Download Problem 23: Let R 1,R2 be rings with 1 and f : R 1 → R2 be a

Problem 23: Let R1 , R2 be rings with 1 and f : R1 → R2 be a ring isomorphism.
Show that f (1) = 1. (Your proof will likely show that the same is true for f a ring
homomorphism that is onto.)
Solution: In the ring R1 , the property x · 1 = x holds for all x. Therefore, f (x · 1) = f (x). Using
that f is a ring homomorphism, we conclude f (x) · f (1) = f (x). Since f is onto, any y ∈ R2 can
be written as f (x), and therefore we have y · f (1) = y for every y ∈ R2 . This means f (1) is a unity
in R2 . By uniqueness of the unity, f (1) = 1.
Problem 24:
Show that R = {0, 2, 4, 6, 8} with addition and multiplication modulo 10 is a field. (May
use results from Pblm 13). I also claim that R is (ring-)isomorphic to Z5 . Give a specific
mapping φ : R → Z5 (by value table) that is a ring isomorphism. Hint: Use Pblm 23
to find φ. Find φ(1), then φ(1 + 1), and φ(1 + 1 + 1) etc., if φ is a ring homormorphism.
Then show that the φ so constructed is indeed a ring isomorphism.
Solution: We have already seen the multiplication table in #13 and concluded that 6 is the unity
in the ring. The multiplication table also reveals 2 · 8 = 6 and 4 · 4 = 6. So every non-zero element
in this ring has a multiplicative inverse. Alternatively, we can argue that the multiplication table
shows the absence of 0 divisors, so R is a finite integral domain, hence a field.
If there is an isomorphism f : R → Z5 , we must have f (6) = 1 by the previous problem. Then
f (6 + 6) = f (6) + f (6) = 1 + 1, i.e., f (2) = 2. Continuing, we get f (8) = f (2 + 6) = f (2) + f (6) =
2+1 = 3 and f (4) = f (2+2) = f (2)+f (2) = 2+2 = 4. [In any ring and for any ring homomorphism
f , we conclude f (n · 1) = n · f (1) this way, where n is a positive integer]. We also define f (0) = 0.
By construction f (n · 6) = n · f (6) for n = 1, 2, 3, 4. It is also true for n = 5 since in both rings,
5 · x = 0 for all x. Since in both rings, (n + 5) · x = n · x, we conclude f (n · 6) = n · f (6) for all
positive integers by strong induction.
For the f constructed, we now want to show that it is indeed a ring homomorphism, i.e., f (x + y) =
f (x) + f (y) and f (xy) = f (x)f (y).
Since all elements x, y in the ring R are of the form x = n·6, y = m·6, the claim f (x+y) = f (x)+f (y)
reduces to f (n · 6 + m · 6) = f (n · 6) + f (m · 6). The left side of this equation is f ((n + m) · 6) =
(n + m) · f (6) = (n + m) · 1. The right side is n · f (6) + m · f (6) = (n + m) · f (6) = (n + m) · 1. So
the additive homomorphism property has been verified.
Similarly, the multiplicative property f (xy) = f (x)f (y) reduces to f ((n·6)(m·6)) = f (n·6)f (m·6).
But the left side is f (nm · 6) = nm · f (6) = nm] · 1, and the right side is (n · f (6))(m · f (6)) =
(n · 1)(m · 1) = nm · 1 so they are both equal.
NOTE: The isomorphism f actually is the reduction modulo 5.
Problem 25: Suppose a and b belong to an integral domain. Show: If a5 = b5 and
a3 = b3 , then a = b.
Now assume m, n are positive integers that are relatively prime, i.e., gcd(m, n) = 1.
Show: If am = bm and an = bn then a = b. (Hint: How would you do it if m = n + 1;
then remember the gcd can be expressed as a linear combination).
Solution: Part (a) a3 = b3 implies a6 = b6 by squaring, and a5 = b5 implies a6 = ab5 . So we
combine this to b6 = ab5 , or (b − a)b5 = 0. Since we are working in an integral domain, either
b − a = 0 (hence a = b) or b = 0. If b = 0, then a3 = b3 = 0 also implies a = 0, hence again a = b.
The same reasoning (with n instead of 5, n + 1 instead of 6) shows that an = bn and an+1 = bn+1
imply a = b.
Part (b): We may assume m, n > 1 as otherwise the claim would be trivially true. Since gcd(m, n) =
1, we can write 1 as a linear combination: 1 = km − ln with k, l ∈ Z. Neither k nor l are 0. If k > 0
then l > 0; if k < 0 then l < 0. We assume the former with no loss of generality, b/c otherwise we
swap m, n.
From am = bm we conclude akm = bkm , i.e., aln+1 = bln+1 (1). From an = bn , we conclude aln = bln
and then aln+1 = abln (2). Combining the equations (1) and (2), we get b bln = a bln , hence b = 0
or b = a. But if b = 0, then a = 0 also (since am = bm ); so b = a in any case.
Problem 26: How many solutions does the equation x2 − 5x + 6 = 0 have in each of
the following rings?:
(a) Z7
(b) Z8
(c) Z12
(d) Z14
Solution: With the moduli being small numbers, it is probably easiest to check all by plugging
in. Factoring the equation into (x − 2)(x − 3) = 0 helps if the modulus is a prime, because then
(and only then) can we conclude x − 2 = 0 or x − 3 = 0. We can enlist the result for Z7 in our
study for Z14 .
(a) In Z7 : Two solutions x = 2 and x = 3 by factorization since Z7 is free of zero divisors.
(b) In Z8 : Again two solutions. They are x = 2, x = 3, by plugging them all in:
(−2)(−3), (−1)(−2), 0(−1), 1 · 0, 2 · 1, 3 · 2, 4 · 3, 5 · 4. While zero divisors might have allowed
for further solutions, this doesn’t happen here, b/c when x − 2 is a zero divisor then x − 3 doesn’t
happen to be a ‘partner zero divisor’ for x − 2.
(c) In Z12 : Four solutions. Namely x ∈ {2, 3, 6, 11}, found by plugging them all in:
(−2)(−3), (−1)(−2), 0(−1) = 0, 1 · 0 = 0, 2 · 1, 3 · 2, 4 · 3 = 0, 5 · 4, 6 · 5, 7 · 6, 8 · 7, 9 · 8 = 0.
If I were asked the question in something like Z252 , I’d look for the divisors of 252 and check only those
x for which x − 2 or x − 3 divides 252, b/c only zero divisors need to be considered beyond x − 2 = 0
or x − 3 = 0. But for 12, saving a few calcs is hardly worth the hassle of explaining the selections.
(d) Four solutions: Namely x ∈ {2, 3, 9, 10}. — If x2 − 5x − 6 = 0 in Z14 , then in Z, 14 must divide
x2 − 5x − 6; so in particular 7 must divide the same expression, which means x2 − 5x − 6 = 0 in Z7
also. This leaves only the cases x ∈ {2, 3, 2 + 7, 3 + 7} to check in Z14 . Indeed, 7 · 6 and 8 · 7 are 0
in Z14 .
Problem 27:
(a) Let n = 7 and define
R :=
!"
a
b
#$
%
− b $$
a, b ∈ Zn
a $
Show that R is a field. How many elements does it have?
(b) Let n = 5 and take the same definition for R. Show that this R has zero&divisors.
'
(c) Let n be a positive integer, n ≥ 2. Show that the mapping φ : a + bi %→ ab −b
a is a
ring isomorphism from Zn [i] → R.
Solution:
(a) Clearly R has 49 elements, any pair of a and b, each in Z7 gives rise to an element of R. We
know that R is a subset of the matrix ring M2 (Z7 ). To show that R is a ring, we need to check the
closure properties:
&
'
&
'
'
&
−(b−d)
If M = ab −ba and N = cd −dc are in R, then M − N = a−c
is also in the ring. Before
b−d
a−c
checking the& multiplicative
closure as well, I’d rather introduce a less clumsy notation: We’ll denote
'
a −b
the matrix b a by Φ(a, b). We have thus shown that M − N = Φ(a, b) − Φ(c, d) = Φ(a − c, b − d).
Now let’s calculate M N = Φ(a, b)Φ(c, d).
Φ(a, b)Φ(c, d) =
"
a −b
b a
#"
c −d
d c
#
=
"
ac − bd −(ad + bc)
bc + ad −bd + ac
#
= Φ(ac − bd, ad + bc)
The same calculation also shows (by swapping (a, b) and (c, d) and using commutativity in Zn ) that
Φ(a, b)Φ(c, d) = Φ(c, d)Φ(a, b). So R is a commutative ring. [We mark this up in our experience: a
subring of a non-commutative ring may well happen to be commutative.]
Clearly R has a unity, Φ(1, 0). To show that R is a field, we have two choices: Either we show it has
no zero divisors and use that a finite integral domain is a field; or else we obtain a multiplicative
inverse for each Φ(a, b) other than the zero element Φ(0, 0). We’ll give an inverse.
'
& ' &
d −b
In real matrix algebra, we have the formula that the inverse of ac db is −c
a /(ad − bc) provided
the denominator (which is the determinant) doesn’t vanish; if it does vanish, then the matrix
doesn’t have an inverse. We can borrow this formula and verify it directly for M2 (F ) over any
field F . For our purposes, we claim specifically:
If Φ(a, b) ∈ M2 (F ) and a2 +b2 &= 0 in F , then Φ(qa, −qb) with q = (a2 +b2 )−1 is a (the) multiplicative
inverse of Φ(a, b). Proof by calculating Φ(a, b)Φ(qa, −qb) is immediate.
To apply this specifically to the field F = Z7 , we only need to verify that a2 +b2 &= 0 if (a, b) &= (0, 0).
If so, then every nonzero element Φ(a, b) has an inverse by the above formula.
02 + b2 &= 0 if b &= 0 because Z7 is a field.
12 + 12 = 2, 12 + 22 = 5, 12 + 32 = 3; 22 + 22 = 1, 22 + 32 = 6; 32 + 32 = 4. b ∈ {4, 5, 6} didn’t have
to be checked since 4 = −3, 5 = −2, 6 = −1 and the minus is eaten by the square.
If you use the result from #18b that in a field, a2 + b2 = 0 has a nontrivial solution if and only if
x2 + 1 has a solution, you only need to check x ∈ {1, 2, 3}.
We have proved that R is a field when n = 7.
(b) The same calculation shows that for n = 5 (or any n for that matter), the set R is still a
commuative ring with 1, and that Φ(a, b)Φ(a, −b) = Φ(a2 + b2 , 0). But now 22 + 12 = 0 in Z5 ,
hence Φ(2, 1) and Φ(2, −1) are zero divisors.
(c) In R[i] of Pblm. 18’s fame, we had the pairs (a, b) of elements of R, written more conventionally as
a+bi after the construction of Pblm 18, with the multiplication (a+bi)(c+di) := (ac−bd)+(ad+bc)i
and the addition (a + bi) + (c + di) := (a + c) + (b + d)i. Here we are using R = Zn .
(Here, I have used a different font R to distinguish the ring R of #18 from the present ring R ⊂
M2 (R). )
To show that φ : Zn [i] → R , a + bi %→ Φ(a, b) is a ring homomorphism, we have to show:
φ((a + bi) + (c + di))= φ(a + bi) + φ(c + di)
i.e.,
Φ(a + c, b + d) = Φ(a, b) + Φ(c, d)
and
φ((a + bi)(c + di))= φ(a + bi) φ(c + di)
i.e.,
Φ(ac − bd, ad + bc) = Φ(a, b) Φ(c, d)
This is exactly what we have shown already in part (a).
It is obvious that φ is injective (one-to-one) and surjective (onto).
Comment: This example is well known in complex variables for the ring (field) R = R of real
numbers. It provides a geometric interpretation for the multiplication of complex numbers. Adding
complex numbers (represented by points in the plane) is obvious of course: it’s like vector addition:
The mapping z %→ (a + bi) + z is a translation by the vector (a, b).
As for multiplication, the mapping z %→ (a + bi)z represents a combined rotation by an angle α and
dilation (enlargement) by a factor r, which are determined from (a, b) in terms of polar coordinates
a = r cos α and b &= r sin α. 'This combined rotation and dilation is represented by the matrix
√
α −sin α
Φ(a, b) = a2 + b2 cos
sin α cos α .
The algebra part of this observation is however not specific to R and C = R[i], but works (at least)
with any commutative ring with 1. Go figure, maybe, not even commutativity is needed.