# Cyclotomic Fields (Lec. 15, Saturday, March 13) :::{.remark} This is chapter 14 continued. ::: :::{.definition title="Cyclotomic Fields"} A **cyclotomic field** is a number field $\QQ( \zeta_m)$ where \( \zeta_m \da e^{2\pi i / m} \), a primitive $m$th root of 1. ::: :::{.remark} The Kronecker-Webber theorem: any *abelian extension* $K/\QQ$ (so $\Gal(K/\QQ) \in \Ab$) is contained in a cyclotomic extension, and every cyclotomic field is an abelian extension. Given such a number field $K = \QQ( \zeta_m)$, what is $\ZZ_K$? ::: :::{.theorem title="The ring of integers of a cyclotomic field is given by adjoining a primitive root of unity"} For $K = \QQ( \zeta_m)$, \[ \ZZ_K = \ZZ[ \zeta_m ] .\] ::: :::{.remark} The degree of any such $K/\QQ$ is $\phi(m)$, and here $\phi(p) = p-1$. Also recall Eisenstein's criterion: if $p$ divides all of the coefficients of a polynomial $f(x) \da \sum a_i x^i$ but $p^2\notdivides a_0$, then $f$ is irreducible over $\QQ$. ::: :::{.lemma title="The minimal polynomials of roots of unity"} The minimal polynomial of $\zeta_p$ over $\QQ$ is \[ \Phi_p(x) \da x^{p-1} + x^{p-2} + \cdots + x + 1 ,\] and so $[\QQ (\zeta_p) : \QQ] = p-1$. ::: :::{.proof title="of lemma, a sketch"} Note that $\zeta_p$ is a root of $\Phi_p$, since \[ \Phi_p(x) = {x^p-1 \over x - 1} ,\] and $\zeta_p$ is a root of the numerator of the right-hand side and not of the denominator. This is irreducible by Eisenstein's criterion at $p$, using $x\mapsto x+1$. ::: :::{.proposition title="Eisenstein primes don't divide the extension degree"} Let \( \alpha\in \bar{\ZZ} \) be an algebraic integer such that \[ \min_ \alpha (x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 && \in \ZZ[x] \] is Eisenstein at the prime $p$. Let $K \da \QQ( \alpha)$, a number field of degree $n$. Then \[ p\notdivides [\ZZ_K : \ZZ[ \alpha] ] .\] ::: :::{.proof title="of proposition"} We first observe that \( \alpha^n \) is a multiple of $p$ in $\ZZ_K$. To see this, plug \( \alpha \) into the minimal polynomial to get $0 = \alpha^n + \cdots$ and solve for \( \alpha^n \) to obtain \[ \alpha^n = -(a_{n-1} \alpha^{n-1} + \cdots + a_1 \alpha + a_0) \equiv 0 \mod p \text{ in } \ZZ_K ,\] and this is a multiple of $p$ by the assumption on Eisenstein's criterion. We want to show $p$ doesn't divide $\# \ZZ_K/ \ZZ[ \alpha]$ as \(\ZZ\dash\)modules, identify the index as the size of this quotient. It suffices to show that $\ZZ_K/\ZZ[ \alpha]$ has no elements of order $p$, by applying Cauchy's theorem. If \( \beta\in \ZZ_K \) represents an element of order $p$ in the quotient, then \( p \beta\in \ZZ[ \alpha] \) and so \( p \beta = b_0 + b \alpha + \cdots + b_{n-1} \alpha^{n-1} \) for some $b_i \in \ZZ$. The order of \( \beta \) to be exactly $p$, so not all of the \( b_i \) are multiples of $p$: otherwise one could divide through by $p$ and conclude \( \beta\in \ZZ[ \alpha] \), making it zero in the quotient (and in particular, not of order $p$ as assumed). Suppose toward a contradiction that $i$ is the smallest index such that $p$ does not divide $b_i$. Then take this last equation mod $p$: \[ p \beta \equiv 0 \equiv b_i \alpha^{i} + \cdots + b_{n-1} \alpha^{n-1} \mod p .\] Now multiply by \( \alpha^{n-1-i} \) to obtain \[ 0 \equiv b_i \alpha^{n-1} + \cdots \equiv b_i \alpha^{n-1} \mod p ,\] where $p$ divides all of the other terms since they all contain a factor of \( \alpha^n \equiv 0 \mod p\). So \( b_i \alpha^{n-1} /p \in \ZZ_K \), and by a previous theorem, this forces \( N( b_i \alpha^{n-1} / p ) \in |ZZ \). But we can write \[ N \qty{ b_i \alpha^{n-1} \over p } &= N \qty{ b_i \over p} N( \alpha^{n-1} ) \\ &= \qty{b_i \over p}^{n} N( \alpha)^{n-1} \\ &= \qty{b_i \over p}^{n} \pm a_0 \\ &= \pm {b_i^n a_0^{n-1} \over p^n } \not\in \ZZ .\] where we've used that all embeddings fix rational numbers. But this is not an integer, since by Eisenstein $p^2$ does not divide $a_0$. So $a_0^{n-1}$ contributes exactly $n-1$ copies of $p$, leaving a $p$ in the denominator, and $p\notdivides b_i$ since we choose $i$ precisely to arrange for this. $\contradiction$ ::: :::{.remark} Recall some facts about the discriminant: let $F$ be a field and $f(x) \in F[x]$ monic. Then factor $f(x) = \prod_{i=1}^n (x - \alpha_i)$ over some splitting field. We then define \[ \discriminant(f) \da \prod_{i < j} ( \alpha_j - \alpha_i )^2 .\] We won't discuss the theory, but we'll use a few facts. ::: :::{.fact} For each fixed $n$ and all polynomials $f$ of degree $n$, $\discriminant(f)$ is given by a universal polynomial in the coefficients of $f$ with integer coefficients. For example, for $n=2$ and $f(x) = x^2 + bx + c$, we have \( \discriminant(f) = b^2 - 4c \in \ZZ[b,c] \). If $n=3$ and $f(x) = x^3 + bx^2 + cx + d$, we have \[ \discriminant(f) = 18bcd - 4b^3d + b^2c^2 - 4c^3 - 27d^2\in \ZZ[b,c,d] .\] So the discriminant is some polynomial expression in the coefficients, which (more importantly) have *integer* coefficients. Some consequences: - \( \discriminant(f) \in F \), despite the fact that the roots are generally not in $F$ and are instead in some splitting field. - If $F =\QQ$ and $f\in \QQ[x]$ *and* in fact $f\in \ZZ[x]$, then \( \discriminant(f) \in \ZZ \). - If $F = \QQ$ and $f\in \ZZ[x]$ with $q$ some prime, \[ \discriminant( f) \mod q = \discriminant(f \mod q) ,\] where we first take the discriminant to land in $\ZZ$ and then reduce to $\FF_q$, or we reduce $f\in \ZZ[x]$ to $f\mod q \in \FF_q[x]$ and take the discriminant using some algebraic close of $\FF_q$. ::: :::{.proof title="That $\ZZ_K = \ZZ\adjoin{\zeta_p}$ for $K = \QQ(\zeta_p)$."} To save space, we'll write $\zeta \da \zeta_p$. We want to show \( 1, \zeta, \cdots, \zeta^{p-2} \) forms an integral basis, from last time we have \[ \discriminant( 1, \zeta, \cdots, \zeta^{p-2}) = \discriminant_K [\ZZ_K : \ZZ[ \zeta ] ]^2 \implies [\ZZ_K : \ZZ[ \zeta ] ]^2 \divides \discriminant( 1, \zeta, \cdots, \zeta^{p-2}) .\] :::{.claim} The right-hand side is a power of $p$ (up to a sign), and hence so is the left-hand side. ::: We'll proceed by showing that the only prime that could divide the right-hand side is $p$. Suppose $q$ divides the right-hand side, i.e. $q \divides \discriminant(x^{p-1} + \cdots + x + 1)$. So this is zero mod $q$, and thus $\discriminant( x^{p-1} + \cdots + x + 1 \mod q) \equiv 0$. The discriminant was a product of roots, so it can only be zero if two roots coincide, so there is a multiple root of $x^{p-1} + \cdots + x + 1 \mod q$ and thus also of $x^p - 1$. So $x^p-1$ and its derivative $px^{p-1}$ have a root in common, and (check!) this can only happen if $q=p$. \ So $[\ZZ_K : \ZZ[\zeta] ] = p^\ell$ for some $\ell$. Using that fact that $\ZZ[ \zeta] \cong \ZZ[ \zeta - 1]$, we have \( [ \ZZ_K : \ZZ[ \zeta] ] = [ \ZZ_K: \ZZ[\zeta - 1] ] \). But by the previous lemma, we know that the minimal polynomial of $\zeta_p - 1$ is $\Phi(x+1)$, which is $p\dash$Eisenstein. So by that lemma, $p\notdivides [\ZZ_K: \ZZ[ \zeta - 1]]$, which forces $\ell = 0$ and $\ZZ_K = \ZZ[ \zeta_p ]$. ::: :::{.proof title="Sketch of the same proof for $K = \QQ(\zeta_m)$"} \envlist 1. Do roughly the same proof for prime powers $m = p^\ell$ 2. Show that if $a, b\in \ZZ^{\geq 0}$ are coprime then $\discriminant_a \da \discriminant_{\QQ( \zeta_a)}, \discriminant_b \da \discriminant_{\QQ(\zeta_b)}$ are coprime. These are defined in terms of integral bases, and we're trying to prove that something *is* an integral basis, so how do you show this if you don't know your basis is integral to begin with? Without knowing the exact values of the discriminants, you can show $\discriminant_a \divides a^?$ divides some power of $a$, and the same for $b$, and so $a, b$ coprime will make $a^?, b^?$ coprime as well. This can be shown by computing the discriminant of a *candidate* integral bases rather than an actual one. 3. Use a key lemma: if $K_1, K_2$ are number fields with coprime discriminants, then considering the composite field, we have $\ZZ_{K_1 K_2} = \ZZ_{K_1} \ZZ_{K_2}$, a composite ring. 4. If $a, b$ are coprime, check that $\QQ(\zeta_a) \QQ(\zeta_b) = \QQ(\zeta_{ab})$ and $\ZZ[ \zeta_a] \ZZ[ \zeta_b] = \ZZ [ \zeta_{ab} ]$. 5. Factor $m = \prod_{i} p_i^{\ell_i}$ and apply steps (3) and (4) inductively. ::: :::{.remark} The hard part is the lemma in (3). Also, questions about discriminants tend to come up during oral exams that include algebraic number theory. ::: ## Ideal Theory in General Number Rings (Ch. 15) :::{.remark} Here "number rings" means $\ZZ_K$ for $K$ a general number field. Let $K$ be a number field with $[K: \QQ] = n$. We'd want 1. $\Id(\ZZ_K)$ to be a UFM as a monoid, 2. $\Cl(\ZZ_K)$ is a finite group, Recall that we proved (1) and used it to deduce (2) for quadratic fields, whereas for the general case we'll prove (2) and deduce (1). The approach we'll take here is somewhat idiosyncratic -- the standard treatment involves the theory of Dedekind domains, which uses a lot of commutative algebra. This approach is more classic (circa 19th century, very concrete), and we'll skip over less important details (e.g. those that are unlikely to show up on oral exams). ::: :::{.definition title="Class Group of a Number Ring"} \[ \Cl(\ZZ_K) \da \Id(\ZZ_K)/ \sim ,\] where $\sim$ denotes dilation equivalence. ::: :::{.remark} Our strategy: - Prove $\Cl(\ZZ_K)$ is finite. - Prove $\Cl(\ZZ_K)$ is actually a group, i.e. there are inverses, so that for every ideal there is another ideal such that their product is principal (the "Principal Multiple Lemma"). - The remaining proofs from the quadratic field case go through almost word-for-word. ::: :::{.proposition title="Dilations of elements are always close to integers"} There is a constant $T = T(K)$ that only depends on $K$ such that for every \( \theta\in K \), there is a positive integer $t \leq T$ and a $\xi \in \ZZ_K$ such that \[ \abs{N(t \theta - \xi ) } < 1 .\] ::: :::{.remark} I.e. anything in the field can be multiplied by a bounded integer to make it close to something in the ring of integers. This proposition came up for imaginary quadratic fields in the Rabinowitz criterion, crucial for proving that the class group was generated by prime ideals which lie above small primes. ::: :::{.proof title="of proposition"} Omitted! See book, this proof wouldn't show up on an oral exam. This uses Dirichlet's approximation criterion again, although in a different way. ::: :::{.theorem title="The class group is finite"} \[ \# \Cl(\ZZ_K) < \infty .\] ::: :::{.proof title="that the class group is finite"} Very similar to how it goes for quadratic fields. As before, let $I \in \Id(\ZZ_K)$ be nonzero and $\beta \in I$ nonzero with $\abs{N \beta}$ minimal. \ :::{.claim} Let $T$ be as in the proposition, then $T! I \subseteq \gens{ \beta }$. ::: This follows from exactly the same argument as before. \ Now define \( J \da {T! \over \beta} I \subseteq \ZZ_K \), which is a dilation of $I$ and thus $J \normal \ZZ_K$ as well. By definition, $I\sim J$, i.e. $[I] = [J] \in \Id(\ZZ_K)$, and it's now enough to show that there are only finitely many possibilities for $J$, since then every class is equal to the class of one of finitely many such $J$. Since \( \beta\in I \), we can deduce that \( T! \in J \) and thus \( \gens{ T! } \subseteq J \). We'd like to say "to contain is to divide" (as in the case of unique factorization) and conclude $J\divides T!$, which only has finitely many divisors. However, we haven't proved this yet! We can use an algebra fact instead: \[ \correspond{ \text{Ideals of $\ZZ_K$ } \\ \text{containing } \gens{ T! } } &\mapstofrom \correspond{ \text{Ideals of } \ZZ_K / \gens{ T! } } ,\] so it's enough to show that the right-hand side is finite. This is "obvious", since $\# \ZZ_K / \gens{ T! } = (T!)^n$. This comes from the fact that $\ZZ_K \cong_{\Ab} \ZZ^n$, so as a \(\ZZ\dash\)module this is isomorphic to $\ZZ^n / T! \ZZ^n \cong (\ZZ/T! \ZZ)^n$, so this is a finite ring and can thus only have finitely many ideals.[^only_finite] [^only_finite]: In fact, we've already proved that $\ZZ_K / I$ for any nonzero ideal $I$ is finite. ::: :::{.remark} We now want to establish the cancellation law in $\Id(\ZZ_K)$, then the principal multiple lemma, and then everything else will follow as in the quadratic case. :::