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My purpose is to decide whether two different field extensions $K_1, K_2$ of $\mathbb Q_p$ have same ramification index or not, provided $[K_1:\mathbb Q_p]=[K_2: \mathbb{Q}_p]$ and $K_2=\mathbb{Q}_p(\sqrt[n]{-p})$ for some natural number $n$.

For example, consider two field extensions $K_1:=\mathbb{Q}_3(\sqrt[9]{1})$ and $K_2:=\mathbb{Q}_3(\sqrt[6]{-3})$ over the $3$-adic field $\mathbb Q_3$.

Note that $K_1$ is cyclotomic field with degree of extension $\varphi(9)=6$ and it is totally ramified since $9=3^2$.

In other word, $K_1/\mathbb{Q}_p$ has ramification index $6$.

Now consider the field extension $K_2$. The minimal polynomial of $\sqrt[6]{-3}$ is $x^6+3$, which is Eisenstein at prime $p=3$. So it is also totally ramified.

Therefore, the ramification index of $K_2/\mathbb Q_3$ is also $6$.

So I have mentioned two field extensions $K_1$ and $K_2$ of $\mathbb Q_3$ which are different but both has same degree of extension and same ramification index.

Certainly, the information that the degree of extensions of $K_1$ and $K_2$, being equal, was not enough for conclude their ramification index same.

So the question: Why $K_1$ and $K_2$ has same ramification index ? What made it possible here ? I mean, was there any general criterion of looking at the ramification index without doing calculations as I did here ?

More generally,

how to decide whether two different fields extensions $K_1, K_2$ of $\mathbb Q_p$ have same ramification index or not ?

provided that we know $[K_1: \mathbb{Q}_p]=[K_2: \mathbb{Q}_p]$ and $K_2$ is always of the form $\mathbb{Q}_p(\sqrt[n]{-p})$.

So what is the general criteria that their ramification inde will be same ? What might be the possible forms of the other field $K_1$ ?

Thanks

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2 Answers 2

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Since @PaulGarrett has given me the prompt, I’ll say something about higher ramification. This is a way of examining how close to each other the various roots $\lambda,\lambda'$ of the defining equation $f=0$ may be, when you choose $f$ so that one root $\lambda$ generates the ring of integers of the extension field $K$, $\mathcal O_K=\Bbb Z_p[\lambda]$. In the case of an unramified extension, as I recall, the differences between two roots will all be units. It’s the totally ramified extensions where higher ramification theory comes into its own.

Then let $\lambda$ be a uniformizing element of $\mathcal O_K$, $K$ totally ramified over $k$, with $v_k(\lambda)=1/e$ where $e$ is the ramification index (and the degree of the extension). We take $v_k$ normalized so that $v_k(k^\times)=\Bbb Z$. And let $f(x)$ be the minimal polynomial for $\lambda$, necessarily an Eisenstein polynomial over $\mathcal O_k$. Then put $g(x)=f(x+\lambda)$, whose roots are the quantities $\lambda'-\lambda$. We want a good way of measuring the size(s) of the roots of $g$.

Well, you measure the size of the roots of a polynomial with the Newton Polygon, but if you encode the same information in a kind of dual convex body that’s sometimes called the Newton Copolygon, you get something a little more useful. You can stretch this out horizontally by a factor of $e$ and consider the boundary to be the graph of a function, and thereby get the Hassse-Herbrand Transition Function $\psi^K_k$, which also has the wonderful property of being functorial: if $L\supset K\supset k$ is a chain of totally ramified extensions, then $\psi^L_k=\psi^K_k\circ\psi^L_K$.

Note that there are no Galois groups in this story. We can apply it to the two extensions mentioned at the beginning, $\Bbb Q_3(\zeta_9)$ and $\Bbb Q_3((-3)^{1/6})$. The two uniformizers are, respectively, $\tau=\zeta_9-1$ and $\pi=(-3)^{1/6}$. The minimal polynomial for $\zeta_9$ is $\frac{x^9-1}{x^3-1}$, so the minimal polynomial for $\tau$ is $f_1(x)=\frac{(x+1)^9-1}{(x+1)^3-1}$. The numerator and denominator have no constant, while the first-degree coefficient upstairs is $9$, downstairs is $3$, thus the quotient, known to be a polynomial, has constant coefficient $9/3=3$. Looking at the quotient modulo $3$, you see that there, the quotient is $x^6$. Thus $f_1$ is Eisenstein, which we already knew, didn’t we? Forming $g_1(x)=f_1(x+\tau)$, as above, we need to look closely at $g_1$ and its Newton Polygon (and Copolygon). It’s somewhat of a pain to do the computation this way, but you find that the $v_p$-value of the first coefficient is $3/2$, the value of the third is $1/2$ and of the sixth is zero. This gives vertices of the Polygon at $(1,3/2)$, $(3,1/2)$, and $(6,0)$. The Copolygon (in the Cartesian $(\xi,\eta)$-plane) is described by the corresponding inequalities $\eta\le\xi+3/2$, $\eta\le3\xi+1/2$, and $\eta\le6\xi$. The vertices are where two half-planes’ boundaries intersect, namely at $(1/6,1)$ and $(1/2,2)$. Stretch by a factor of six to get the points $(1,1)$ and $(3,2)$ (the “breaks” in traditional terminology), exactly what you’ll find in Serre’s Corps Locaux, except that my coordinatization of the Cartesian plane is off from Serre’s by one in each direction.

If you do the corresponding computation for $\sqrt[6]{-3}$, you’ll see that it’s much less messy, giving $g_2(x)=(x+\pi)^6+3$ with the Polygon’s vertices at $(1,11/6)$, $(3,1/2)$ and $(6,0)$. The Copolygon is defined by the corresponding inequalities $\eta\le\xi+11/6$, $\eta\le3\xi+1/2$, and $\eta\le6\xi$. Here, the half-planes’ boundaries intersect at $(1/6,1)$ and $(2/3,5/2)$. Stretching by six gives the vertices of the polygonal Hasse-Herbrand function at $(1,1)$ and $(4,5/2)$. Here, the “breaks” are not at integer points, which is fine, because the Galois group of the normal closure of our field is not abelian.

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  • $\begingroup$ Thank you Professor @Lubin $\endgroup$
    – MAS
    Commented Aug 9, 2022 at 2:51
  • $\begingroup$ Great! Very illuminating! Thanks! :) $\endgroup$ Commented Aug 9, 2022 at 22:56
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For abelian (Galois) extensions, understanding ramification is part of "local classfield theory". As a special case of that local classfield theory, the cyclic degree $n$ extensions $k$ are in natural bijection with the subgroups of index $n$ of $\mathbb Q^\times_p$, with Galois group isomorphic to the cyclic quotients $\mathbb Q_p^\times/N_{k/\mathbb Q^\times_p}(k^\times)$. In that case, the ramification degree is the index of the norms from the units of the bigger field in the units $\mathbb Z_p^\times$. So, in that special case, if we can "understand" the finite-index subgroups of $\mathbb Z_p^\times$ with cyclic quotient, we "understand" the ramifications of cyclic extensions.

Of course, the ground field doesn't have to be $\mathbb Q_p$, and abelian extensions are treatable as part of classfield theory.

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  • $\begingroup$ Thanks a lot. Does the isomorphism between the fields $K_1$ and $K_2$ preserve the ramification filtrations of their corresponding Galois groups ? In simple word, is ramification degree an invariant under isomorphism of fields ? I think it is not true as in my example, the two fields are not isomorphic but their ramification index is same. So the opposite question is-if two fields are isomorphic, can their ramification index be unequal ? $\endgroup$
    – MAS
    Commented Aug 7, 2022 at 3:49
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    $\begingroup$ Ramification degree is an isomorphism invariant of local field extensions ... Your example is sort of a contrapositive, that two not-isomorphic field extensions of a common base field ($\mathbb Q_p$) have the same ramification index. Further, the ramification filtrations (upper and lower indices?) are preserved under isomorphism, since they, too, are instrinsically defined. $\endgroup$ Commented Aug 7, 2022 at 3:56
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    $\begingroup$ Oop, typo, let me repair it. :) $\endgroup$ Commented Aug 7, 2022 at 4:02
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    $\begingroup$ The ramification filtrations of those two extensions are different. The really interesting conundrum is to distinguish between two extensions with the same (for instance) Hasse-Herbrand transition function. $\endgroup$
    – Lubin
    Commented Aug 8, 2022 at 2:07
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    $\begingroup$ Some of the audience here would be quite interested to hear more from Professor @lubin about further details and examples! :) I myself am just an amateur at algebraic number theory, an "end user"... :) $\endgroup$ Commented Aug 8, 2022 at 2:53

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