Notesale: Turn your study into money

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Exercise and Solution 1

1 Let f (x) be a nonzero real-coefficient polynomial such that f (x) = f (f (x))
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Solution: Suppose f (x) = an xn + · · · + a1 x + a0 with an 6= 0
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2 Find the polynomial f (x) of the lowest degree such that the reminder of f (x)/(x − 1)2 is
2x and the reminder of f (x)/(x − 2)3 is 3x
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Definition 1
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With respect to a fixed element c ∈ R
(an integer or a polynomial), any two elements a, b ∈ R are said to be congruent with each
other, denoted by a ≡ b (mod c), if a − b = kc for some k ∈ R
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For instance, we have
28 ≡ 16 (mod 3) in Z, of course, you also have 28 ≡ 16 (mod 12),
3x + 2 ≡ x2 + 5x + 3 (mod (x + 1)2 ) in R[x]
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If a1 ≡ b1 (mod c) and a2 ≡ b2 (mod c), then a1 + a2 ≡ b1 + b2 (mod c)
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Suppose a ≡ b (mod c), then for any d ∈ R, ad ≡ bd (mod c)
One might think that the notion of congruence can be defined for general rings
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General
definition will be given when you study abstract algebra
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1

We first try to solve a simple case of congruent equations:
p(x) ≡ 1 (mod (x + 1)2 ),
p(x) ≡ 0 (mod (x + 2)3 )
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The polynomial
β(x) := b(x)(x + 2)3 is just a solution
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Actually, α(x) := a(x)(x + 1)2 is just a solution to this equations
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Now we are left to show how to find a(x) and b(x), but the procedure is already given in
the lecture notes: let us apply the Euclidean algorithm to (x + 2)3 and (x + 1)2
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1
2
1
(x − 1)2 = (3x − 4)( x − ) + or simply, 9(x − 1)2 = (3x − 4)(3x − 2) + 1
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Substituting (3x2 − 14x + 17) and −(3x − 2) for a(x) and b(x), we conclude that p(x) =
3x5 − 20x4 + 48x3 − 48x2 + 19x is one solution to the original equations
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After knowing this, we try to divide p(x) by (x − 1)2 (x − 2)3 and obtain
p(x) = 3(x + 1)2 (x + 2)3 + 4x4 − 27x3 + 66x2 − 65x + 24
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The total process of solving this problem is known as the celebrated Chinese
reminder theorem (a formal proof in given in the following for people familiar with the
language of rings and ideals)
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Suppose R is a commutative ring with a collection of
ideals I1 ,
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For any elements a1 ,
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Equivalently, this says that the map
φ:R→

R
R
× ··· ×
I1
I1

defined by natural projections is surjective
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I1 ∩ · · · ∩ In
I1
I1

This further implies that for any element b ∈ R, it satisfies the above congruent equations if and only
if b ≡ a (mod I1 ∩ · · · ∩ In )
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We first try to solve a simple case of congruent equations:
x ≡ 1 (mod I1 ),
x ≡ 0 (mod Ii ) for 2 ≤ i ≤ n
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Q i > 1, we can find some ai Q
Within the product 1 = i>1 (ai + bi ), the part β := i>1 bi ∈ i>1 Ii and the reminder α
of this product is contained in I1
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By a similar procedure, for each j = 2, 3
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Let us denote these solutions by β1 ,
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It is immediately
to see that a is a solution of the original congruent equations

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