# Decomposition and Inertia Fields (April 15) ## Ramification in Composite Fields :::{.remark} Recall the setup: - $L/K$ is a Galois extension of number fields. - $P\in \spec K$ nonzero, and in fact we'll usually mean $P \in \mspec \ZZ_K$. - $Q\in \spec L$. - $\FF_Q \da \ZZ_L/Q, \FF_P \da \ZZ_K/P$ When $Q$ lies above $P$, we defined $D(Q/P)$ as all $\sigma\in G(L/K)$ preserving $Q$. We have a reduction map \[ \red_{Q/P}: D(Q/P) &\to G(\FF_Q / \FF_P) \\ \sigma &\mapsto \bar\sigma ,\] which we saw was a surjective group morphism with kernel $I(Q/P) \da \ker (\red_{Q/P})$. Since $D, I$ are subgroups of $G(\FF_Q/\FF_P)$, we can consider the corresponding intermediate fixed fields, the *inertia field* and *decomposition field*. Using that the Galois correspondence is inclusion-reversing, we get the following: \begin{tikzcd} L \\ \\ {L^{I(Q/P)}} \\ \\ {L^{D(Q/P)}} \\ \\ K \arrow[no head, from=7-1, to=5-1] \arrow[no head, from=5-1, to=3-1] \arrow[no head, from=3-1, to=1-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJMIl0sWzAsNiwiSyJdLFswLDIsIkxee0koUS9QKX0iXSxbMCw0LCJMXntEKFEvUCl9Il0sWzEsMywiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFszLDIsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMiwwLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d) ::: :::{.remark} Suppose we have a tower $L/M/K$ with prime ideals $Q/R/P$ with $R\da Q \intersect \ZZ_M$. We showed that with respect to containment, a. $L^{I(Q/P)}$ is the largest intermediate field in which $e(R/P) = 1$. b. $L^{D(Q/P)}$ is the largest intermediate field in which $e(R/P) = f(R/P) = 1$. Note that (b) is a stronger condition, so this matches the containment in the previous diagram. ::: :::{.theorem title="?"} Let $M/K$ be an extension of number fields, not necessarily Galois. Let $L$ be the **Galois closure** of $M/K$, the smallest field extension containing $M$ which is Galois over $K$. Then a. $P\in \spec K$ is unramified in $M$ $\iff$ $P$ is unramified in $L$. b. $P\in \spec K$ splits completely in $M$ $\iff$ $P$ splits completely in $L$. ::: :::{.proof title="?"} We'll just prove (a), and a very similar argument will yield (b). $\impliedby$: This is clear: if $L/M/K$ is a tower and $P$ is unramified in $L$, why is it unramified in $M$? This is multiplicativity of $e$ in towers. If we have $R$ in $M$ with $R/P$, we can choose $Q$ in $L$ with $Q/R/P$. If $e(R/P) > 1$, we have $e(Q/P) = e(Q/R) e(R/P) > 1$, forcing $P$ to ramify in $L$. Note that we haven't used that $L$ is the Galois closure. $\implies$: By the primitive element theorem, write $M = K(\theta)$, how can we describe the Galois closure $L$? Listing $\elts{\theta}{r}$ the roots of $\min_{\theta}(x)$ over $K$, then $L = K(\elts{\theta}{r})$: \begin{tikzcd} &&&& L = K(\elts{\theta}{r}) \\ \\ {K(\theta_1)} && {K(\theta_2)} && \cdots && {K(\theta_{r-1})} && {K(\theta_{r})} \\ \\ &&&& K \arrow[no head, from=3-1, to=1-5] \arrow[no head, from=3-3, to=1-5] \arrow[dashed, no head, from=3-5, to=1-5] \arrow[no head, from=3-7, to=1-5] \arrow[no head, from=5-5, to=3-1] \arrow[no head, from=3-3, to=5-5] \arrow[dashed, no head, from=3-5, to=5-5] \arrow[no head, from=3-7, to=5-5] \arrow[no head, from=3-9, to=5-5] \arrow[no head, from=3-9, to=1-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) By assumption, $P$ is unramified in $M$, which is one $K(\theta_i)$. There is always an isomorphism between any of the two $K(\theta_i)$ which preserves $K$, so abstractly they're all equivalent. So $P$ is unramified in $K(\theta_i)$ for all $i$. Take any $Q \in \spec L$ over $P$. We can consider the tower involving $K(\theta_i)$ for any $i$ and produce a corresponding prime $R_i$: \begin{tikzcd} L && Q \\ \\ {K(\theta_i)} && {R_i} \\ \\ K && P \arrow[no head, from=1-1, to=3-1] \arrow[no head, from=3-1, to=5-1] \arrow[no head, from=1-3, to=3-3] \arrow[no head, from=3-3, to=5-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJMIl0sWzAsMiwiSyhcXHRoZXRhX2kpIl0sWzAsNCwiSyJdLFsyLDQsIlAiXSxbMiwwLCJRIl0sWzIsMiwiUl9pIl0sWzAsMSwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFsxLDIsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbNCw1LCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzUsMywiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dXQ==) Since $P$ is unramified in all $K(\theta_i)$, we have $e(R_i/P) = 1$. Now use the characterization of the inertia field: $K(\theta_i) \subset L^{I(Q/P)}$, which is true for every $i$. So their composite is also contained in the inertia field, so we have \[ L = K(\elts{\theta}{r}) \subseteq L^{I(Q/P)} \subseteq L ,\] yielding equality. Now note that the inertia field was the largest intermediate field for which $e=1$, and we've just conclude $L = L^{I(Q/P)}$, so $e(Q/P) = 1$. Since $Q/P$ was arbitrary, $P$ is unramified. ::: :::{.theorem title="?"} Let $M_1/K, M_2/K$ be any two extensions of $K$. Then a. $P$ is unramified in $M_1$ and $M_2$ $\iff$ $P$ is unramified in the composite field $M_1 M_2$. b. $P$ splits completely in $M_1$ and $M_2$ $\iff$ $P$ splits completely in the composite field $M_1 M_2$. ::: :::{.proof title="?"} As before, we'll only prove (a). $\impliedby$: Again clear by multiplicativity in towers. $\implies$: Note that everything involving inertia only makes sense for Galois extensions, and we haven't assumed that here. So let $L$ be the Galois closure of $M_1M_2/K$, and let $L_1, L_2$ be the Galois closures of $M_1/K, M_2/K$ respectively. By the previous theorem, $P$ is unramified in $L_1, L_2$. How is $L$ related to $L_1$ and $L_2$? You can convince yourself that $L = L_1 L_2$, since the right-hand side is a Galois extension of $K$, and it contains $M_1 M_2$. We'll forget the $M_i$ and work with the following diagram: \begin{tikzcd} & {} \\ & L \\ {L_1} && {L_2} & {} \\ & K \arrow[no head, from=2-2, to=3-1] \arrow[no head, from=2-2, to=3-3] \arrow[no head, from=3-1, to=4-2] \arrow[no head, from=3-3, to=4-2] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMSwwXSxbMCwyLCJMXzEiXSxbMiwyLCJMXzIiXSxbMywyXSxbMSwxLCJMIl0sWzEsMywiSyJdLFs0LDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbNCwyLCIiLDIseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzEsNSwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFsyLDUsIiIsMix7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=) We have a prime in $K$ which is unramified in the $L_i$, and we want to show it's also unramified in $L$, which is equivalent to being unramified in $M_1 M_2$ (which was our goal). Strat: let $Q \in \spec L$ with $Q/P$ and let $R_1, R_2$ be the corresponding primes in $\spec L_1, \spec L_2$ between $Q$ and $P$. Since $P$ is unramified in the $L_i$, we have $e(R_1/P) = e(R_2/P) = 1$. Thus $L_i \subseteq L^{I(Q/P)}$, and so is their composite $L_1 L_2 \subseteq L^{I(Q/P)}$. But $L = L_1 L_2$, and so $e(Q/P) = 1$. ::: ## Frobenius :::{.remark} Let $E/F$ be an extension of *finite* fields, so $E/F$ is automatically Galois. Taking $F \da \FF_q$, then $E$ is of the form $E = \FF_{q^m}$ for some $m$. Then $G(E/F)$ is cyclic of order $m$, and is generated by $x\mapsto x^q$, the **Frobenius map** on the finite field $E/F$. For number fields, we'll want a slightly different notion of Frobenius. ::: :::{.remark} Let $L/K$ be a Galois extension of number fields, $P$ a prime of $K$ unramified in $L$, and $Q$ a prime of $L$ above $P$. Then $\FF_Q/\FF_P$ is an extension of finite fields, so we know $\red_{Q/P}$ is a surjective group morphism, and the inertia group has size $e(Q/P)$, which is 1 in this situation. So this is an isomorphism between the "global" object $D(Q/P)$ and the "local" object $G(\FF_Q/\FF_P)$ on residue fields. We can pull back the generator on the right-hand side to define the **Frobenius element** \[ \Frob_{Q/P} \da \red_{Q/P}\inv(x\mapsto x^q)\in D(Q/P) && q\da \# \ZZ_K/P \da N(P) .\] ::: :::{.observation} Note that $\Frob_{Q/P}$ is an element of $G(L/K)$ where for every \( \alpha\in \ZZ_L, \Frob_{Q/P}( \alpha) \equiv \alpha^q \mod Q \) where $q\da N(P)$. So $\Frob_{Q/P}$ acts by the $q$th power map in $\ZZ_L/Q$, and in fact this is a characterization of the Frobenius element. ::: :::{.lemma title="?"} Suppose \( \sigma\in \Gal(L/K) \) satisfies \( \sigma( \alpha) \equiv \alpha^q \mod Q \) for all \( \alpha\in \ZZ_L \) where \( q \da N(P) \). Then \( \sigma = \Frob_{Q/P} \). ::: :::{.remark} Why this is useful: you can check if something is the Frobenius element by just checking this congruence. ::: :::{.proof title="?"} Take any \( \alpha\in Q \), then \( \sigma(\alpha) \equiv \alpha^q\mod Q \equiv 0^q \mod Q \) and so \( \sigma( \alpha) \in Q \). So \( \sigma(Q) \subseteq Q \), but these are maximal ideals, forcing equality. Then $\sigma(Q) = Q$ implies $\sigma \in D(Q/P)$ by definition, and now applying $\red_{Q/P}( \sigma) = (x\mapsto x^q)$ is the $q$th power map and $\sigma = \red_{Q/P}\inv(x\mapsto x^q) \da \Frob_{Q/P}$. ::: :::{.proposition title="?"} Suppose $L/K$ is a Galois extension of number fields with $L/M/K$ where we additionally assume that $M/K$ is Galois. For the tower $L/M/K$, take primes $Q/R/P$ where $P$ is unramified in $L$. Then \[ \Frob_{R/P} = \ro{\Frob_{Q/P}}{M} .\] ::: :::{.remark} Note that the right-hand side is an automorphism of $L$ restricted to $M$, which is only an automorphism of $M$ when $M/K$ is Galois. ::: :::{.proof title="?"} Definition chasing and using characterization of Frobenius. It's enough to show that for all \( \alpha\in \ZZ_M \), we have \[ \ro{\Frob_{Q/P}}{M}( \alpha) \equiv \alpha^q \mod R && q = N(P) .\] Since \( \alpha\in \ZZ_M \) and $M \subseteq L$ so we can think of \( \alpha\in L \) and it makes sense to compute $\Frob_{Q/P}( \alpha) \equiv \alpha^q \mod Q$ since this is how Frobenius acts upstairs. So we just need to show that this congruence that holds for $Q$ also holds for $R$. Consider the difference: it's in $Q$ by the modular condition, and using that $M/K$ is Galois, the Frobenius restricts to an automorphism of $M$ and thus $\Frob_{Q/P}( \alpha) \in M$ (and is in fact still in $\ZZ_M$). Thus we have a difference of two things in $\ZZ_M$ and in $Q$, so \[ \Frob_{Q/P}( \alpha) - \alpha^q \in Q \intersect \ZZ_M = R .\] ::: :::{.remark} We'll see a nice example later of how to get the law of quadratic reciprocity from this! :::