Download Solutions to coursework 6 File

MTH4104
Introduction to Algebra
Solutions 6
Semester B, 2017
Question 1 Write out addition and multiplication tables for Z5 , like the ones for Z4
in the lecture notes.
That is, let S := {0, 1, 2, 3, 4} be the canonical set of representatives for Z5 . For
each pair of elements a, b ∈ S, your tables should give the integers c, d ∈ S such that
[a]5 + [b]5 = [c]5
and
[a]5 [b]5 = [d]5 .
Solution Let us make the addition table first. The definition of addition in modular
arithmetic implies that
[a]5 + [b]5 = [a + b]5
where a + b inside the square brackets on the right hand side is ordinary addition of
ordinary integers a and b. But we cannot simply take c = a + b because then c will not
in general be an element of S, and I requested elements of S in the tables.
Now since a ≤ 4 and b ≤ 4 always, a + b is at most 8. If a + b happens to be in S
already then we can take c = a + b. Otherwise 5 ≤ a + b ≤ 8, so the integer quotient of
a + b by 5 will be 1, and the division will give the remainder a + b − 5, which does lie
in S:
5 ≤ a + b ≤ 8 ⇒ a + b = 1 · 5 + (a + b − 5).
If you prefer a formula to the above discussion, we can take
a+b
if
a+b ≤ 5
c=
a+b−5
if
a + b ≥ 5.
This leads to the following addition table for Z5 :
+
[0]5
[1]5
[2]5
[3]5
[4]5
[0]5
[0]5
[1]5
[2]5
[3]5
[4]5
[1]5
[1]5
[2]5
[3]5
[4]5
[0]5
[2]5
[2]5
[3]5
[4]5
[0]5
[1]5
[3]5
[3]5
[4]5
[0]5
[1]5
[2]5
[4]5
[4]5
[0]5
[1]5
[2]5
[3]5
The second case of the rule above, where c = a + b − 5, is visible as the lower-right
triangular chunk of the table, starting at the diagonal stripe of [0]5 .
For the multiplication table, we use the definition
[a]5 [b]5 = [ab]5 ,
1
and again we need to find a representative of [ab]5 that is in S. There is less we can say
in general about what the integer quotient of each product is, but we don’t need to; it’s
enough to compute the remainder of each product a · b modulo 5 and record these in a
table. Eventually the table we produce is the following.
·
[0]5
[1]5
[2]5
[3]5
[4]5
[0]5
[0]5
[0]5
[0]5
[0]5
[0]5
[1]5
[0]5
[1]5
[2]5
[3]5
[4]5
[2]5
[0]5
[2]5
[4]5
[1]5
[3]5
[3]5
[0]5
[3]5
[1]5
[4]5
[2]5
[4]5
[0]5
[4]5
[3]5
[2]5
[1]5
I have been pedantic and inserted brackets and subscripts [c]5 around all the numbers in my tables, but the result is quite visually busy, and of course repetitive to write.
If you left out the brackets and/or the subscripts, that isn’t cricket technically — 3 times
4 does not equal 2 — but in this context I don’t mind.
Question 2
(a) Explain why [76]103 has a multiplicative inverse in Z103 .
(b) Find a non-negative integer b < 103 such that [76]−1
103 = [b]103 .
Solution (a) 103 and 76 are coprime, that is they have no common factors. This
can be justified by inspection, without an involved computation: 76 = 22 · 19 and these
primes do not divide 103. So the gcd of these integers is 1. By the extended Euclidean
algorithm, there are integers x, y such that 103x + 76y = 1. Hence 76y ≡103 1, which
implies that [76]103 [y]103 = [1]103 . So this [y]103 is the multiplicative inverse for [76]103
sought.
(b) To actually compute the inverse, we have to find integer solutions to 103x + 76y = 1.
This is done by applying the Euclidean algorithm, as follows:
103 = 1 · 76 + 27
76 = 2 · 27 + 22
27 = 1 · 22 + 5
22 = 4 · 5 + 2
5 = 2·2+1
2 = 2 · 1 + 0.
⇒ 27 = 103 − 76
⇒ 22 = 76 − 2 · 27
⇒ 5 = 27 − 22
⇒ 2 = 22 − 4 · 5
⇒ 1 = 5−2·2
Now we back-substitute starting from 1 = 5 − 2 · 2, remembering to open up brackets
and to simplify before re-substituting:
1
= 5−2·2
= 5 − 2 · (22 − 4 · 5)
= −2 · 22 + 9 · 5
= −2 · 22 + 9 · (27 − 22)
= 9 · 27 − 11 · 22
= 9 · 27 − 11 · (76 − 2 · 27)
= −11 · 76 + 31 · 27
= −11 · 76 + 31 · (103 − 76) = 31 · 103 + −42 · 76.
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Hence 1 = 31(103) + (−42)(76) gives the solution x = 31 and y = −42. However the
question asks for y in the range 0 ≤ y < 103! To arrange this, we simply add 103 to
−42 to get another solution, representing the same congruence class,
y = 61.
It is always a good idea to check that you haven’t made a mistake by calculating that
the answer (y = 61) really does what it should do, namely satisfy the congruence
76 · 61 ≡103 1.
This is true if and only if 76 · 61 − 1 is an integer multiple of 103. But 76 · 61 − 1 = 4635
and 4635/103 = 45 which is an integer.
Question 3 Which elements x ∈ Z11 satisfy the equation
[6]11 x + [1]11 =
[10]11
?
[6]11
Justify your answer.
Solution It is possible to do this by trial and error — after all, Z11 only has 11 elements. But it’s better to just solve the linear equation as usual, temporarily forgetting
about the fact that the coefficients may look a little unfamiliar.
[10]11
The first thing to deal with is the expression
: we have to work out what that
[6]11
means. Division by [6]11 means multiplying by its multiplicative inverse; it has one,
since it was proved in the Lectures that Z11 forms a field. By inspection, we spot the
fact that
[2]11 · [6]11 = [12]11 = [1]11
so that [2]11 is the multiplicative inverse of [6]11 , and what we have on the right hand
side of the equation is [10]11 · ([6]11 )−1 = [10]11 · [2]11 = [20]11 = [9]11 .
Now, we can solve the linear equation
[6]11 x + [1]11 = [9]11
just as we always do:
x=
[9]11 − [1]11 [8]11
=
= [8]11 · ([6]11 )−1 .
[6]11
[6]11
One more inverse calculation to do. Again we spot
[6]11 · [2]11 = [12]11 = [1]11
so the inverse of [6]11 is [2]11 , and we can finish:
x = [8]11 · [2]11 = [16]11 = [5]11 .
Let’s check that this really works: 6 · 5 + 1 = 31 which is congruent to 9 modulo 11.
3
Alternatively, if you don’t like finding modular inverses, you could have cleared
denominators first by multiplying the whole equation by [2]11 , yielding
[6]11 ([6]11 x + [1]11 ) = [10]11 ,
i.e.
[36]11 x + [2]11 = [10]11
which using standard representatives is
[3]11 x + [2]11 = [10]11
and then you’d only have one inverse-finding step to do. (Check that this yields the
same solution!)
Question 4 Find X,Y ∈ Z7 that satisfy the simultaneous system of linear equations
[2]7 X + [5]7 Y = [1]7
[3]7 X +
Y = [2]7 .
Solution Since Z7 is a field (as 7 is a prime number), we can solve this system of
equations by any of the methods you know from School. For example, the second
equation implies that
Y = [2]7 − [3]7 X
which we can substitute into the first equation to give
[2]7 X + [5]7 ([2]7 − [3]7 X) = [1]7 .
Expanding the brackets and rearranging gives
([2]7 − [15]7 )X = [1]7 − [10]7
[−13]7 X = [−9]7
that is to say
[1]7 X = [5]7
and by a stroke of good luck we have X alone on the left side, with no scalar to invert,
so that X = [5]7 . Back-substituting into Y = [2]7 − [3]7 X gives
Y = [2]7 − [3]7 [5]7 = [2]7 − [15]7 = [−13]7 = [1]7 .
So X = [5]7 and Y = [1]7 is the solution. (Check this by substituting in again, if you
need practice!)
Question 5 Without using a calculator, write [280 ]19 as a class [a]19 where 0 ≤ a < 19.
Try to be efficient in your calculations.
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Solution The really inefficient way to proceed would be to work out 280 (which is a
twenty-five digit number), then divide by 19 and take the remainder. Somewhat better
is to work out the powers of 2 in sequence by successive doubling, taking the remainder
modulo 19 whenever doubling gives you a value greater than 19 — to say that another
way, this is working out the canonical representatives of [2]19 , ([2]19 )2 , ([2]19 )3 , etc. in
sequence. This computation would begin thus:
20 = 1,
21 = 2,
22 = 4,
23 = 8,
24 = 16,
25 = 32 ≡ 32 − 19 = 13,
26 ≡ 26 ≡ 26 − 19 = 7,
27 ≡ 14,
etc.
It’s still tedious to do that 80 times. If you were dedicated and pressed ahead, though,
you would soon notice that these values repeat, starting from 218 ≡19 1. So to compute
280 mod 19 we only need to know what the 80th term of this sequence is, and for that
we only need to know 80 mod 18, which is 80 − 72 = 8. This is just one step beyond
where I stopped in my table above: 28 ≡ 2 · 14 ≡ 9, so the answer is [9]19 .
In fact, this repetition is nothing but Fermat’s little theorem, the fact that a p−1 ≡ p 1
whenever p is a prime and a is not a multiple of p. So a good answer is “use Fermat’s
little theorem to factor 280 = 272 · 28 and ignore the first factor”.
Another efficient approach you could have used is to factor the exponent 80 as
5·2·2·2·2, which implies that 280 = ((((25 )2 )2 )2 )2 . You can take the remainder mod 19
after each exponentiation. This plays out in the following way, where at intermediate
stages I’m allowing myself negative numbers to make the squares smaller.
25 = 32 ≡ 13 ≡ −6,
210 ≡ (−6)2 = 36 ≡ 17 ≡ −2,
220 ≡ (−2)2 = 4,
240 ≡ 42 = 16 ≡ −3,
280 ≡ (−3)2 = 9.
Again we get the same answer [9]19 .
If you didn’t know in advance when the sequence of powers would repeat (for instance if you couldn’t factor the modulus m, which would stop you from using Fermat’s
little theorem), procedures of this general kind, based on repeated squaring, are the
most efficient known.
Question 6 How many of the elements of Z12 have multiplicative inverses? What
about Z30 ?
Describe a general method that would allow you to calculate how many elements
of Zm have multiplicative inverses, without simply counting them all.
Solution Recall that [a]m has a multiplicative inverse in Zm if and only if a and m
have no common factor. Now Z12 has twelve elements [a]12 , namely
[0]12 , [1]12 , . . . , [11]12 ,
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so we have to find all a in the range from 0 to 11 that are coprime to 12 (meaning
gcd(a, 12) = 1). With just twelve elements it is not such an ordeal to go through the
whole list and check each a for common factors. Doing this shows that
a = 1, 5, 7, 11
are the only possibilities; all the other values of a either have a factor of 2 or 3 in
common with 12. So exactly 4 elements of Z12 have multiplicative inverses, and the
remaining elements do not.
As m gets larger, writing down every element of Zm starts to become a more arduous
and error-prone task, and we should look for another way. Here is another approach for
Z30 . We note that the prime factorisation of 30 is 2 · 3 · 5. So these primes are the
ones whose multiples we have to avoid when looking for numbers a for which [a]30 is
invertible. Multiples of 2 are the same as even numbers, so right away we can discard
them and focus on the odd numbers only. 15 out of the 30 congruence classes are odd.
Next we wish to get rid of the multiples of 3. In each group of three odd numbers, one
is a multiple of 3, so this gets rid of one third of the remaining elements, and 10 are left.
Finally, we have to get rid of the multiples of 5 that are still on the list, i.e. odd and not
multiples of 3. The only such numbers less than m = 30 are
5 = 5 · 1 and 25 = 5 · 5
so we remove these two more elements from the list of 10, leaving 8. These 8 elements
have no common factors with 30, i.e. they are the elements with multiplicative inverses
in Z30 .
Later, in lecture, we will see a formula that generalises the argument above to any
value of m, using the Euler totient function. If p1 , . . . , pk are the distinct primes dividing
m, then the number of invertible elements in Zm is
φ (m) := m(1 −
1
1
) · · · (1 − ).
p1
pn
Though this is not a proof, the formula should seem plausible in light of the above
analysis: starting with m elements of Zm , we remove as non-invertible those that are
multiples of p1 , that is one p1 -th of them; then of the remainder we remove those that
are multiples of p2 , that is one p2 -th of what’s left; etc.
Question 7 This question compares a naı̈ve way to take the “sum” and “product” of
two sets of integers to the definitions that we actually use in modular arithmetic.
(a) Prove that, for any integer m > 0, if X and Y are congruence classes of ≡m , then
the set
{x + y : x ∈ X, y ∈ Y }
is their sum X +Y within Zm .
(b) Give an example of an integer m > 0 and two congruence classes X,Y of ≡m such
that the set
{xy : x ∈ X, y ∈ Y }
is not the product XY within Zm .
Write down a general statement about how the above set is related to XY .
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Solution (a) First of all, the definitions of addition and multiplication in Zm are
framed in terms of representatives. So let’s pick representatives for the congruence
classes X and Y : let X = [a]m and Y = [b]m . Now X +Y = [a + b]m , and what we have
to prove is
{x + y : x ∈ [a]m , y ∈ [b]m } = [a + b]m .
This is an equality of sets, so we will prove it by showing each set is a subset of
the other. Let’s think about the ⊆ direction first. A little reflection shows that this is a
statement we have already proved, when we were proving that addition in Zm is welldefined. What we showed is that if you took “different” representatives a0 ∈ [a]m and
b0 ∈ [b]m , then the sum a0 + b0 was a representative of the same class that a + b was a
representative of, that is a0 + b0 ∈ [a + b]m . Renaming a0 to x and b0 to y, this is the
content of the ⊆ inclusion.
For ⊇, what we have to show is that every element z of the class [a + b]m can be
written as x + y for some x ∈ [a]m and y ∈ [b]m . Well, we get to pick x and y here, so
there is lots of freedom: we may as well begin by taking x = a. That implies we must
take y = z − a, so we have succeeded if z − a is in the congruence class [b]m .
In fact, we have succeeded. Since z ∈ [a + b]m , we have z ≡m a + b, which means
m | z − (a + b). What we wish to show is that z − a ∈ [b]m , that is z − a ≡m b, which
means m | (z − a) − b; but this is exactly what we just assumed. This completes the ⊇
containment, and the proof.
(b) There are many counterexamples. One of them occurs when m = 5, X = [2]5 and
Y = [3]5 . Then
XY = [2]5 [3]5 = [6]5 = [1]5 ,
so 1 ∈ XY . However, 1 is not an element of the set
{xy : x ∈ [2]5 , y ∈ [3]5 }
since the only ways to factor 1 as a product of two integers are 1 = 1 · 1 and 1 =
(−1)(−1), and neither of these factorisations consists of an element of [2]5 times an
element of [3]5 .
The general statement that can be made relating the two is that
{xy : x ∈ X, y ∈ Y } ⊆ XY .
This is proved in a fashion exactly parallel to the proof of the ⊆ inclusion in part (a).
Question 8
(a) Prove that, for any integer a,
[a]6 = [a]2 ∩ [a]3
is a true equation of sets.
(b) Let m and n be positive integers. Is there always an integer ` that makes the
equation of sets
[a]` = [a]m ∩ [a]n
true for any value of a? How do you find ` in terms of m and n?
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Solution (a) To prove an equality of two sets, we need to show containment of each
set in the other. So let b be an element of [a]6 . Then b − a is a multiple of 6, that is
b − a = 6k for some integer k. But then a fortiori b − a is a multiple of 2 and of 3:
indeed, b − a = 2(3k) = 3(2k). So b ∈ [a]2 ∩ [a]3 .
Conversely, suppose b ∈ [a]2 ∩ [a]3 . This implies b − a is divisible, separately, by
both 2 and 3. We can write, e.g. b − a = 2` = 3q, where ` and q are integers. Therefore
3q is a multiple of 2. This implies that q itself is a multiple of 2 — you could prove this
e.g. by a case analysis going through every possible residue modulo 2, which amounts
just to the two cases of whether q is even or odd, or by considering prime factorisations.
We may thus write q = 2r for some integer r, so that b − a = 3(2r) = 6r, which implies
b ∈ [a]6 as desired.
(b) There is, indeed, always such an integer `.
Based on part (a), you might guess that ` should be m · n. But this does not always
work. For example, if m = n = 2, then the assertion amounts to [a]4 = [a]2 ∩ [a]2 . But
since [a]2 intersected with itself is still [a]2 , this is false.
In fact the correct choice is ` = lcm(m, n). By calling the variable ` I was trying to
give you a subtle hint! Elements b of [a]` are characterised by the property that b − a is
a multiple of `. So what the equation
[a]` = [a]m ∩ [a]n
asserts is that b − a is a multiple of ` if and only if it is both a multiple of m and a
multiple of n. And this is nearly the definition of the least common multiple.
What our definition of the least common multiple actually says is that ` is the least
natural number such that ` is both a multiple of m and a multiple of n. But the analogue
for lcm of Proposition 4.4 from the notes is also true, and is exactly the fact we want.
Try to prove this yourself!
Commentary. This exercise is a case of an important theorem in number theory or ring
theory known as the Chinese remainder theorem. You will encounter the statement of
this theorem if you do further modules in algebra.
Question 9 Let F be the set {a + bI : a, b ∈ Z3 }, where I is a formal symbol. Define
operations of addition and multiplication on F by
(a + bI) + (c + dI) = (a + c) + (b + d)I,
(a + bI) · (c + dI) = (ac − bd) + (ad + bc)I.
That is, I is meant to behave like a square root of [−1]3 .
(a) How many elements does F have?
(b) Prove the distributive law in F.
(c) Prove the multiplicative inverse law in F.
[In fact, F is a field. I didn’t want to make you prove all the axioms, as that would
be too much to write. The multiplicative inverse law, part (c), is the most “interesting”
one.]
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Solution (a) The cardinality of F is 32 = 9. Because Z3 has three elements, there are
three choices for a, and three choices for b, and therefore 3 × 3 elements altogether.
Or to say the same thing in a slightly fancier way, you can put aside the fancy a + bI
notation and just regard the elements of F as being ordered pairs (a, b), in which case
F is the cartesian product (Z3 )2 , whose cardinality is 32 = 9 by the formula for the
cardinality of a cartesian product.
Here are all 9 elements:
F = { [0]3 + [0]3 I,
[0]3 + [1]3 I,
[0]3 + [2]3 I,
[1]3 + [0]3 I,
[1]3 + [1]3 I,
[1]3 + [2]3 I,
[2]3 + [0]3 I,
[2]3 + [1]3 I,
[2]3 + [2]3 I }.
This is interesting because F is a field, but the number system we have already built
with 9 elements, namely Z9 , is not a field, since 9 is not prime. So F must be something
fundamentally different to Z9 .
(b) Each of the field laws aside from the multiplicative inverse law has a proof in F
which is visually identical to the proof we gave earlier for it in C. The only difference
is that instead of the symbols a, b, etcetera denoting real numbers, they now denote
elements of Z3 . But Z3 is a field, so everything we did with them still works!
Here, then, is a proof of the distributive law in F. I have literally copied and pasted
it from the course notes for week 2, changing only a few symbols.
Let z1 = a1 + b1 I, z2 = a2 + b2 I, and z3 = a3 + b3 I be elements of F, so that
a1 , a2 , a3 , b1 , b2 , b3 ∈ Z3 . Now
z1 (z2 + z3 ) = (a1 + b1 I)((a2 + a3 ) + (b2 + b3 )I)
= (a1 (a2 + a3 ) − b1 (b2 + b3 )) + a1 (b2 + b3 ) + b1 (a2 + a3 ))I,
and
z1 z2 + z1 z3 = ((a1 a2 − b1 b2 ) + (a1 b2 + a2 b1 )I) + ((a1 a3 − b1 b3 ) + (a1 b3 + a3 b1 )I)
= (a1 a2 − b1 b2 + a1 a3 − b1 b3 ) + (a1 b2 + a2 b1 + a1 b3 + a3 b1 )I,
and a little bit of rearranging, using the field laws in Z3 , shows that the two expressions
are the same.
(c) Since F is finite, we could prove the multiplicative inverse law by brute force: write
down the times table, and notice that every row other than the zero row contains the
multiplicative identity, [1]3 + [0]3 I.
But there is a more conceptual way to do it. Recall that in the complex numbers,
we had a formula for the multiplicative inverse that used the conjugate:
(a + bi)−1 =
a − bi
.
a2 + b2
This worked as long as a2 + b2 was not zero, which it never was if a + bi was a nonzero
complex number, because then one of a2 or b2 was positive and the other was nonnegative.
Does the same formula work in F? Well, we no longer have the notion “positive” to
work with. But it turns out still to be true that a2 + b2 = [0]3 if and only if a = b = [0]3 ,
for a and b elements of Z3 ! Why is this? Let’s square all the elements of Z3 :
([0]3 )2 = [0]3 ,
([1]3 )2 = [1]3 ,
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([2]3 )2 = [1]3 .
The only way to get a sum of [0]3 from two of these squares is if a = b = [0]3 ; otherwise
the sum is [1]3 or [2]3 . So if a + bI is nonzero, we can in fact define (a + bI)−1 using
the formula above. Let’s check that it really is a multiplicative inverse:
(a + bI) ·
a − bI
(a + bI)(a − bI) (a2 + b2 ) + [0]3 I
=
=
=
a2 + b2
a2 + b2
a2 + b2
= ((a2 + b2 ) + [0]3 I)(a2 + b2 )−1 = 1.
There is a subtlety in the above. It might look like circular reasoning that we are using the inverse (a2 + b2 )−1 to define the inverse (a + bI)−1 . But the former inverse is
computed in Z3 , which is a subset of F since we should regard a and a + [0]3 I to be the
same for all a ∈ Z3 . That is, (a2 + b2 ) + [0]3 I is just an element of Z3 , and we know that
elements of Z3 have inverses, so this computation goes through.
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