1 Thursday, January 14

See website for notes on books, intro to class.

Paul’s description of the course:

“This course is an introduction to arithmetic”beyond \({\mathbb{Z}}\)“, specifically arithmetic in the ring of”integers" in a finite extension of \({\mathbb{Q}}\). (Among many other things) we’ll prove three important theorems about these rings:

1.1 Motivation

Solving Diophantine equations, i.e. polynomial equations over \({\mathbb{Z}}\).

Consider \(y^2 = x^3 + x\).

\((x, y) = (0, 0)\) is the only solution.

To see this, write \(y^2 = x(x^2+1)\), which are relatively prime, i.e. no \(D\in {\mathbb{Z}}\) divides both of them. Why? If \(d \mathrel{\Big|}x\) and \(d \mathrel{\Big|}x+1\), then \(d\mathrel{\Big|}(x^2+1) + (-x) = 1\). It’s also the case that both \(x^2+1\) and \(x^2\) are squares (up to a unit), so \(x^2, x^2 + 1\) are consecutive squares in \({\mathbb{Z}}\). But the gaps between squares are increasing: \(1, 2, 4, 9, \cdots\). The only possibilities would be \(x=0, y=1\), but in this case you can conclude \(y=0\).

Consider \(y^2 = x^3-2\).

\((3, \pm 5)\) are the only solutions.

Rewrite \begin{align*} x^3 = y^2+2 &= (y+ \sqrt{-2})(y - \sqrt{-2}) \\ &\in {\mathbb{Z}}[\sqrt{-2}] \mathrel{\vcenter{:}}=\left\{{a+b\sqrt{-2} {~\mathrel{\Big|}~}a,b,\in {\mathbb{Z}}}\right\} \leq {\mathbb{C}} .\end{align*} This is a subring of \({\mathbb{C}}\), and thus at least an integral domain. We want to try the same argument: showing the two factors are relatively prime. A little theory will help here:

For \(\alpha\in {\mathbb{Z}}[\sqrt{-2}]\) define \(N \alpha = \alpha\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu\).

Let \(\alpha, \beta \in {\mathbb{Z}}[\sqrt{-2}]\). Then

  1. \(N(\alpha \beta) = N(\alpha) N(\beta)\)

  2. \(N( \alpha) \in {\mathbb{Z}}_{\geq 0}\) and \(N(\alpha) = 0\) if and only if \(\alpha= 0\).

  3. \(N(\alpha) = 1 \iff \alpha\in R^{\times}\)

  1. Missing, see video (10:13 AM).

  2. \(N(\alpha) = a^2 + 2b^2 \geq 0\), so this equals zero if and only if \(\alpha= \beta= 0\)

  3. Write \(1 = \alpha\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu\) if \(N(\alpha) = 1 \in R^{\times}\). Conversely if \(\alpha\in R^{\times}\) write \(\alpha \beta = 1\), then \begin{align*} 1 = N(1) = N(\alpha \beta) = N(\alpha ) N(\beta ) \in {\mathbb{Z}}_{\geq 0} ,\end{align*} which forces both to be 1.

The two factors \(y \pm \sqrt 2\) are coprime in \({\mathbb{Z}}[\sqrt{-2}]\), i.e. every common divisor is a unit.

Suppose \(\delta\mathrel{\Big|}y\pm \sqrt{-2}\), then \(y + \sqrt{-2} = \delta \beta\) for some \(\beta\in {\mathbb{Z}}[\sqrt{-2}]\). Take norms to obtain \(y^2 + 2 = N \delta N \beta\), and in particular

In the original equation \(y^2 = x^3-2\), if \(y\) is even then \(x\) is even, and \(x^3 - 2 \equiv 0-2 \pmod 4 \equiv 2\), and so \(y^2 \equiv 2 \pmod 4\). But this can’t happen, so \(y\) is odd, and we’re done: we have \(N \delta\mathrel{\Big|}8\) which is even or 1, but \(N \delta\mathrel{\Big|}y^2 +2\) which is odd, so \(N \delta = 1\).

We can identify the units in this ring: \begin{align*} {\mathbb{Z}}[\sqrt{-2} ]^{\times}= \left\{{ a + b \sqrt{-2} {~\mathrel{\Big|}~}a^2 + 2b^2 = 1}\right\} \end{align*} which forces \(a^2 \leq 1, b^2 \leq 1\) and thus this set is \(\left\{{\pm 1}\right\}\).

So we have \(x^3 = ab\) which are relatively primes, so \(a,b\) should also be cubes. We don’t have to worry about units here, since \(\pm 1\) are both cubes. So e.g. we can write \begin{align*} y + \sqrt{-2} = (a + b \sqrt{-2} )^3 = (a^3-6ab^2) + (3a^2b -2b^3) \sqrt{-2} .\end{align*} Comparing coefficients of \(\sqrt{-2}\) yields \begin{align*} 1 = b(3a^2b - 2b^2) \in {\mathbb{Z}}\implies b \mathrel{\Big|}1 ,\end{align*} and thus \(b\in {\mathbb{Z}}^{\times}\), i.e. \(b\in \left\{{\pm 1}\right\}\). By cases:

Consider \(y^2 = x^3 - 26\). Rewrite this as \begin{align*} x^3 = y^2 + 26 = (y + \sqrt{-26} )(y - \sqrt{-26} ) ,\end{align*} then the same lemma goes through with \(2\) replaced by \(26\) everywhere where the RHS factors are still coprime. Setting \(y + \sqrt{-26} = (a + b \sqrt{-26} )^3\) and comparing coefficients, you’ll find \(b=1, a = \pm 3\). This yields \(x=35, y=\pm 207\). But there are more solutions: \((x, y) = (3, \pm 1)\)! The issue is that we used unique factorization when showing that \(ab\) is a square implies \(a\) or \(b\) is a square (say by checking prime factorizations and seeing even exponents). In this ring, we can have \(ab\) a cube with neither \(a,b\) a cube, even up to a unit.

When does a ring admit unique factorization? Do you even need it?

This will lead to a discussion of things like the class number, which measure the failure of unique factorization. In general, the above type of proof will work when the class number is 3!

2 Lecture 2 (Tuesday, January 19)

Today: Ch.2 of the book, “Cast of Characters.” Note that all rings will be commutative and unital in this course.

Last time: looked at factorization in \({\mathbb{Z}}[\sqrt 2], {\mathbb{Z}}[\sqrt{26}]\). Where do rings like this come from?

A number field is a subfield \(K \subseteq {\mathbb{C}}\) such that \([K: {\mathbb{Q}}] < \infty\).

Some authors don’t require \(K \subseteq {\mathbb{C}}\), but any finite extension of \({\mathbb{Q}}\) will embed into \({\mathbb{C}}\) so there’s no harm in this extra requirement.

\({\mathbb{Q}}[\sqrt[3]{2}, {\mathbb{Q}}[\sqrt 2, \sqrt[5]{7}]\) or \({\mathbb{Q}}(\theta)\) where \(\theta\) is a root of \(x^5 - x - 1\) (which you can check is irreducible. Now that the round vs. square brackets here won’t make a difference, since we’re adjoining algebraic numbers.

Let \(K_{/{\mathbb{Q}}}\) be a finite extension, say of degree \(n\mathrel{\vcenter{:}}=[K: {\mathbb{Q}}]\). Then there are \(n\) distinct embeddings1 of \(K\) into \({\mathbb{C}}\)

We have \(K_{/{\mathbb{Q}}}\), which is necessarily separable since \(\operatorname{ch}({\mathbb{Q}}) = 0\). By the primitive element theorem, we can write \(K = {\mathbb{Q}}(\theta)\) where \(\theta\) is a root of some degree \(n\) irreducible polynomial \(f(x) \in {\mathbb{Q}}[x]\). Since \({\mathbb{C}}\) is algebraically closed, \(f\) splits completely over \({\mathbb{C}}\) as \(f = \prod_{i=1}^n (x- \theta_i\) which each \(\theta_i \in CC\) distinct since \(f\) was irreducible and we’re in characteristic zero. Then for each \(i\) there is an embedding \(K = {\mathbb{Q}}[\theta]\) given by \begin{align*} \iota_i: {\mathbb{Q}}[\theta] &\hookrightarrow{\mathbb{C}}\\ g(\theta) &\mapsto g(\theta_i) .\end{align*} There are some easy things to check:

Let \(K_{/{\mathbb{Q}}}\) be a finite extension of degree \(n = [K : {\mathbb{Q}}]\). We’ll say an embedding \(\sigma:K \to {\mathbb{C}}\) is real if \(\sigma(K) \subseteq {\mathbb{R}}\) , otherwise we’ll say the embedding is nonreal.

If \(\sigma\) is a nonreal, then \(\mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu\) is a nonreal embedding, so this embeddings come in pairs. As a consequence, the total number of embeddings is given by \(n = r_1 + 2r_2\), where \(r_1\) is the number of real embeddings and \(r_2\) is the number of nonreal embeddings.

Let \(K = {\mathbb{Q}}(\sqrt[3]{2})\). Here \(n=3\) since this is the root of a degree 3 irreducible polynomial. Using the proof we can find the embeddings: factor \begin{align*} x^3 - 2 = (x - \sqrt[3]{2})(x - \omega \sqrt[3]{2}) (x - \omega^2 \sqrt[3]{2}) .\end{align*} where \(\omega = e^{2\pi i / 3}\) is a complex cube root of unity. We can form an embedding by sending \(\sqrt[3]{2} \to \omega^j \sqrt[3]{2}\) for \(j=0,1,2\). The case \(j=0\) sends \(K\) to a subset of \({\mathbb{R}}\) and yields a real embedding, but the other two will be nonreal. So \(r_1 = 1, r_2 = 1\), and we have \(3 = 1 + 2(1)\) and this is consistent.

We’ve only been talking about fields, since unique factorization is trivial since there are no primes. There are thus “too many” units, compared to the rings we were considering before, so we’ll restrict to subrings. The question is: where is the arithmetic? Given a number field \(K\), we want a ring \({\mathbb{Z}}_K\) that fits this analogy:

Given \(\alpha\in {\mathbb{C}}\) we say \(\alpha\) is an algebraic number if and only if \(\alpha\) is algebraic over \({\mathbb{Q}}\), i.e. the root of some polynomial in \({\mathbb{Q}}[x]\).

We know that if we define \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\mathrel{\vcenter{:}}=\left\{{\alpha\in {\mathbb{C}}{~\mathrel{\Big|}~}\alpha \text{ is algebraic over } {\mathbb{Q}}}\right\}\), we can alternatively describe this as \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu= \left\{{ \alpha\in {\mathbb{C}}{~\mathrel{\Big|}~}[{\mathbb{Q}}(\alpha) : {\mathbb{Q}}] < \infty }\right\}\). This is convenient because it’s easy to see that algebraic numbers are closed under sums and products, just using the ways degrees behave in towers.

\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\hookrightarrow{\mathbb{C}}\) is a subfield and every number field is a subfield of \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\).

These are still fields, so lets define some interesting subrings.

Define \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\mathrel{\vcenter{:}}=\left\{{ \alpha\in {\mathbb{C}}{~\mathrel{\Big|}~}\alpha\text{ is the root of a monic polynomial }f\in {\mathbb{Z}}[x]}\right\}\).

\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\) is a ring, and in fact a domain since it’s a subring of \({\mathbb{C}}\).

We’ll use an intermediate criterion to prove this:

Let \(\alpha\in {\mathbb{C}}\) and suppose there is a finitely generated \({\mathbb{Z}}{\hbox{-}}\)submodule of \({\mathbb{C}}\) with \(\alpha M \subseteq M \neq 0\). Then \(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\), i.e. \(\alpha\) is the root of a monic polynomial with integer coefficients.

Chasing definitions, take \(M\) and choose a finite list of generators \(\beta_1, \beta_2, \cdots, \beta_m\) for \(M\). Then \(\alpha M \subseteq M \implies \alpha \beta_i \in M\) for all \(M\), and each \(\alpha \beta_i\) is a \({\mathbb{Z}}{\hbox{-}}\)linear combination of the \(\beta_i\) . I.e. we have \begin{align*} \alpha \begin{bmatrix} \beta_1 \\ \vdots \\ \beta_n \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & \cdots \\ a_{21} & a_{22} & \\ \vdots & &\ddots \end{bmatrix} \begin{bmatrix} \beta_1 \\ \vdots \\ \beta_n \end{bmatrix} \mathrel{\vcenter{:}}= A \vec{\beta} ,\end{align*} where \(A \in \operatorname{Mat}(n\times m, {\mathbb{Z}})\). We can rearrange this to say that \begin{align*} \qty{ \alpha \operatorname{id}- A} \begin{bmatrix} \beta_1 \\ \vdots \\ \beta_n \end{bmatrix} = \mathbf{0} .\end{align*} Not all of the \(\beta_i\) can be zero since \(M\neq 0\), and thus \(\alpha \operatorname{id}- A\) is singular and thus has determinant zero, so \(\det(x \operatorname{id}- A)\Big|_{x=a} = 0\). We have \begin{align*} x\operatorname{id}- A = \begin{bmatrix} x - a_{1,} & & & \\ & x - a_{2, 2} & & \\ & & \ddots & \\ & & & x - a_{m, m} \end{bmatrix} ,\end{align*} where the off-diagonal components are constants in \({\mathbb{Z}}\) coming from \(A\). Taking the determinant yields a monic polynomial: the term of leading degree comes from multiplying the diagonal components, and expanding over the remaining minors only yields terms of smaller degree. So \(\det (x\operatorname{id}- A) \in {\mathbb{Z}}[x]\) is monic.

We want to show that \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\) is a ring, and it’s enough to show that

Note that the first property generalizes to \({\mathbb{Z}}\subseteq \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\), since \(x-n\) is monic for any \(n\in {\mathbb{Z}}\). For the second, let \(\alpha, \beta \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\). Define \(M \mathrel{\vcenter{:}}={\mathbb{Z}}[\alpha, \beta]\), then it’s clear that \((\alpha + \beta)M \subseteq M\) and \((\alpha \beta)M \subseteq M\) since \({\mathbb{Z}}[\alpha, \beta]\) are polynomials in \(\alpha, \beta\) and multiplying by these expression still yields such polynomials. It only remains to check the following:

\(M\) is finitely-generated.

Let \(\alpha\) be a root of \(f \in {\mathbb{Z}}[x]\) and \(\beta\) a root of \(g\), both monic with \(\deg f = n, \deg g = m\). We want to produce a finite generating set for \(M\mathrel{\vcenter{:}}={\mathbb{Z}}[\alpha, \beta]\), and the claim is that the following works: \(\left\{{ \alpha^i \beta^j}\right\} _{\substack{0\leq i < n \\ 0 \leq j < m} }\), i.e. every element of \(M\) is some \({\mathbb{Z}}{\hbox{-}}\)linear combination of these.

Note that this is clearly true if we were to include \(n, m\) in the indices by collecting terms of any polynomial in \(\alpha, \beta\), so the restrictions are nontrivial. It’s enough to show that for any \(0 \leq I, J \in {\mathbb{Z}}\), the term \(\alpha^I \beta^J\) is a \({\mathbb{Z}}{\hbox{-}}\)linear combination of the restricted elements above. Divide by \(f\) and \(g\) to obtain \(x^I = f(x) q(x) + r(x)\) and \(x^J = g(x) \tilde q(x) \tilde r(x)\) where \(r(x) = 0\) or \(\deg r < n\) and similarly for \(\tilde r\), where (importantly) all of these polynomials are in \({\mathbb{Z}}[x]\).

We’re not over a field: \({\mathbb{Z}}[x]\) doesn’t necessarily have a division algorithm, so why is this okay? The division algorithm only requires inverting the leading coefficient, so in general \(R[x]\) admits the usual division algorithm whenever the leading coefficient is in \(R^{\times}\).

Now plug \(\alpha\) into the first equation to obtain \(\alpha^I = r(\alpha)\) where \(\deg r < n\), which rewrite \(\alpha^I\) as a sum of lower-degree terms. Similarly writing \(\beta^J = r(\beta)\), we can express \begin{align*} \alpha^I \beta^J = r(\alpha) r(\beta) ,\end{align*} which is what we wanted.

We’ve just filled in another part of the previous picture:

Define \({\mathbb{Z}}_K = \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\cap K\), the ring of integers of \(K\). Note that this makes sense since the intersection of rings is again a ring.

Why not just work in \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)? It doesn’t have the factorization properties we want, e.g. there are no irreducible elements. Consider \(\sqrt 2\), we can factor is into two non-units (noting that \(\sqrt 2\) is not a unit) as \(\sqrt{\sqrt 2} \cdot \sqrt{\sqrt 2}\), and it’s easy to check that if \(a\) is not a unit then \(\sqrt a\) is not a unit. So this would yield arbitrarily long factorizations, and is thus not Noetherian.

The following is a reality check, and certainly a property we would want:

\({\mathbb{Z}}_{\mathbb{Q}}= {\mathbb{Z}}\).

\(\subseteq\): easy, since \({\mathbb{Z}}\subseteq \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\) and \({\mathbb{Z}}\subseteq {\mathbb{Q}}\), and is thus in their intersection \({\mathbb{Z}}_{\mathbb{Q}}\) .

\(\supseteq\) : Let \(\alpha\in {\mathbb{Z}}_{\mathbb{Q}}= {\mathbb{Q}}\cap\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\) , so \(\alpha\) is a root of \(x^n - a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \in {\mathbb{Z}}[x]\). We know \(\alpha= a/b\) with \(a,b \in {\mathbb{Z}}\), and we can use the rational root test which tells us that \(a\mathrel{\Big|}a_0, b\mathrel{\Big|}1\), so \(b = \pm 1, \alpha = a/\pm 1 = \pm a \in {\mathbb{Z}}\) and thus \(\alpha \in {\mathbb{Z}}\).

We’ll want to study \({\mathbb{Z}}_K\) for various number fields \(K\), but we’ll need more groundwork.

Let \(\alpha \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\), then \begin{align*} \alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\iff \min_ \alpha \in {\mathbb{Z}}[x], \end{align*} where \(\min_ \alpha(x)\) is the unique monic irreducible polynomial in \({\mathbb{Q}}[x]\) which vanishes at \(\alpha\).

\(\impliedby\): Trivial, if the minimal polynomial already has integer coefficients, just note that it’s already monic and thus \(\alpha \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\) by definition.

\(\implies\): Why should the minimal polynomial have integer coefficients? Choose a monic \(f(x) \in {\mathbb{Z}}[x]\) with \(f(\alpha) = 0\), using the fact that \(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\) , and factor \(f(x) = \prod_{i=1}^n (x- \alpha_i) \in {\mathbb{C}}[x]\). Note that each \(\alpha_i \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\) since they are all roots of \(f\) (a monic polynomial in \({\mathbb{Z}}[x]\)). Use the fact that \(\min_ \alpha(x)\) divides every polynomial which vanishes on \(\alpha\) over \({\mathbb{Q}}\), and thus divides \(f\) (noting that this still divides over \({\mathbb{C}}\)). Moreover, every root of \(\min_ \alpha(x)\) is a root of \(f\), and so every such root is some \(\alpha_i\).

Now factor \(\min_ \alpha(x)\) over \({\mathbb{C}}\) to obtain \(\min_ \alpha(x) = \prod_{i=1}^m (x - \beta_i)\) with all of the \(\beta_i \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\). What coefficients appear after multiplying things out? Just sums and products of the \(\beta_i\), so all of the coefficients are in \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\) . Thus \(\min_ \alpha(x) \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu[x]\). But the coefficients are also in \({\mathbb{Q}}\) by definition, so the coefficients are in \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\cap{\mathbb{Q}}= {\mathbb{Z}}\) and thus \(\min_ \alpha(x) \in {\mathbb{Z}}[x]\).

\(\sqrt{5}/ 3 \not\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\) since \(\min_ \alpha(x) = x^2 - 5/9 \not\in {\mathbb{Z}}[x]\), so this is not an algebraic integer.

  1. \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\) has \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\) as its fraction field, and
  2. For any number field \(K\), \({\mathbb{Z}}_K\) has \(K\) as its fraction field.

Moreover, both of these statements follow from:

  1. If \(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\) then \(d \alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\) for some \(d\in {\mathbb{Z}}^{\geq 0}\)

Thus the subring is “big” in the sense that if you allow taking quotients, you recover the entire field. That \(c\implies a,b\): suppose you want to write \(\alpha \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\) as \(\alpha=p/q\) with \(p,q \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\). Use \(c\) to produce \(d\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\), then just take \(d\alpha /d\). The same argument works for \(b\).

Prove the proposition!

Suppose \(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{C}}\mkern-1.5mu}\mkern 1.5mu\) and \(\alpha\) is a root of a monic polynomial in \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu [x]\). Then \(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\).

This says that if a number \(\alpha\) is the root of a monic polynomial whose coefficients are algebraic integers, then \(\alpha\) itself is an algebraic integer coefficients. This corresponds to the fact that integral over integral implies integral in commutative algebra.

Prove this! Can use the integrality criterion (slightly challenging), can also use Galois theory.

3 Lecture 3 (Thursday, January 21)

Today: roughly corresponds to chapter 3 in the book. Goal: do all of the big theorems in the setting of quadratic number fields, then redo everything for general number fields.

3.1 Quadratic Number Fields

Simplest case: \({\mathbb{Q}}\), a degree 1 number field, so the next simplest case is degree 2.

A field \(K\) is a quadratic number field if and only if \(K\) is a number field and \([K: {\mathbb{Q}}] = 2\).

Some notation: if \(d\in {\mathbb{R}}^{\times}\), then \(\sqrt d\) means the positive square root of \(d\) if \(d \geq 0\), and if \(d<0\) this denotes \(i\sqrt{{\left\lvert {d} \right\rvert}}\).

If \(K\) is a quadratic number field, then \(K = {\mathbb{Q}}(\sqrt{d})\) for some squarefree2 \(d\in {\mathbb{Z}}\). Moreover, this \(d\) is uniquely determined by \(K\), so all quadratic number fields are parameterized by the set of squarefree integers.

Existence: Since \([K: {\mathbb{Q}}] = 2\), we have \(K\supsetneq {\mathbb{Q}}\) so pick \(\alpha\in K\setminus{\mathbb{Q}}\) then \(K = {\mathbb{Q}}(\alpha)\). Note that we could also furnish this \(\alpha\) from the primitive element theorem, although this is overkill here. So \(\alpha\) is a root of some degree 2 \(p\in {\mathbb{Q}}[x]\), and by scaling coefficients we can replace this by \(p\in {\mathbb{Z}}[x]\). So write \(p(x) = Ax^2 + Bx + C\), in which case we can always write \(\alpha = {-B \pm \sqrt{B^2 - 4AC} \over 2A}\) where \(A\neq 0\) since this would imply that \(\alpha\in{\mathbb{Q}}\). Writing \(\Delta\mathrel{\vcenter{:}}= B^2 - 4AC\), we have \(K = {\mathbb{Q}}(\alpha) = {\mathbb{Q}}(\sqrt{\Delta})\). This is close to what we want – it’s \({\mathbb{Q}}\) adjoin some integer – but we’d like it to be squarefree.

Now let \(f\in {\mathbb{Z}}^{\geq 0}\) be chosen such that \(f^2 \mathrel{\Big|}\Delta\) and \(f\) is as large as possible, i.e. the largest square factor of \(\Delta\). Writing \(\Delta = f^2 - d\) where \(d\) is whatever remains. Then \(d\) must be squarefree, otherwise if \(d\) had a square factor bigger than 1, say \(d = r^2 d'\), in which case \(f^2 r^2 > f^2\) would be a larger factor of \(\Delta\). So \(d\) is squarefree, and \(\Delta = f \sqrt d\) and thus \({\mathbb{Q}}(\Delta) = {\mathbb{Q}}(\sqrt{d})\).

Uniqueness: Well use some extra machinery.

Let \(K\) be a number field with \(K_{/{\mathbb{Q}}}\) Galois. For each \(\alpha\in K\) define \begin{align*} N(\alpha) &\mathrel{\vcenter{:}}=\prod_{\sigma\in \operatorname{Gal}(K_{/{\mathbb{Q}}})} \sigma(\alpha) && \text{the norm} \\ \operatorname{Tr}(\alpha) &\mathrel{\vcenter{:}}=\sum_{\sigma\in \operatorname{Gal}(K_{/{\mathbb{Q}}})} \sigma(\alpha) && \text{the trace} .\end{align*}

Why use these kind of sum at all? Applying any element in the Galois group just permutes the elements. Note that \(N( \alpha), \operatorname{Tr}( \alpha)\) are \(G(K_{/{\mathbb{Q}}}){\hbox{-}}\)invariant, and thus rational numbers in \({\mathbb{Q}}\). The norm is multiplicative, and the trace is additive and in fact \({\mathbb{Q}}{\hbox{-}}\)linear: \(\operatorname{Tr}(a \alpha + b \beta) = a \operatorname{Tr}( \alpha) + b \operatorname{Tr}( \beta)\) for all \(\alpha, \beta\in K\) and all \(a,b \in {\mathbb{Q}}\).

What do the norm and trace look like for a quadratic field? We can write \(K = \left\{{a + b \sqrt d {~\mathrel{\Big|}~}a,b \in {\mathbb{Q}}}\right\}\) and there is a unique (non-identity) element \(g\in \operatorname{Gal}(K_{/{\mathbb{Q}}})\) with \(\sigma(a + b \sqrt d = a - b \sqrt{d}\). We’ll refer to this automorphism as conjugation. We can compute \begin{align*} N(a + b \sqrt{d} ) &= a^2 - db^2 \\ \operatorname{Tr}(a + b \sqrt{d} ) &= 2a .\end{align*}

Returning to the proof, suppose otherwise that \(K = {\mathbb{Q}}(\sqrt{d_1} ) = {\mathbb{Q}}( \sqrt{d_2} )\) with \(d_1\neq d_2\) squarefree integers. Note that they must have the same sign, otherwise one of these extensions would not be a subfield of \({\mathbb{R}}\). We know \(\sqrt{d_1} \in {\mathbb{Q}}( \sqrt{d_2} )\) and thus \(\sqrt{d_1} = a + b \sqrt{d_2}\) for some \(a, b\in {\mathbb{Q}}\). Taking the trace of both sides, the LHS is zero and the RHS is \(2a\) and we get \(a=0\) and \(\sqrt{d_1} = b \sqrt{d_2}\). Write \(b = u/v\) with \(u,v\in {\mathbb{Q}}\). Squaring both sides yields \(v^2 d_1 = u^2 d_2\). Let \(p\) be a prime dividing \(d_1\); then since \(d_1\) is squarefree there is only one copy of \(p\) occurring in its factorization. Moreover there are an even number of copies of \(p\) coming from \(v^2\), thus forcing \(d_2\) to have an odd power of \(p\). This forces \(p\mathrel{\Big|}d_2\), and since this holds for every prime factor \(p\) of \(d_1\), we get \(d_1 \mathrel{\Big|}d_2\) since \(d_1\) is squarefree. The same argument shows that \(d_2 \mathrel{\Big|}d_1\), so they’re the same up to sign: but the signs must match and we get \(d_1 = d_2\).

Note that this results holds for every squarefree number not equal to 1.

If \(K = {\mathbb{Q}}( \sqrt{d} )\), what is the ring of integers \({\mathbb{Z}}_K\)? Some more machinery will help here.

Assume \(K_{/{\mathbb{Q}}}\) is a Galois number field and for \(\alpha\in K\) define \begin{align*} \varphi_{\alpha}(x) \mathrel{\vcenter{:}}=\prod_{ \sigma\in \operatorname{Gal}(K_{/{\mathbb{Q}}})} \qty{ x - \sigma(\alpha)} .\end{align*}

For the same reasons mentioned for the norm/trace, we get \(\varphi_{\alpha} \in {\mathbb{Q}}[x]\), and moreover \(\varphi_{ \alpha } (\alpha) = 0\).

When is \(\alpha\in {\mathbb{Z}}_K\)? We have the following criterion:

\begin{align*} \alpha\in {\mathbb{Z}}_K \iff \varphi_{ \alpha } (x) \in {\mathbb{Z}}[x] .\end{align*}

\(\impliedby\): This is easy, since if \(\varphi_\alpha\) is a monic polynomial with integer coefficients, meaning that \(\alpha\) is an algebraic integer and thus in \({\mathbb{Z}}_K\).

\(\implies\): If \(\alpha \in {\mathbb{Z}}_K\) then it’s the root of some monic polynomial in \({\mathbb{Z}}[x]\), and the same is true for \(\sigma(\alpha)\) and thus each \(\sigma(\alpha) \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\). So \(\varphi_{ \alpha}(x) \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu[x]\). We said \(\varphi_{ \alpha}\) has coefficients in \({\mathbb{Q}}\) too, and thus in \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\cap{\mathbb{Q}}= {\mathbb{Z}}\). So the problem is reduced to finding out when \(\varphi_{\alpha}(x)\) has integer coefficients.

If \(\deg(K_{/{\mathbb{Q}}}) = n\), then \begin{align*} \varphi_{ \alpha}( \alpha) = \prod x- \sigma(\alpha) = x^n - \operatorname{Tr}(\alpha)x^{n-1} + \cdots + (-1)^n N( \alpha) .\end{align*} If \(n=2\), these are the only terms, and so if \(K\) is a quadratic number field then \(\alpha\in K\) is in \({\mathbb{Z}}_K\) if and only if \(\operatorname{Tr}( \alpha), N(\alpha) \in {\mathbb{Z}}\).

Let \(K = {\mathbb{Q}}( \sqrt{5} )\), then is it true that \({\mathbb{Z}}_K = {\mathbb{Z}}[\sqrt{5} ]\)? Since \(1, \sqrt{5} \in {\mathbb{Z}}_K\), we have \(\supseteq\) since \(1, \sqrt{5}\) are algebraic. The answer is no: take \(\alpha\mathrel{\vcenter{:}}={1 + \sqrt{5} \over 2}\), then \(N( \alpha) -4/4 = -1\) and \(\operatorname{Tr}( \alpha) = 1\). These are integers, so \(\alpha\in {\mathbb{Z}}_K\), and in fact \(\alpha\) is a root of \(x^2 - x - 1 \in {\mathbb{Z}}[x]\).

Let \(K = {\mathbb{Q}}( \sqrt{d} )\) be a quadratic number field. Then if \(d = 2,3 \pmod 4\), then \({\mathbb{Z}}_K = \left\{{ a + b \sqrt{d} {~\mathrel{\Big|}~}a, b\in {\mathbb{Z}}}\right\}\). If \(d=1 \pmod 4\), then \({\mathbb{Z}}_K = \left\{{ {1 + b \sqrt{d} \over 2} {~\mathrel{\Big|}~}a,b\in {\mathbb{Z}},\, a\equiv b \pmod 2}\right\}\).

For \(d=1\), if \(a, b\) are even then we just recover the \(d=2,3\) case, so we’re picking up extra elements from when \(a,b\) both odd.

Let \(\alpha\in K\) and write \(\alpha = A + B \sqrt{d}\) with \(A, B\in {\mathbb{Q}}\).

Check that \(N( \alpha), \operatorname{Tr}( \alpha) \in {\mathbb{Z}}\) for both cases.

Assuming now that \(N( \alpha), \operatorname{Tr}( \alpha) \in {\mathbb{Z}}\), then \(A^2 - dB^2 \in {\mathbb{Z}}\). Multiply this by 2 to get \((2A)^2 - d(2B)^2 \in 4{\mathbb{Z}}\). Recalling that \(\operatorname{Tr}( \alpha) = 2 A\), we have \((2A)^2 \in {\mathbb{Z}}\) and thus \(d(2B)^2 \in {\mathbb{Z}}\) as well. The claim now is that \(2B \in {\mathbb{Z}}\): we know \(2B\in {\mathbb{Q}}\). If \(2B\not\in {\mathbb{Z}}\), then the denominator has some prime factor. This prime factor appears twice in \((2B)^2\), and \(d(2B)^2 \in {\mathbb{Z}}\) then means that two copies of \(p\) appear in \(d\) in order to cancel – however, we assumed \(d\) was squarefree. We now know that \(A, B \in {1\over 2}{\mathbb{Z}}\), so write \(A = (1/2)a'\) and \(B = (1/2)b'\). Writing \(\alpha= (1/2)a' + (1/2)b' \sqrt{d}\), we find that \(N( \alpha) = ((a')^2 - d(b')^2) / 4 \in {\mathbb{Z}}\). So the numerator is a multiple of 4, which yields \((a')^2 \equiv d(b')^2 \pmod 4\). We proceed by cases.

Case 1: \(d = 2,3 \pmod 4\). If \(b'\) is odd then \((b')^2 = 1\pmod 4\), which holds for any odd number. But then \((a')^2 = d(b')^2 = d \pmod 4\), which is a problem – squares modulo 4 can only be \(0\) or \(1\). This is a contradiction, so \(b'\) must be even. Then \((b')^2 \pmod 4 = 0\), which forces \(a' \equiv 0 \pmod 4\) and \(a'\) must be even. But if \(a', b'\) are both even, \((1/2)a', (1/2)b'\in {\mathbb{Z}}\) and we obtain \(\alpha\in {\mathbb{Z}}+ \sqrt{d} {\mathbb{Z}}\) .

Case 2: If \(d\equiv 1 \pmod 4\), then \((a')^2 \equiv (b')^2 \pmod 4\). We can conclude that \(a', b'\) are either both odd or both even, otherwise we’d get \(0\equiv 1 \pmod 4\), and thus we can write \(a' \equiv b' \pmod 2\). But this was exactly the condition appearing in the theorem.

Let \(K\) be a quadratic number field. Then we can reformulate the previous results as:

\begin{align*} {\mathbb{Z}}_K = \begin{cases} {\mathbb{Z}}[ \sqrt{d} ] & d \equiv 2,3 \pmod 4 \\ {\mathbb{Z}}\left[{1 + \sqrt{d} \over 2}\right] & d \equiv 1 \pmod 4. \end{cases} \end{align*}

We’ve also shown that \({\mathbb{Z}}_K\) is a free \({\mathbb{Z}}{\hbox{-}}\)module of rank 2, with basis either \(\left\{{ 1, \sqrt{d} }\right\}\) or \(\left\{{ 1, {1 + \sqrt{d} \over 2 } }\right\}\).

What is true for general number fields? Important theorem: \({\mathbb{Z}}_K\) is always a free \({\mathbb{Z}}{\hbox{-}}\)module, i.e. there always exists an integral basis. Surprisingly, the it’s not always true that \({\mathbb{Z}}_K = {\mathbb{Z}}[\ell]\) for \(\ell\) a single element.

4 Lecture 4 (Wednesday, January 27)

Today: the failure of unique factorization. Roughly corresponds to chapter 4: “Paradise Lost!”

Setup: \(K\) is a quadratic field, a degree 2 extension of \({\mathbb{Q}}\), which can be written as \(K = {\mathbb{Q}}(\sqrt{d})\) with \(d\) squarefree. Last time, we completely described \({\mathbb{Z}}_K\) (the algebraic integers in \(K\)): \begin{align*} {\mathbb{Z}}_K = \begin{cases} {\mathbb{Z}}[ \sqrt{d} ] & d \equiv 2,3 \pmod 4 \\ {\mathbb{Z}}\left[{1 + \sqrt{d} \over 2}\right] & d \equiv 1 \pmod 4. \end{cases} \end{align*} We saw that the second admitted a different description as \(\left\{{ {a + b \sqrt{d} \over 2}}\right\}\) where \(a,b\) are either both even or both odd. Note that we can do interesting arithmetic in \({\mathbb{Z}}_K\), but it’s not necessarily well-behaved: \({\mathbb{Z}}_K\) is not always a UFD. Letting \(d=-5\), we have \({\mathbb{Z}}_K = {\mathbb{Z}}[ \sqrt{-5} ]\) where \(6\) factors in two ways: \(6 = (1 + \sqrt{5} )(1 - \sqrt{-5} ) = (2)(3) = 6\).

Note that this isn’t quite enough to show failure of unique factorization, e.g. we can factor \(16 = (4)(4) = (2)(8)\). Here you should check that all 4 factors are irreducible, and that the factors on the right aren’t unit multiples of the ones on the left. For example, \(21 = (-7)(-3) = (7)(3)\), but the factors only differ by the unit \(-1\in {\mathbb{Z}}^{\times}\). The key to checking all of those: the norm map: \begin{align*} N(a + b \sqrt{-5} ) = (a + b \sqrt{-5} ) (a - \sqrt{-5} ) = a^2 + 5b^2 .\end{align*} where the second factor was the conjugate, i.e. the image of the element under the nontrivial element of the Galois group of \(K_{/{\mathbb{Q}}}\). If \(a + b \sqrt{-5} \in {\mathbb{Z}}_K\), then \(N(a + b \sqrt{-5} \in {\mathbb{Z}}_{\geq 0}\) and is equal to zero if and only if \(a + b \sqrt{-5} = 0\). Moreover, this is a unit if and only if its norm is 1,3 i.e. \(a^2 + 5b^2 = 1\), which forces \(b=0\) and \(a=\pm 1\). So \(U({\mathbb{Z}}[ \sqrt{-5} ] ) = \left\{{\pm 1}\right\}\).

We’ll show one of the factors is irreducible, \(1 + \sqrt{-5}\). Recall that \(x\in R\) a domain is irreducible if and only if whenever \(x = ab\), one of \(a,b\) is a unit. It itself is not a unit, since \(N(1 + \sqrt{-5 }) = 6 \neq 1\). So suppose \(1 + \sqrt{-5} = \alpha \beta\). Then \begin{align*} 6 = N(\alpha \beta ) = N( \alpha) N( \beta) ,\end{align*} and so up to reordering, we have \(N \alpha = 2, N \beta= 3\). Writing