Download School of Mathematics and Statistics The University of Sydney

School of Mathematics and Statistics
The University of Sydney
MATH3062 NUMBER THEORY AND ALGEBRA, 2012, TERRY GAGEN
Tutorial 9
1. Find all integers x such that x2 − 6x + 8 = 0 ∈ Z105 using the Chinese Remainder Theorem.
(See Question 5 Tutorial 1)
Solution
So for example, the Question 5 Tutorial 1 Solve x2 − 6x + 8 = 0 ∈ Z15 etc should have
asked for the solution in Z105 .
Solve the equation in Z3 , Z5 and Z7 to get:
x ≡ 2, 4
(mod 3),
x ≡ 2, 4
(mod 5),
x ≡ 2, 4
(mod 7).
We will use the CRT to find x ∈ Z105 in each of the 8 possible cases. I prefer to do this
using the Primary School method. Note that CRT Mark 1 asserts that there is exactly one
solution modulo 105 and so if we find one, we can stop because we know there is only one
solution modulo the product of the moduli 105 in this case.
1.
x ≡ 2 (mod 3),
x≡2
(mod 5)
x≡2
(mod 7).
x≡2
(mod 3),
x≡2
(mod 5),
x≡4
(mod 7).
So x = 2 ∈ Z105 .
2.
Solving the first two congruences we have x = 2 (mod 15) and so x = 2, 17, 32, 47, 62, 77, 92.
We need x ≡ 4 (mod 7) and so x = 32 ∈ Z105 .
3.
x ≡ 2 (mod 3),
x≡4
(mod 5),
x≡2
(mod 7).
Solving the first and third congruences we have x = 2 (mod 21) and so x = 2, 23, 44, 65, 86.
We need x ≡ 4 (mod 5) and so x = 44 ∈ Z105 .
4.
x ≡ 2 (mod 3),
x≡4
(mod 5),
x≡4
(mod 7).
x ≡ 4 (mod 35) and x = 4, 39, 74 but x ≡ 2 (mod 3) and so x = 74 ∈ Z1 05.
5.
x≡4
(mod 3),
x≡2
(mod 5),
x≡2
(mod 7).
x≡4
(mod 3),
x≡2
(mod 5),
x≡4
(mod 7).
x≡4
(mod 3),
x≡4
(mod 5),
x≡2
(mod 7).
x≡4
(mod 3),
x≡4
(mod 5),
x≡4
(mod 7).
Leads to x = 37 ∈ Z105 .
6.
Leads to x = 67 ∈ Z105 .
7.
Leads to x = 79 ∈ Z105 .
8.
Leads to x = 4 ∈ Z105 .
So the equation x2 − 6x + 8 = 0 ∈ Z105 has solution x = 2, 4, 32, 37, 44, 67, 74, 79.
2. Consider Z[i]4+i .
(i ) Find an irredundant set of remainders and calculate the number of elements in Z[i]4+i .
(ii ) Note that 4 = −i ∈ Z[i]4+i and so 17 = 0 ∈ Z[i]4+i . Hence show that Z[i]4+i ∼
= Z17
∗
and hence find the generators for Z[i]4+i .
Solution
(i )
Draw the Gaussian integer multiples of 4 + i in the complex plane. Then
0
i 1+i 2+i 3+i
2i 1 + 2i 2 + 2i 3 + 2i
3i 1 + 3i 2 + 3i 3 + 3i
4i 1 + 4i 2 + 4i 3 + 4i
(ii )
We will set up an isomorphism between Z[i]4+i and Z17 in the following way:
0→0
1 → 4i = 1 − 1 + 4i
2 → 1 + 4i = 2 + −1 + 4i
3 → 2 + 4i = 3 + −1 + 4i
4 → 3 + 4i = 4 + −1 + 4i
5 → 3i = 5 − 1 + 4i − (4 + i)
6 → 1 + 3i = 6 − 1 + 4i − (4 + i)
7 → 2 + 3i = 7 − 1 + 4i − (4 + i)i
8 → 3 + 3i = 8 − 1 + 4i − (4 + i
9 → 2i = 9 − 1 + 4i − 2(4 + i)
10 → 1 + 2i = 10 − 1 + 4i − 2(4 + i)
11 → 2 + 2i = 11 − 1 + 4i − 2(4 + i)
12 → 3 + 2i = 12 − 1 + 4i − 2(4 + i)
13 → i = 13 − 1 + 4i − 3(4 + i)
14 → 1 + i = 14 − 1 + 4i − 3(4 + i)
15 → 2 + i = 15 − 1 + 4i − 3(4 + i)
16 → 3 + i = 16 − 1 + 4i − 3(4 + i).
The map above is obviously an isomorphism.
Since we know (if we have written and thought about the multiplication table for
Z∗17 - if we haven’t too bad!) that 3 is a generator for Z∗17 . So one generator for
Z[i]4+i is 2 + 4i. In fact we would know, if we had etc etc, that the generators are
3, 10, 5, 11, 14, 7, 12, 6. So generators for Z[i]4+i are
2 + 4i, 1 + 2i, 3i, 2 + 2i, 1 + i, 2 + 3i, 3 + 2i, 1 + 3i.
3. (i)
(ii )
Show that R ⊕ S ∼
= S ⊕ R by defining ϕ(r, s) = (s, r), for all r ∈ R and s ∈ S.
Show that Z ⊕ Z2 is not isomorphic to Z, or Z ⊕ Z2 ⊕ Z2 . (Hint: Think about the
elements x in each ring such that 2x = x + x = 0.)
Solution
(i)
Define ϕ : R ⊕ S → S ⊕ R by ϕ(r, s) = (s, r), for all r ∈ R and s ∈ S. Then ϕ is obviously a one to one correspondence. It also remembers addition and multiplication.
ϕ((r1 , s1 ) + (r2 , s2 )) = ϕ(r1 + r2 , s1 + s2 )
= (s1 + s2 , r1 + r2 )
= (s1 , r1 ) + (s2 , r2 )
= ϕ(r1 , s1 ) + ϕ(r2 , s2 )
and
ϕ((r1 , s1 )(r2 , s2 )) = ϕ(r1 r2 , s1 s2 )
= (s1 s2 , r1 r2 )
= (s1 , r1 )(s2 , r2 )
= ϕ(r1 , s1 )ϕ(r2 , s2 ).
Hence ϕ is an isomorphism.
(ii )
Consider
T = {(x, y) ∈ Z ⊕ Z2 : 2(x, y) = (0, 0)}
= {(0, 0), (0, 1)},
and so T has two elements.
But the corresponding subset of Z, namely, {z ∈ Z : 2z = 0} = {0} has only one
element. Hence these two rings cannot be isomorphic.
The same argument shows that Z ⊕ Z2 is not isomorphic to Z ⊕ Z2 ⊕ Z2 because
the same subset of Z ⊕ Z2 ⊕ Z2 is {(0, 0, 0), (0, 1, 0), (0, 0, 1), (0, 1, 1)} and has four
elements.
4. Show that Z12 ∼
6= Z2 ⊕ Z6 , but that Z2 ⊕ Z2 ⊕ Z3 ∼
= Z2 ⊕ Z6 . Thus we know two rings with
12 elements, namely Z12 and Z2 ⊕ Z6 .
Solution
We just need to find one property that Z2 ⊕ Z6 has that Z12 doesn’t have. Well, notice
that if we add any element of Z2 ⊕ Z6 to itself 6 times we get zero, that is
6(x, y) = (x, y) + . . . + (x, y) = (6x, 6y) = (0, 0) ∈]Z2 ⊕ Z6 .
But clearly 6.1 = 1 + 1 + . . . + 1 = 6 6= 0 ∈ Z12 .
Therefore Z2 ⊕ Z6 ∼
6= Z12 .
On the other hand, Z2 ⊕ Z3 ∼
= Z6 by CRT Theorem 46, since (2, 3) = 1. Hence Z2 ⊕ Z6 ∼
=
Z2 ⊕ Z2 ⊕ Z3 .
5. How many different rings with 4 elements can you think of? For each ring you consider,
decide if it is or is not different from the other rings in your list.
Solution
1. Z4
2. Z2 ⊕ Z2 - not isomorphic to ring 1 because every element of Z2 ⊕ Z2 added to itself is
zero, while 1 + 1 = 2 6= 0 ∈ Z4 .
3. Z2 [x]x2 - not isomorphic to ring 1 for the same reason as ring 2 is not isomorphic to
1. It is not isomorphix to ring 2 because there is an element of ring 3, namely x which
squares to give zero. The square of any non-zero element of ring 2 is never 0.
4. Z2 [y]y2 +1 - since y 2 +1 = (y+1)2 , this ring is the same as ring 3. There is an isomorphism
from ring 3 to ring 4 defined by a + bx → a + b(y + 1). Check that.
5. Z2 [x]x2 +x = Z2 [x]x(x+1) ∼
= Z2 [x]x ⊕ Z2 [x]x+1 by the CRT - Since Z2 [x]x ∼
= Z2 and
∼
Z2 [x]x+1 = Z2 , it follows that Z2 [x]x2 +x ∼
= Z2 ⊕ Z − 2 - that is ring 2.
6. Z2 [x]x2 +x+1 . This is a field and so is different form any of the rings 1 to 5. In fact
X
X −1
0 1
x
x+1
∗ 1 x+1
x
7, Z[i]2 - This is isomorphic to Z2 [x]x2 because (1 + i)2 = 0 ∈ Z[i]2 . So the map a + bx →
a + b(1 + i) is an isomorphism from Z2 [x]x2 to Z[i]2 . Check that.
Thus we have four different rings with 4 elements. Can you think of any more?
6. (i )
(ii )
Given a field F = Zp [x]f (x) , for some polynomial f (x) ∈ Zp , show that
(a + b)p = ap + bp , for all a, b ∈ F. (Hint: Use the binomial Theorem.)
Consider the field K = Z2 [x]x4 +x+1 . The polynomial X 4 + X + 1 has a root x ∈ K.
Use part (i) to find three other roots in K.
(iii ) The polynomial X 4 + X 3 + 1 is prime in Z2 [x]. It has four roots in K[X]. Can you
find them?
Solution
(i)
p p−1
p p−2 2
(a + b) = a +
a b+
a b + . . . + bp .
1
2
Now each term
p
p!
=
.
k
k!(p − k)!
The coefficients in every case are integers and, since
p is a prime, if 1 ≤ k ≤ p − 1,
p
the denominators are not multiples of p. Hence
is an integer multiple of p and
k
so is 0 ∈ F. Hence (a + b)p = ap + bp .
p
(ii )
p
If f (x) = x4 + x + 1,
f (x2 ) = (x4 )2 + x2 + 1
= (x4 + x + 1)2 , by part (i)
= 0.
hence f (x) = f (x2 ) = f (x4 ) = f (x8 ) = . . . = 0.
Hence the roots of x4 + x + 1 are
x, x2 , x4 = x + 1, x8 = (x + 1)2 = x2 + 1.
(iii ) Check that x3 + 1 solves X 4 + X 3 + 1:
(x3 + 1)4 + (x3 + 1)3 + 1 = x12 + 1 + x9 + x6 + x3 + 1 + 1
= (x + 1)3 + x(x + 1)2 + x2 (x + 1) + x3 + 1
= x3 + x2 + x + 1 + x3 + x + x3 + x2 + x3 + 1
= 0.
Hence by part (i), the roots of X 4 + X 3 + 1 = 0 ∈ K[x] are (x3 + 1), (X 3 + 1)2 , (X 3 +
1)4 , (x3 + 1)8 . These are:
x3 + 1, x3 + x2 + 1, x3 + x2 + x, x3 + x + 1.