### gateway to the p-adics

today’s writing project.

i played guitar (and even talked a little)

in church today, too, so it’s been

a pretty productive day for a sunday.

(i suppose. now that i think about it,

monday morning deadlines have led me

to many a *highly* productive sunday

here at the grading table. and grading

is the actual *work*…)

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Let f(x) = x^3 + x + 57.

(Find all x s.t.

f(x)\equiv 0 (mod 125).

)

Since 125=5^3, we begin by working “mod 5”:

f(0) = 57 == 2

f(1) = 59 == 4

f(2) = 67 == 2

f(3) = 87 == 4

f(4) == f(-1) == 50 == 0 (mod 5).

So f has exactly one “mod-5 root”

(namely 4) and we consider

f'(4) = 3(4)^2 + 5 = 21 == 1 \not == 0 (mod 5).

This means that 4 is a *non-singular* root.

Hensel’s Lemma (HL)

now tells us that there is

exactly one root of f (mod 125).

But now, since

f(4) = 4^3 + 4 + 57 = 125,

we are done:

x=4

is the *only* solution to f(x)==0 (mod 125).

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Let g(x) = x^3 + 10x^2 + x + 3.

(Find all x s.t.

g(x)\equiv 0 (mod 27).

)

Since 27=3^3, we begin by working “mod 3”:

g(0)==0

g(1)==0

g(2)==2 (mod 3).

Next, we’ll compute g'(x) = 3x^2 + 20x + 1

and evaluate it at each of our “mod 3” roots:

g'(0) =1 and g'(1) = 24 == 0 (mod 3).

The root at 1 is *singular*

(so “HL” isn’t helpful).

If “1” were lifted to a mod-9 root, x,

we would have x \in {1 +k*3} = {1, 4, 7}.

But

g(1) = 15 == 6,

g(4) = 231 == 1, and

g(7) = 843 == 6 (mod 9),

so there is *no* mod-9 root of f

satisfying x == 1 (mod 3)

(and so certainly no such mod-27 root).

The root at 0 is *non*-singular.

Since g'(0) = 1, its “mod-3 inverse” is

\bar{g'(0)} = 1.

“Plugging in” on “Hensel’s Formula”

(we have a_j = a_1 = a = 0;

our “f” is called “g”)

a_{j+1} = a_j – f(a_j)\bar{f'(a)}

(Display 2.6 of [NZM]) gives us

a_2 = 0 – 3(1) == -3 == 6 (mod 9).

Repeating the process,

a_3 = a_2 – g(a_2)\bar{g'(a)}

a_3 = 6 – 585(1)

a_3 = -579

a_3 = -(21*27 + 12).

a_3 == -12 == 15 (mod 27).

Our one-and-only solution

to f(x)==0 (mod 27) is x = 15.

(The result checks readily

[a calculator is helpful]:

f(15) = 5643 = 209*27 == 0 (mod 27).)

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