(14, 52, 78, 133, 248\). Compared with the classification of finite simple groups (Chapter 16), which took decades and thousands of pages, this one fits in a semester, and its tool, root systems, is a gem of combinatorial geometry.

In physics. Emmy Noether's theorem (1918) says each continuous symmetry of a physical system gives a conserved quantity: time translation gives energy, space translation momentum, rotation (\(SO(3)\)) angular momentum. The Standard Model of particle physics is built on the group \(SU(3) \times SU(2) \times U(1)\), and its particles are representations (Chapter 21) of it.

22.5 The Langlands programme¶

Level: expert. In January 1967 Robert Langlands, then 30, wrote a 17-page letter to André Weil proposing a set of conjectures that would connect number theory (Galois groups) with harmonic analysis (automorphic forms). It is now the largest unifying project in mathematics. Here is the honest one-page picture.

Reciprocity. Take a polynomial with integer coefficients and ask: how does it factor mod \(p\), for each prime \(p\)? For \(x^2 + 1\), the answer is "it has roots mod \(p\) exactly when \(p = 2\) or \(p \equiv 1 \pmod 4\)" (Lemma 18.5 and its converse): a rule given by congruences. For any polynomial whose Galois group is abelian, the answer is always a congruence rule; that is class field theory (22.3). For \(x^3 - 2\), Galois group \(S_3\), no congruence rule exists, and the Langlands programme predicts what replaces it.

What replaces it: modular forms. Here is a famous example you can check yourself. Take the elliptic curve

\[ E: \quad y^2 + y = x^3 - x^2 . \]

For each prime \(p\), count its solutions \((x, y)\) mod \(p\), add one "point at infinity", and call the total \(N_p\); put \(a_p = p + 1 - N_p\). Separately, expand the infinite product

\[ q\prod_{n=1}^{\infty}(1 - q^n)^2(1 - q^{11n})^2 = q - 2q^2 - q^3 + 2q^4 + q^5 + 2q^6 - 2q^7 - 2q^9 - 2q^{10} + q^{11} - 2q^{12} + 4q^{13} + \cdots \]

and call the coefficient of \(q^n\) \(b_n\). This series is a modular form of level 11: a function with an enormous symmetry group of transformations of the upper half-plane. The surprise:

\(p\)	2	3	5	7	13	17	19	23	29	31	37	41
\(N_p\)	5	5	5	10	10	20	20	25	30	25	35	50
\(a_p = p + 1 - N_p\)	\(-2\)	\(-1\)	\(1\)	\(-2\)	\(4\)	\(-2\)	\(0\)	\(-1\)	\(0\)	\(7\)	\(3\)	\(-8\)
\(b_p\)	\(-2\)	\(-1\)	\(1\)	\(-2\)	\(4\)	\(-2\)	\(0\)	\(-1\)	\(0\)	\(7\)	\(3\)	\(-8\)

\(a_p = b_p\) for every prime \(p \ne 11\) (the one prime where the curve has bad reduction). A question about solving an equation mod \(p\), for all \(p\) at once, is answered by one analytic function.

Try it in Python (standard library only):

def points(p):                       # solutions of y^2 + y = x^3 - x^2 mod p, plus the point at infinity
    return 1 + sum((y*y + y - x**3 + x*x) % p == 0 for x in range(p) for y in range(p))

N = 50
c = [0] * N; c[1] = 1                # start with the series q
for n in range(1, N):
    for k in (n, n, 11*n, 11*n):     # multiply by (1 - q^n)^2 (1 - q^11n)^2, one factor at a time
        if k < N:
            c = [c[i] - (c[i - k] if i >= k else 0) for i in range(N)]

for p in (2, 3, 5, 7, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47):
    print(p, p + 1 - points(p), c[p])  # the two columns agree

The modularity theorem. Every elliptic curve over \(\mathbb{Q}\) matches a modular form in this way. Andrew Wiles, with Richard Taylor, proved it for the curves needed for Fermat's last theorem (1995); Christophe Breuil, Brian Conrad, Fred Diamond and Richard Taylor completed it for all elliptic curves over \(\mathbb{Q}\) (2001). In the language of Chapter 21: the 2-dimensional Galois representation of the curve corresponds to an automorphic representation of \(GL_2\).

The conjectures. Langlands's reciprocity conjecture says, roughly: every \(n\)-dimensional Galois representation (of the right kind) corresponds to an automorphic representation of \(GL_n\), with matching data at almost every prime, as the \(a_p\) and \(b_p\) match above. His functoriality conjecture says automorphic representations of different groups are related whenever their "dual groups" are. Each verified case has been a major theorem: class field theory is the case \(n = 1\); the Langlands-Tunnell theorem (1980-1981), for 2-dimensional representations with solvable image, was a key input to Wiles's proof; Laurent Lafforgue proved the conjecture for \(GL_n\) over function fields (Fields Medal 2002); Ngô Bảo Châu proved the "fundamental lemma", a technical statement needed everywhere in the programme (Fields Medal 2010). A parallel geometric Langlands programme replaces numbers by curves over the complex numbers; in 2024 a team of nine mathematicians led by Dennis Gaitsgory and Sam Raskin posted a proof of its central conjecture in a series of papers running to several hundred pages. The original number-theoretic conjectures remain mostly open.

22.6 Computer algebra and Gröbner bases¶

Level: expert. How does a computer solve a system of polynomial equations exactly? For linear equations, Gaussian elimination (Chapter 3): eliminate variables until the system is triangular. For one variable, Euclid's algorithm: the gcd of several polynomials generates the ideal they generate. Gröbner bases do both at once, for any number of variables and any degrees.

Monomial orders. To divide polynomials in several variables we must say which term "leads". The lexicographic order with \(x > y > z\) compares monomials \(x^a y^b z^c\) by the exponent of \(x\) first, then \(y\), then \(z\), like a dictionary. So in \(f = x^2 + y^2 - 1\), the leading term is \(x^2\); in \(g = x - y^2\), it is \(x\).

The problem with division. Dividing by several polynomials in several variables gives remainders that depend on the order of the divisors, and a polynomial in the ideal \((f_1, \ldots, f_m)\) can leave a nonzero remainder. A Gröbner basis of an ideal \(I\) is a generating set \(G\) such that the leading term of every nonzero element of \(I\) is divisible by the leading term of some element of \(G\). With a Gröbner basis, division has a unique remainder, and \(f \in I\) exactly when the remainder is 0: ideal membership is decided by computation.

Buchberger's algorithm. Bruno Buchberger, in his 1965 PhD thesis under Wolfgang Gröbner (after whom he named them), gave an algorithm to compute them. For two polynomials, form the S-polynomial: multiply each by the monomial needed to make their leading terms equal, and subtract, so the leading terms cancel. Reduce it by the current basis; if a nonzero remainder is left, add it to the basis, and repeat. Buchberger's criterion says the process has finished (the basis is Gröbner) when every S-polynomial reduces to 0, and Hilbert's basis theorem (through Dickson's lemma on monomials) guarantees it finishes.

Worked example: the circle and the parabola of 22.2. \(f = x^2 + y^2 - 1\), \(g = x - y^2\), lex order \(x > y\).

Leading terms \(x^2\) and \(x\). The S-polynomial: \(f - x \cdot g = x^2 + y^2 - 1 - x^2 + xy^2 = xy^2 + y^2 - 1\).
Reduce by \(g\): its leading term \(xy^2\) is \(y^2 \cdot x\), so subtract \(y^2 g = xy^2 - y^4\), leaving \(y^4 + y^2 - 1\). Not zero: add \(h = y^4 + y^2 - 1\).
All new S-polynomials reduce to 0, and \(f\) itself is now redundant: \(f = (x + y^2)\,g + h\).

The reduced Gröbner basis is

\[ \{\, x - y^2,\ \ y^4 + y^2 - 1 \,\} . \]

It is triangular: the last polynomial involves only \(y\), and the first gives \(x\) from \(y\). That is the elimination we did by hand in 22.2, now as an algorithm. (The elimination theorem: with lex order, the basis elements free of \(x\) generate the ideal of equations in the remaining variables.)

A bigger example. Solve \(x^2 + y^2 + z^2 = 4\), \(x^2 + 2y^2 = 5\), \(xz = 1\). A lex Gröbner basis (\(x > y > z\)) is

\[ \{\, x + 2z^3 - 3z,\ \ y^2 - z^2 - 1,\ \ 2z^4 - 3z^2 + 1 \,\} . \]

The last factors as \((2z^2 - 1)(z^2 - 1)\), so \(z^2 = 1\) or \(z^2 = \tfrac12\). Then \(y^2 = z^2 + 1\) and \(x = 3z - 2z^3 = z(3 - 2z^2)\). With \(z = \pm 1\): \(y = \pm\sqrt2\), \(x = z\). With \(z = \pm\tfrac{1}{\sqrt2}\): \(y = \pm\sqrt{3/2}\), \(x = 2z = \pm\sqrt2\). Eight solutions. Check one: \(z = 1\), \(x = 1\), \(y = \sqrt2\): \(1 + 2 + 1 = 4\), \(1 + 4 = 5\), \(1 \cdot 1 = 1\).

Try it in Python (with sympy):

from sympy import symbols, groebner
x, y, z = symbols("x y z")
print(groebner([x**2 + y**2 - 1, x - y**2], x, y, order="lex"))
# GroebnerBasis([x - y**2, y**4 + y**2 - 1], ...)
print(groebner([x**2 + y**2 + z**2 - 4, x**2 + 2*y**2 - 5, x*z - 1], x, y, z, order="lex"))
# GroebnerBasis([x + 2*z**3 - 3*z, y**2 - z**2 - 1, 2*z**4 - 3*z**2 + 1], ...)

Uses and limits. Gröbner bases decide ideal membership, solve systems, compute the equation of a parametrized curve (implicitization: from \((t, t^2, t^3)\) they produce \(y = x^2\), \(z = x^3\)), prove geometry theorems automatically, and appear in robotics (inverse kinematics of arms) and in cryptanalysis. They can be very expensive: Ernst Mayr and Albert Meyer showed in 1982 that the worst case is doubly exponential in the number of variables, though typical problems are far better. Every computer algebra system (Macaulay2, Singular, Magma, Maple, Mathematica, SageMath, SymPy) implements them, along with the factoring algorithms of Chapter 18.

22.7 Open problems¶

Algebra is not finished. Here are some problems that are easy to state and unsolved as of this writing.

The inverse Galois problem. Is every finite group the Galois group of some extension of \(\mathbb{Q}\) (equivalently, of some polynomial with rational coefficients)? Known for \(S_n\) and \(A_n\) (Hilbert, 1892, via his irreducibility theorem), for all solvable groups (Igor Shafarevich, 1954), and for 25 of the 26 sporadic simple groups; the Mathieu group \(M_{23}\) is the one still open.
The Jacobian conjecture (Ott-Heinrich Keller, 1939). If a polynomial map \(F: \mathbb{C}^n \to \mathbb{C}^n\) has Jacobian determinant equal to a nonzero constant, then \(F\) has a polynomial inverse. Open for every \(n \ge 2\). Hyman Bass, Edwin Connell and David Wright showed in 1982 that it suffices to prove it for maps of degree 3 (in all dimensions). Many published "proofs" have turned out to be wrong.
Class numbers. Are there infinitely many real quadratic fields with class number 1? (Gauss.)
The Burnside problem. If a finitely generated group has \(x^n = 1\) for every element \(x\), must it be finite? No in general (Evgeny Golod and Shafarevich, 1964, without a fixed \(n\); Pyotr Novikov and Sergei Adian, 1968, for large odd \(n\)). Efim Zelmanov solved the "restricted" version in 1991 (Fields Medal 1994). But whether the free Burnside group \(B(2, 5)\), two generators and exponent 5, is finite is still unknown.
Kaplansky's conjectures on group rings. For a torsion-free group \(G\) and a field \(k\): the zero-divisor conjecture says \(k[G]\) has no zero divisors, and the unit conjecture said all its units are of the form \(\lambda g\). Giles Gardam found a counterexample to the unit conjecture in 2021; the zero-divisor conjecture remains open.
Hilbert's twelfth problem. Find explicit generators of all abelian extensions of a number field, as roots of unity do for \(\mathbb{Q}\) (the Kronecker-Weber theorem) and special values of elliptic functions do for imaginary quadratic fields. Recent work of Samit Dasgupta and Mahesh Kakde gives a \(p\)-adic answer for totally real fields; the general problem is open.
The Birch and Swinnerton-Dyer conjecture, one of the Clay Millennium Problems: the rank of the group of rational points on an elliptic curve equals the order of vanishing of its \(L\)-function (built from the \(a_p\) of 22.5) at \(s = 1\).
The Hadamard conjecture. A Hadamard matrix (entries \(\pm 1\), rows orthogonal) of order \(n > 2\) requires \(4 \mid n\). Is the converse true for every multiple of 4? The smallest open case is \(n = 668\).
And from Chapter 19: is \(e + \pi\) irrational?

Check yourself¶

1. Explain how a group \(G\) is a category with one object. What is a functor between two such categories?

One object \(\ast\); \(\operatorname{Hom}(\ast, \ast) = G\); composition is the group product (associative); the identity morphism is \(1 \in G\). Every morphism is an isomorphism (it has an inverse). A functor must send \(\ast\) to \(\ast\) and preserve composition and identity: that is exactly a group homomorphism.

2. In the poset of positive integers ordered by divisibility (\(a \to b\) when \(a \mid b\)), what is the product of \(a\) and \(b\)?

A product is an object \(P\) with \(P \to a\) and \(P \to b\) (so \(P \mid a\), \(P \mid b\)) such that any \(X\) with \(X \mid a\), \(X \mid b\) has \(X \mid P\). That is the definition of \(\gcd(a, b)\). (The coproduct, with arrows reversed, is the lcm.)

3. Show \(V(I) \cup V(J) = V(IJ)\) for ideals \(I, J\) in \(k[x_1, \ldots, x_n]\).

If \(a \in V(I)\), every product \(fg\) (\(f \in I\), \(g \in J\)) vanishes at \(a\), so \(a \in V(IJ)\); same for \(V(J)\). Conversely, if \(a \notin V(I) \cup V(J)\), pick \(f \in I\) with \(f(a) \ne 0\) and \(g \in J\) with \(g(a) \ne 0\); then \(fg \in IJ\) and \((fg)(a) \ne 0\), so \(a \notin V(IJ)\).

4. Over \(\mathbb{R}\), show that the Nullstellensatz fails for \(f = x^2 + y^2 + 1\). What happens over \(\mathbb{C}\)?

Over \(\mathbb{R}\), \(f > 0\) everywhere, so \(V(f) = \emptyset\), but \(1 \notin (f)\) (every nonzero multiple of \(f\) has degree at least 2), so the weak form fails. Over \(\mathbb{C}\), \(f\) has many zeros, for example \((i, 0)\) and \((0, i)\): \(V(f)\) is a complex curve.

5. Is \(\frac{1 + \sqrt{-5}}{2}\) an algebraic integer? Is \(\frac{1 + \sqrt{5}}{2}\)?

\(\alpha = \tfrac{1 + \sqrt{-5}}{2}\) has minimal polynomial \(x^2 - x + \tfrac{6}{4} = x^2 - x + \tfrac32\), not with integer coefficients; since the minimal polynomial of an algebraic integer has integer coefficients (by Gauss's lemma), \(\alpha\) is not one. \(\tfrac{1 + \sqrt5}{2}\) is a root of \(x^2 - x - 1\): yes. This is why \(\mathcal{O} = \mathbb{Z}[\sqrt{-5}]\) but \(\mathbb{Z}\big[\tfrac{1+\sqrt5}{2}\big]\) for \(\mathbb{Q}(\sqrt5)\) (\(5 \equiv 1 \pmod 4\)).

6. Show that \(2 + \sqrt3\) is a unit of \(\mathbb{Z}[\sqrt3]\) and find two solutions of \(a^2 - 3b^2 = 1\) with \(a > 2\).

\((2 + \sqrt3)(2 - \sqrt3) = 4 - 3 = 1\). Powers are units: \((2 + \sqrt3)^2 = 7 + 4\sqrt3\), giving \(49 - 48 = 1\); \((2 + \sqrt3)^3 = (7 + 4\sqrt3)(2 + \sqrt3) = 26 + 15\sqrt3\), giving \(676 - 675 = 1\).

7. Count the points of \(y^2 + y = x^3 - x^2\) mod 5 by hand and check that \(a_5 = 1\).

Values of \(y^2 + y\) for \(y = 0, \ldots, 4\): \(0, 2, 1, 2, 0\). Values of \(x^3 - x^2\) for \(x = 0, \ldots, 4\): \(0, 0, 4, 18 \equiv 3, 48 \equiv 3\). Matches: \(x = 0\) with \(y = 0, 4\); \(x = 1\) with \(y = 0, 4\); none for \(x = 2, 3, 4\) (values 4 and 3 never occur). Four affine points plus infinity: \(N_5 = 5\), and \(a_5 = 5 + 1 - 5 = 1\), the coefficient of \(q^5\).

8. Compute \(e^{X}\) for \(X = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\), and the commutator \([X, Y]\) with \(Y = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\).

\(X^2 = 0\), so \(e^X = I + X = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\) (a shear). \(XY = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\), \(YX = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\), so \([X, Y] = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\). With \(H = [X, Y]\), these are the standard basis \(E, F, H\) of \(\mathfrak{sl}_2\), the Lie algebra of \(SL_2\) (trace-zero matrices).

9. Check that \([L_x, L_y] = L_z\) for the matrices of 22.4.

\(L_x L_y\): row 2 of \(L_x\) is \((0, 0, -1)\), row 3 is \((0, 1, 0)\); the product is \(\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\). \(L_y L_x = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\). The difference is \(\begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = L_z\).

10. Find a lex Gröbner basis (\(x > y\)) of the ideal \((xy - 1,\ y^2 - 1)\), and solve the system.

S-polynomial: \(y \cdot (xy - 1) - x \cdot (y^2 - 1) = -y + x\), so \(x - y\) is in the ideal. Then \(xy - 1 = y(x - y) + (y^2 - 1)\) is redundant. Basis \(\{x - y,\ y^2 - 1\}\). Solutions: \(y = \pm 1\), \(x = y\): \((1, 1)\) and \((-1, -1)\).

11. Verify that \(n^2 + n + 41\) is prime for \(n = 0, 1, 2, 3\) and fails at \(n = 40\). Which class number fact is behind it?

\(41, 43, 47, 53\) are prime; at \(n = 40\), \(1600 + 40 + 41 = 1681 = 41^2\). (Also at \(n = 41\), \(41 \cdot 43\).) By Rabinowitsch's theorem the run of primes for \(n = 0, \ldots, 39\) is equivalent to \(\mathbb{Q}(\sqrt{-163})\) having class number 1, since \(1 - 4 \cdot 41 = -163\).

12. Why does Hilbert's basis theorem imply that the circle \(x^2 + y^2 = 1\) and any other variety can be described by finitely many equations, even if we start from an infinite list?

If \(X = V(S)\) for an infinite set \(S\), then \(X = V(I)\) where \(I\) is the ideal generated by \(S\). By Hilbert's theorem \(I = (f_1, \ldots, f_m)\) for finitely many \(f_i\), and \(V(I) = V(f_1, \ldots, f_m)\). (Each \(f_i\) is a combination of finitely many elements of \(S\), so even finitely many of the original equations suffice.)

What you learned¶

Categories, functors and natural transformations name the patterns shared by all algebraic structures; universal properties define products, free objects and quotients once for all.
Varieties are zero sets of polynomials; Hilbert's basis theorem makes them finitely defined, and the Nullstellensatz (over an algebraically closed field) matches radical ideals with varieties and maximal ideals with points.
Rings of integers of number fields are Dedekind domains with unique factorization of ideals; the class group measures the failure for elements, and the unit group is described by Dirichlet's theorem.
Lie groups are continuous symmetry groups; their Lie algebras, with the bracket \(XY - YX\), capture them near the identity, and the simple ones are classified.
The Langlands programme predicts that Galois representations match automorphic forms; for elliptic curves this is the modularity theorem, and you can watch it hold by counting points.
Gröbner bases turn polynomial systems into triangular ones by algorithm; many easy-to-state questions in algebra remain open.

Next: this is the last chapter. Return to the book's contents to revisit any chapter, or go on to the other books on the shelf.