# Number Fields (Lec. 2,Tuesday, January 19) ## Embeddings :::{.remark} 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 $\ZZ[\sqrt 2], \ZZ[\sqrt{26}]$. Where do rings like this come from? ::: :::{.definition title="Number Field"} A **number field** is a subfield $K \subseteq \CC$ [^dontrequireKC] such that $[K: \QQ] < \infty$. [^dontrequireKC]: Some authors don't require $K \subseteq \CC$, but any finite extension of $\QQ$ will embed into $\CC$ so there's no harm in this extra requirement. ::: :::{.example title="of number fields"} Examples of number fields include - $\QQ[\sqrt[3]{2}]$, - $\QQ[\sqrt 2, \sqrt[5]{7}]$, or - $\QQ(\theta)$ where $\theta$ is a root of $x^5 - x - 1$, which one can check is irreducible. Note that the round vs. square brackets here won't make a difference, since we're adjoining *algebraic* numbers. ::: :::{.proposition title="Degree equals number of embeddings for finite extensions"} Let $K_{/\QQ}$ be a finite extension, say of degree $n\da [K: \QQ]$. Then there are $n$ distinct embeddings [^embedding_means] of $K$ into $\CC$ [^embedding_means]: An **embedding** is an injective ring morphism. ::: :::{.proof title="of proposition"} We have $K_{/\QQ}$, which is necessarily separable since $\ch(\QQ) = 0$. By the primitive element theorem, we can write $K = \QQ(\theta)$ where \( \theta \) is a root of some degree $n$ irreducible polynomial $f(x) \in \QQ[x]$. Since $\CC$ is algebraically closed, $f$ splits completely over $\CC$ as \[ f = \prod_{i=1}^n (x- \theta_i) \] with each \( \theta_i \in \CC \) distinct since $f$ was irreducible and we're in characteristic zero. Then for each $i$ there is an embedding $K = \QQ[\theta]$ given by \[ \iota_i: \QQ[\theta] &\injects \CC \\ g(\theta) &\mapsto g(\theta_i) .\] There are some easy things to check: - This is well-defined: elements in $K$ are polynomials in \( \theta \) but they all differ by a multiple of the minimal polynomial of \( \theta \), - This is an injective homomorphism and thus an embedding, and - For distinct $i$ you get distinct embeddings: just look at the image $\iota_i(\theta)$, these are distinct numbers in $\CC$. ::: :::{.definition title="Real and Nonreal embeddings"} Let $K_{/\QQ}$ be a finite extension of degree $n = [K : \QQ]$. We'll say an embedding \( \sigma:K \to \CC \) is **real** if \( \sigma(K) \subseteq \RR \) , otherwise we'll say the embedding is **nonreal**. ::: :::{.remark} If \( \sigma \) is a nonreal, then \( \bar \sigma \) is a nonreal embedding, so these 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. ::: :::{.example title="of computing the number of real and nonreal embeddings"} Let $K = \QQ(\sqrt[3]{2})$. Here $n=3$ since this is the root of a degree 3 irreducible polynomial. Using the proof we can find the embeddings: factor \[ x^3 - 2 = (x - \sqrt[3]{2})(x - \omega \sqrt[3]{2}) (x - \omega^2 \sqrt[3]{2}) .\] 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 $\RR$ and yields a real embedding, but the other two will be nonreal. So $r_1 = 1, r_2 = 1$, and we have $3 = 1 + 2(1)$, which is consistent. ::: ## Algebraic Closures :::{.remark} We've only been talking about fields, where unique factorization is trivial since there are no primes. There are thus "too many" units in fields when compared to the rings we were considering before, so we'll restrict to subrings of fields. The question is: where is the arithmetic? Given a number field $K$, we want a ring $\ZZ_K$ that fits this analogy: \begin{tikzcd} \QQ \ar[dd] & \leadsto & K\ar[dd] \\ \\ \ZZ & \leadsto & \ZZ_K = ? \end{tikzcd} ::: :::{.definition title="Algebraic Numbers"} Given \( \alpha\in \CC \) we say \( \alpha \) is an **algebraic number** if and only if \( \alpha \) is algebraic over $\QQ$, i.e. the root of some polynomial in $\QQ[x]$. ::: :::{.remark} We know that if we define $\bar \QQ \da \ts{\alpha\in \CC \st \alpha \text{ is algebraic over } \QQ}$, we can alternatively describe this as $\bar \QQ = \ts { \alpha\in \CC \st [\QQ(\alpha) : \QQ] < \infty }$. 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. ::: :::{.corollary title="Every number field is a subfield of $\bar\QQ$"} $\bar \QQ \injects \CC$ is a subfield and every number field is a subfield of $\bar \QQ$. ::: :::{.remark} These are still fields, so lets define some interesting subrings. ::: :::{.definition title="$\bar \ZZ$ "} Define $\bar \ZZ \da \ts{ \alpha\in \CC \st \alpha\text{ is the root of a monic polynomial }f\in \ZZ[x]}$. ::: :::{.theorem title="$\bar \ZZ$ is a ring"} $\bar \ZZ$ is a ring, and in fact a domain since it's a subring of $\CC$. ::: :::{.remark} We'll use an intermediate criterion to prove this: ::: :::{.proposition title="Integrality Criterion"} Let \( \alpha\in \CC \) and suppose there is a finitely generated $\ZZ\dash$submodule of $\CC$ with \( \alpha M \subseteq M \neq 0\). Then \( \alpha\in \bar \ZZ \), i.e. \( \alpha \) is the root of a monic polynomial with integer coefficients. ::: :::{.proof title="of integrality criterion"} 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 $\ZZ\dash$linear combination of the \( \beta_i \) . I.e. we have \[ \alpha \begin{bmatrix} \beta_1 \\ \vdots \\ \beta_n \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & \cdots \\ a_{21} & a_{22} & \\ \vdots & &\ddots \end{bmatrix} \begin{bmatrix} \beta_1 \\ \vdots \\ \beta_n \end{bmatrix} \da A \vec{\beta} ,\] where $A \in \Mat(n\times m, \ZZ)$. We can rearrange this to say that \[ \qty{ \alpha I - A} \begin{bmatrix} \beta_1 \\ \vdots \\ \beta_n \end{bmatrix} = \vector{0} .\] Not all of the $\beta_i$ can be zero since $M\neq 0$, and thus \( \alpha I - A \) is singular and has determinant zero, so $\det(x I - A)\evalfrom_{x=\alpha} = 0$. We have \[ x\id - A = \begin{bmatrix} x - a_{1,1} & & & \\ & x - a_{2, 2} & & \\ & & \ddots & \\ & & & x - a_{m, m} \end{bmatrix} ,\] where the off-diagonal components are constants in $\ZZ$ 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 $\det (x I - A) \in \ZZ[x]$ is monic. ::: :::{.proof title="of theorem"} We want to show that \( \bar \ZZ \) is a ring, and it's enough to show that - \( 1\in \bar \ZZ \), which is true since $x-1$ is monic. - It's closed under addition $(+)$ and multiplication $(\cdot)$. Note that the first property generalizes to $\ZZ \subseteq \bar \ZZ$, since $x-n$ is monic for any $n\in \ZZ$. For the second, let \( \alpha, \beta \in \bar \ZZ \). Define $M \da \ZZ[\alpha, \beta]$, then it's clear that \( (\alpha + \beta)M \subseteq M \) and \( (\alpha \beta)M \subseteq M \) since $\ZZ[\alpha, \beta]$ are polynomials in \( \alpha, \beta \) and multiplying by these expression still yields such polynomials. It only remains to check the following: :::{.claim} $M$ is finitely-generated. ::: :::{.proof title="?"} Let \( \alpha \) be a root of $f \in \ZZ[x]$ and \( \beta \) a root of $g$, both monic with $\deg f = n, \deg g = m$. We want to produce a finite generating set for \( M\da \ZZ[\alpha, \beta] \), and the claim is that the following works: \( \ts{ \alpha^i \beta^j} _{\substack{0\leq i < n \\ 0 \leq j < m} } \), i.e. every element of $M$ is some $\ZZ\dash$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 $0 \leq I, J \in \ZZ$, the term \( \alpha^I \beta^J \) is a $\ZZ\dash$linear combination of the restricted elements above. Divide by $f$ and $g$ to obtain \[ x^I &= f(x) q(x) + r(x) \\ x^J &= g(x) \tilde q(x) \tilde r(x) \] where $r(x) = 0$ or $\deg r < n$ and similarly for $\tilde r$, where (importantly) all of these polynomials are in $\ZZ[x]$. \ We're not over a field: $\ZZ[x]$ 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 $R\units$. 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 \[ \alpha^I \beta^J = r(\alpha) r(\beta) ,\] which is what we wanted. ::: ::: :::{.remark} We've just filled in another part of the previous picture: \begin{tikzcd} \QQ \ar[dd] & K\ar[dd] & \bar \QQ \ar[dd]\\ \\ \ZZ & \ZZ_K & \bar \ZZ \end{tikzcd} ::: ## Rings of Integers and Fraction Fields :::{.definition title="Ring of Integers"} Define $\ZZ_K = \bar \ZZ \intersect K$, the **ring of integers** of $K$. Note that this makes sense since the intersection of rings is again a ring. ::: :::{.remark} Why not just work in $\bar \ZZ$? 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 as $\sqrt{2} = \sqrt{\sqrt 2} \cdot \sqrt{\sqrt 2}$, noting that $\sqrt 2$ is not a unit, 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 a non-Noetherian ring. The following is a reality check, and certainly a property we would want: ::: :::{.proposition title="The ring of integers of $\QQ$ is $\ZZ$ "} $\ZZ_\QQ = \ZZ$. ::: :::{.proof title="of proposition"} \( \subseteq \): Easy, since \( \ZZ \subseteq \bar \ZZ \) and \( \ZZ \subseteq \QQ \), and is thus in their intersection \( \ZZ_\QQ \) . \ \( \supseteq \) : Let \( \alpha\in \ZZ_\QQ = \QQ \intersect\bar \ZZ \) , so \( \alpha \) is a root of \( x^n - a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \in \ZZ[x] \). We know \( \alpha= a/b \) with \( a,b \in \ZZ \), and we can use the rational root test which tells us that \( a\divides a_0 \) and \( b\divides 1 \), so \( b = \pm 1 \) and \( \alpha = a/\pm 1 = \pm a \in \ZZ \) and thus \( \alpha \in \ZZ \). ::: :::{.remark} We'll want to study $\ZZ_K$ for various number fields $K$, but we'll need more groundwork. ::: :::{.proposition title="Easy criterion to check if an integer is algebraic"} Let \( \alpha \in \bar \QQ \), then \[ \alpha\in \bar \ZZ \iff \min_ \alpha \in \ZZ[x], \] where \( \min_ \alpha(x) \) is the unique monic irreducible polynomial in $\QQ[x]$ which vanishes at \( \alpha \). ::: :::{.proof title="?"} $\impliedby$: Trivial, if the minimal polynomial already has integer coefficients, just note that it's already monic and thus \( \alpha \in \bar \ZZ \) by definition. \ $\implies$: Why should the minimal polynomial have *integer* coefficients? Choose a monic $f(x) \in \ZZ[x]$ with \( f(\alpha) = 0 \), using the fact that \( \alpha\in \bar \ZZ \) , and factor \( f(x) = \prod_{i=1}^n (x- \alpha_i) \in \CC[x] \). Note that each \( \alpha_i \in \bar \ZZ \) since they are all roots of $f$ (a monic polynomial in \( \ZZ[x] \)). Use the fact that \( \min_ \alpha(x) \) divides every polynomial which vanishes on \( \alpha \) over $\QQ$, and thus divides $f$ (noting that this still divides over $\CC$). 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 \( \CC \) to obtain \( \min_ \alpha(x) = \prod_{i=1}^m (x - \beta_i) \) with all of the \( \beta_i \in \bar \ZZ \). What coefficients appear after multiplying things out? Just sums and products of the \( \beta_i \), so all of the coefficients are in \( \bar \ZZ \) . Thus \( \min_ \alpha(x) \in \bar \ZZ [x] \). But the coefficients are also in $\QQ$ by definition, so the coefficients are in $\bar \ZZ \intersect\QQ = \ZZ$ and thus \( \min_ \alpha(x) \in \ZZ[x] \). ::: :::{.example title="Showing an integer is not algebraic using minimal polynomials"} $\sqrt{5}/ 3 \not\in \bar \ZZ$ since $\min_ \alpha(x) = x^2 - 5/9 \not\in \ZZ[x]$, so this is not an algebraic integer. ::: :::{.proposition title="$\ff(\ZZ_K) = K$ "} \envlist a. $\bar \ZZ$ has $\bar \QQ$ as its fraction field, and b. For any number field $K$, the fraction field of $\ZZ_K$ is $K$. c. If \( \alpha\in \bar \QQ \) then \( d \alpha\in \bar \ZZ \) for some \( d\in \ZZ^{\geq 0} \) Moreover, both (a) and (b) follow from (c). ::: :::{.remark} 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 $\alpha \in \bar \QQ$ as $\alpha=p/q$ with $p,q \in \bar \ZZ$. Use $c$ to produce $d\alpha\in \bar \ZZ$, then just take $d\alpha /d$. The same argument works for $b$. ::: :::{.exercise title="?"} Prove the proposition! ::: :::{.proposition title="The ring $\bar{\ZZ}$ contains all roots of monic polynomials with integer coefficients"} Suppose \( \alpha\in \bar \CC \) and \( \alpha \) is a root of a monic polynomial in \( \bar{\ZZ} [x] \). Then \( \alpha\in \bar \ZZ \). ::: :::{.remark} 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. ::: :::{.exercise title="Prove the proposition."} Prove this! One can use the integrality criterion (slightly challenging), or alternatively Galois theory. :::