Newton Polygons and Factoring polynomials over Local Fields

4 Pages·2005·0.069 MB·English

by  Andrei Jorza

Newton Polygons and Factoring polynomials over Local Fields
Andrei Jorza
March 8, 2005
1 Generalities
Let K be a number (cid:12)eld. We have seen that a (cid:12)nite place of K is a valuation v : K Z
! [f1g
such that v(xy) = v(x) + v(y);v(x + y) min(v(x);v(y)) and v(0) = . This de(cid:12)nes the
(cid:21) 1
v(x)
metric x v = qv(cid:0) where qv = #kv the size of the residue (cid:12)eld kv = v=}v. Here Kv is the
j j O
completion of K at v, is the ring of integers of K and } is the maximal ideal of the local
v v v
O
ring .
v
O
Let v be a (cid:12)nite place. Let f(x) K [x] be a polynomial f(x) = f +f x+ +f xn.
v 0 1 n
2 (cid:1)(cid:1)(cid:1)
The Newton polygon NP(f) is the lower convex hull of the points (0; );(n; ) P =
i
f 1 1 g[f
(i;v(f )) i = 0;1;:::;n . The NP(f) is a polygonal like formed by two vertical lines together
i
j g
with a set of lines of various slopes.
Proposition 1.1. Let f be a polynomial of degree n. If u is a root of f then there exists a
segment in NP(f) of slope equal to v(u).
(cid:0)
Proof. We have f + f u + + f un = 0. If minv(f ui) is uniquely attained, then the
0 1 n i
(cid:1)(cid:1)(cid:1)
nonarchimedean property of v would imply that v(f +f u + + f un) = min(v(f ui)) =
0 1 n i
(cid:1)(cid:1)(cid:1)
v(0) = which cannot be. So i = j such that v(f ui) = v(f uj). But this corresponds to
i j
1 9 6
the line of slope v(u) through P and P . The valuation v(f ui) is the place where this line
i j i
(cid:0)
intersectstheverticalaxisandthefactthatthisvaluationisminimalimpliesthatallthepoints
on NP(f) are on or above this line. So this line contains a segment of NP(f) which proves
the lemma.
Example 1.2. Let p be a prime number and let f(x) = x3+px2+px+p2 Qp[x]. According
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to the theory this will have one root of valuation 1 and two roots of valuation 1=2.
Proposition 1.3. Let f;g K [x] be two polynomials (f = f + +f xd;g = g + +g xe)
v 0 d 0 e
2 (cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)
such that all the slopes of NP(f) are less or equal to all the slopes of NP(g). Then NP(fg)
is obtained by adjoining NP(f) and NP(g) in the following explicit manner (here we interpret
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NP(f) as a piecewise linear function in x)
NP(f)(x)+NP(g)(0);x [0;d]
NP(fg)(x) = 2
(NP(f)(d)+NP(g)(x d);x [d;d+e]
(cid:0) 2
Example 1.4. f(x) = x3+px2+px+p2;g(x) = px2+x+1 and
(fg)(x) = px5+(p2+1)x4+(p2+p+1)x3+px2+px+p2:
Proof. (fg)(x) = d+eh xi where h = f g . If i [0;d] then
0 i i j i(cid:0)j 2
P P
v(h ) = v(g f + +g f + )
i 0 i j i j
(cid:1)(cid:1)(cid:1) (cid:0) (cid:1)(cid:1)(cid:1)
and v(g f ) = v(g )+v(f ) NP(g)(0)+NP(f)(i) with equality if NP(f)(i) = v(f ). For
0 i 0 i i
(cid:21)
j > 0 we still have v(g f ) NP(g)(j)+NP(f)(i j) > NP(g)(0)+NP(f)(i) because of
j i j
(cid:0) (cid:21) (cid:0)
the slope condition ( NP(g)(j) NP(g)(0) > NP(f)(i) NP(f)(i j)). This takes care
() (cid:0) (cid:0) (cid:0)
of the case i [0;d].
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Now assume that i [d;d+e]. Then h = f g + +f g + and the proof is
i d i d d j i+j d
2 (cid:0) (cid:1)(cid:1)(cid:1) (cid:0) (cid:0) (cid:1)(cid:1)(cid:1)
similar.
2 Factorization
This is a very nice theorem since it tells you that you can compose Newton polygons when
multiplying polynomials. Can we go the other way around? The answer is yes. But (cid:12)rst we
need a technical lemma:
Lemma2.1. Letc R. Writevc(f) = min(v(fi)+ic). Thenvc(fg) = vc(f)+vc(g);vc(f+g)
2 (cid:21)
min(v (f);v (g)).
c c
Proof. Let v (f) = v(f ) + ic;v (g) = v(g ) + jc. So v (f) + v (g) = v(f g ) + (i + j)c
c i c j c c i j
(cid:20)
v( f g )+(i+j)c) v (fg) because v(h ) minv(f )+v(g ). In the other direction,
u i+j u c i j i j
(cid:0) (cid:20) (cid:21) (cid:0)
v (fg) v(h )+(i+j)c = v( f g )+(i+j)c. If k = 0 then v(f g )+(i+j)c >
c i+j i k j+k i k j+k
P (cid:20) (cid:0) 6 (cid:0)
v (f)+v (g) by choice of i and j. So v( f g )+(i+j)c = v(f g )+(i+j)c = v (f)+
c c i k j+k i j c
P (cid:0)
v (g).
c
P
Lemma 2.2. This essentially bounds the quantities in the division with remainder. Let f;h
2
K [x] with degf = d and v (f) = v(f )+dc. Write h = qf +r division with remainder. Then
v c d
v (q) v (h) v (f) which in turn implies that v (r) v (h).
c c c c c
(cid:21) (cid:0) (cid:21)
Proof. If h has degree n and let degf = d; write q = q + + q xn d. If we show by
0 n d (cid:0)
induction on i that v (q xn d i) v (h) v (f) then we (cid:1)a(cid:1)r(cid:1)e done(cid:0).
c n d i (cid:0) (cid:0) c c
(cid:0) (cid:0) (cid:21) (cid:0)
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For each i n d there is no contribution from r in the formula for h so h xn i =
n i n i (cid:0)
f q xn i+(cid:20)f (cid:0)q xn i+ . (cid:0) (cid:0)
d n d i (cid:0) d 1 n d i+1 (cid:0)
N(cid:0)ot(cid:0)e that v (f(cid:0)q (cid:0) (cid:0)xn i) = v ((cid:1)f(cid:1))(cid:1)+v (q xn i d) by the hypothesis on f. Also, from
c d n d i (cid:0) c c n d i (cid:0)(cid:0)
the inductive hypoth(cid:0)esi(cid:0)s we get that v (q (cid:0) (cid:0) xn d (i j)) v (h) v (f) which implies
c n d (i j) (cid:0) (cid:0) (cid:0) c c
that v (f q xn i) v (h). So (cid:0) (cid:0) (cid:0) (cid:21) (cid:0)
c d j n d i+j (cid:0) c
(cid:0) (cid:0) (cid:0) (cid:21)
v (f q xn i) = v (h xn i f q xn i) v (H);
c d n d i (cid:0) c n i (cid:0) d j n d i+j (cid:0) c
(cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:21)
j>0
X
which implies the result for i given the result for i j for j > 0.
(cid:0)
Now v (r)= v (h fq) min(v (h);v (f)+v (q)) = v (h).
c c c c c c
(cid:0) (cid:21)
The reason why this technical lemma is important is that it gives an algorithmic way to
approximate factorizations of polynomials.
Theorem 2.3. Let h K [x] be apolynomialof degree d+e andlet d be a pointof discontinuity
v
2
in NP(h). We saw in Proposition 1.3 that such Newton polygons arise when h is the product
of a polynomial of degree d and one of degree e. This theorem states that each h arise in such
manner.
Proof. As mentioned, the prood will be algorithmic. Let f = h + h xd, the (cid:12)rst d terms
0 0 d
(cid:1)(cid:1)(cid:1)
in the expansion of h and let g = 1. Choose c such that v (h) = v (h xd) and such that d is
0 c c d
the smallest index with this property (v (h) = v (h xi);i > d).
c c i
6
Now v (h f g ) = " > 0 and v (f ) = v (h). We’ll construct f ;g such that degf =
c 0 0 c 0 c i i i
(cid:0)
d;degg n d, v (f ) = v (h), v (f f ) v (h) + i", v (g g ) i" and (cid:12)nally
i c i c c i i+1 c c i i 1
(cid:20) (cid:0) (cid:0) (cid:21) (cid:0) (cid:0) (cid:21)
v (h f g ) v (h)+(i+1)". Clearly this implies that f f;g g;f g fg;h so h = fg.
c i i c i i i i
(cid:0) (cid:21) ! ! !
Write h f g = qf +r, division with remainder. Take f = f +r;g = g +q. Let’s
i i i i+1 i i+1 i
(cid:0)
show that the conditions are satis(cid:12)ed. The conditions on the degrees are clearly satis(cid:12)ed.
Now v (f f ) = v (r) v (h f g ) and v (g g ) = v (q) v (h f g ) v (f ) by
c i+1 i c c i i c i+1 i c c i i c i
(cid:0) (cid:21) (cid:0) (cid:0) (cid:21) (cid:0) (cid:0)
Lemma 2.2. In particular this shows that v (f ) = v (f ) = v (h).
c i+1 c i c
By the inductive hypothesis we have v (h f g ) v (h)+ (i + 1)" so we have v (r)
c i i c c
(cid:0) (cid:21) (cid:21)
v (h)+(i+1)" which implies that v (f f ) v (h)+(i+1)", the (cid:12)rst condition. Also
c c i+1 i c
(cid:0) (cid:21)
v (q) v (h)+(i+1)" v (f ) = (i+1)" and so v (g g ) (i+1)".
c c c i c i+1 i
(cid:21) (cid:0) (cid:0) (cid:21)
Lastly, v (h (f + r)(g + q)) = v (h f g f q rg rq) = v (r rg rq) =
c i i c i i i i c i
(cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0)
v (r)+v (1 g q) but this is v (h)+(i+1)"+min(v (1 g );v (q)). The condition on
c c i c c i c
(cid:0) (cid:0) (cid:21) (cid:0)
g implies that v (1 g ) " and v (q) (i+1)". So we get what we want.
i c i c
(cid:0) (cid:21) (cid:21)
Problem 2.4. Let f K [x] such that NP(f) consists of one segment that contains no other
v
2
lattice points. Then f is irreducible.
3
Proof. Assume it is reducible. Then f = gh and each roots of g;h has to have the same
valuation as f so the NP of g and h have the same slope as that of f. But then we can put
NP(g) at the top of NP(f) and we get a lattice point on NP(f). So f is irreducible.
Problem 2.5. Factor the following polynomials x3+5x+25;x3+5x2+25 Q5[x].
2
3 Galois Groups
Let L =K be a Galois extension.
w v
Lemma 3.1. Let (cid:11);(cid:12) L such that v((cid:12) (cid:27)(cid:11)) < v((cid:27)(cid:11) (cid:11)) for any (cid:27) Gal(L =K )
w w v
2 (cid:0) (cid:0) 2 n
Gal(L =K ((cid:11))). Prove that (cid:11) K ((cid:12)).
w v v
2
Proof. Let (cid:27) Gal(L =K ((cid:12))). We want to show that (cid:27)(cid:11) = (cid:11) which is enough to prove
w v
2
the lemma. Assume that (cid:27)(cid:11) = (cid:11) so v((cid:11) (cid:27)(cid:11)) > v((cid:12) (cid:27)(cid:11)) = v((cid:27)((cid:12) (cid:11))) = v((cid:12) (cid:11)) =
6 (cid:0) (cid:0) (cid:0) (cid:0)
v((cid:12) (cid:27)(cid:11)+(cid:27)(cid:11) (cid:11)) min(v((cid:12) (cid:27)(cid:11));v((cid:11) (cid:27)(cid:11))) = v((cid:11) (cid:27)(cid:11)) which is a contradiction.
(cid:0) (cid:0) (cid:21) (cid:0) (cid:0) (cid:0)
Theorem 3.2 (Krasner). Let f K [x] be a monic irreducible polynomial of degree d. Let
v
2
x ;:::;x be the roots of f and let " = max v(x x )=2 and let C = max(d";v(f )). Assume
1 d i=j i j i
6 (cid:0)
that g is a polynomial of degree d in K [x] such that v (f g) > C. Then g is irreducible and
v 0
(cid:0)
K [x]=(f) = K [x]=(g). (This essentially says that the two Galois groups are equal.)
v (cid:24) v
Proof. SinceC > v(f )theNewtonpolygonsaysthatf = g (mod p )soifg factorsbyHensel’s
i (cid:24) v
lemma f factors. (I won’t prove Hensel’s lemma here.) So assume g is irreducible.
Let y ;:::;y be the roots of g. By dimension count enough to show that y K (x ). For
1 d i v j
2
this it is enough by Lemma 3.1 to show that there exist i and j such that for all k we have
v(y x ) > v(x x ), forthenv(y x ) > v(x x ) min(v(x y );v(y x )) = v(x y )
i j k j i j k j k i i j k i
(cid:0) (cid:0) (cid:0) (cid:0) (cid:21) (cid:0) (cid:0) (cid:0)
which gives the result.
C
Now, v(f(y )) = v(f(y ) g(y )) = v( (f g )yi) = (cid:21) = . But here the
j j (cid:0) j i (cid:0) i j (C +nv(yj)
polynomials are monic so v(y ) 0 so C Pv(f(y )) = v( (y x )) = v(y x ). For at
i j j k i k
(cid:21) (cid:20) (cid:0) (cid:0)
least one k we have v(y x ) C=d > " so the hypotheses are satis(cid:12)ed.
j k
(cid:0) (cid:21) Q P
Example 3.3. Let K=Q3 bethe extensionde(cid:12)ned by the polynomial f(x) = x4 10x2+27x+1.
(cid:0)
Find Gal(K=Q3).
Proof. The idea is that g(x) = x4 10x2+1 has roots p2+ p3 so the Galois group of g is
(cid:0) (cid:6) (cid:6)
(Z=2Z)2.
So now all we need is that f and g satisfy the hypotheses of Theorem 3.2.
Westartoutwithg insteadoff. WouldliketohaveC = v (f g) = 3. ThenC > v(g ) = 0
0 i
(cid:0)
so the only inequality we want to check is that C > 4". But " = 1=2 so the hypotheses are
satis(cid:12)ed.
4