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Contents - Student subdomain for University of Bath

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156 CHAPTER 5. P -ADIC METHODS<br />

written f (1) = g (1) h (1) (mod p 1 ) and our aim is to calculate a corresponding<br />

factorization f (k) = g (k) h (k) (mod p k ) such that p k is sufficiently large.<br />

Obviously, g (2) ≡ g (1) (mod p), and there<strong>for</strong>e we can write g (2) = g (1) +pĝ (2)<br />

where ĝ (2) is a measure <strong>of</strong> the difference between g (1) and g (2) . The same holds<br />

<strong>for</strong> f and h, so that f (2) = g (2) h (2) (mod p 2 ) becomes<br />

Since f (1) = g (1) h (1)<br />

f (1) + p ˆf (2) = (g (1) + pĝ (2) )(h (1) + pĥ(2) ) (mod p 2 ).<br />

(mod p 1 ), this equation can be rewritten in the <strong>for</strong>m<br />

f (1) − g (1) h (1)<br />

+<br />

p<br />

ˆf (2) = ĝ (2) h (1) + ĥ(2) g (1) (mod p). (5.5)<br />

The left hand side <strong>of</strong> this equation is known, whereas the right hand side depends<br />

linearly on the unknowns ĝ (2) and ĥ(2) . Applying the extended Euclidean<br />

Algorithm (4) ) to g (1) and h (1) , which are relatively prime, we can find polynomials<br />

ĝ (2) and ĥ(2) <strong>of</strong> degree less than g (1) and h (1) respectively, which satisfy<br />

this equation modulo p. The restrictions on the degrees <strong>of</strong> ĝ (2) and ĥ(2) are<br />

valid in the present case, <strong>for</strong> the leading coefficients <strong>of</strong> g (k) and h (k) have to be<br />

1. Thus we can determine g (2) and h (2) .<br />

Similarly, g (3) ≡ g (2) (mod p 2 ), and we can there<strong>for</strong>e write g (3) = g (2) +<br />

p 2 ĝ (3) where ĝ (3) is a measure <strong>of</strong> the difference between g (2) and g (3) . The same<br />

is true <strong>for</strong> f and h, so that f (3) = g (3) h (3) (mod p 3 ) becomes<br />

Since f (2) = g (2) h (2)<br />

f (2) + p 2 ˆf (3) = (g (2) + p 2 ĝ (3) )(h (2) + p 2 ĥ (3) ) (mod p 3 ).<br />

f (2) − g (2) h (2)<br />

Moreover, g (2) ≡ g (1)<br />

(mod p 2 ), this equation can be rewritten in the <strong>for</strong>m<br />

p 2 + ˆf (3) = ĝ (3) h (2) + ĥ(3) g (2) (mod p). (5.6)<br />

f (2) − g (2) h (2)<br />

(mod p), so this equation simplifies to<br />

p 2 + ˆf (3) = ĝ (3) h (1) + ĥ(3) g (1) (mod p).<br />

The left hand side <strong>of</strong> this equation is known, whilst the right hand side depends<br />

linearly on the unknowns ĝ (3) and ĥ(3) . Applying the extended Euclidean algorithm<br />

to g (1) and h (1) , which are relatively prime, we can find the polynomials<br />

ĝ (3) and ĥ(3) <strong>of</strong> degrees less than those <strong>of</strong> g (1) and h (1) respectively, which satisfy<br />

this equation modulo p. Thus we determine g (3) and h (3) starting from g (2) and<br />

h (2) , and we can continue these deductions in the same way <strong>for</strong> every power p k<br />

<strong>of</strong> p until p k is sufficiently large.<br />

We should note that Euclid’s algorithm is always applied to the same polynomials,<br />

and there<strong>for</strong>e it suffices to per<strong>for</strong>m it once. In fact, we can state the<br />

algorithm in Figure 5.4.<br />

To solve the more general problem <strong>of</strong> lifting a factorization <strong>of</strong> a non-monic<br />

polynomial, we adopt the same solution as in the g.c.d. case: we impose the<br />

leading coefficient in each factor to be the leading coefficient <strong>of</strong> the orgiinal<br />

polynomial (which we may as well assume to be primitive). We also address the<br />

problem <strong>of</strong> lifting n (rather than just 2) factors in Algorithm 26.

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