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IMMERSE 2010
Algebra Course
Problem Set 7 Solutions
1. Decide which of the following are subrings of Q:
(a) the set of all rational numbers with odd denominators (when written in lowest terms).
(b) the set of all rational numbers with even denominators (when written in lowest terms)
(c) the set of squares of rational numbers
(d) the set of nonnegative rational numbers
Proof (by Derrek, Matthew, Lilith, Linda). (a) To test if it a subring, the subring must be closed
under subtraction and multiplication. Subtraction and multiplication are tested. (a/b) − (x/y) =
(ay−xb)/(by) where b and y are both odd numbers rational numbers. Also, (a/b)·(x/y) = (ax/by)
When multiplying two odd numbers, the product is an odd number. So, dividing out any number
of the factors still leaves the number both odd and rational. Thus (a) is a subring of Q.
(b) Once again, both subtraction and multiplication must be tested for closure. However, (b) is not a
subring as is evident in the following example to test subtraction. (5/6) − (1/6) = (4/6) = (2/3).
As both 5/6 and 1/6 are rational numbers with even denominators in lowest terms, but the
difference is be reduced to a rational number with an odd denominator. Thus the counter example
shows that that (b) is not a subring of Q.
(c) Once again, both subtraction and multiplication must be tested for closure. But, a counter
example is provided in testing subtraction. 42 − 32 = 16 − 9 = 7. In order for this set to be a
subring, 7 should be a square or rational numbers but is not. Hence, (c) is not a subring of Q.
(d) Likewise in the first three problems, subtraction and multiplication must be test for closure. A
counter example is found for this set as can be verified in the following example to test subtraction.
4 − 5 = −1. The difference is not in the set. Hence, (d) is not a subring of Q either.
2. Prove that if R is an integral domain and x2 = 1 for some x ∈ R, then x = ±1.
Proof (by Robert, Caitlyn, Nathan M, Annette). Let R be an integral domain and let x2 = 1 for some
x ∈ R. So
x2 = 1
x2 − 1 = 0
(x + 1)(x − 1) = 0
Since R is an integral domain, we now know that either x + 1 = 0 or x − 1 = 0, which implies that
x = −1 or x = 1 as was to be shown.
3. Show that the nilpotent elements of a commutative ring form a subring.
Proof (by Rob, Anne, Kirsten, Nathan S). Let R be a commutative ring and let N = {n ∈ R|nk =
0 for some k ∈ Z+ }. Also suppose a and b are any two elements in N . In order to show that N is a
subring of R, we must show that ab ∈ N and a − b ∈ N .
Consider ab. We know that ap = 0 and bq = 0 for some positive integers p and q. Then
(ab)pq = apq bpq = (ap )q (bq )p = 0q 0p = 0.
Therefore, (ab)pq = 0, and ab ∈ N .
Now consider a−b. This can be written as a+(−b). We can use the binomial theorem on (a+(−b))p+q
to form
p+q p+q X
X
p + q p+q−k
p+q
(a + (−b))p+q =
a
(−b)k =
−
(ap+q−k bk ).
k
k
k=0
k=0
p+q−k
Thus, whenever k ≤ q, p + q − k ≥ p. This gives us a
= 0. Moreover, when q ≤ k ≤ p + q,
the exponent of b is at least q. In this case, bk = 0. Hence, for each term in the binomial expansion,
ap+q−k = 0 or bk = 0 for every k. Then every term reduces to 0, and so (a − b)p+k = 0, and a − b ∈ N .
Therefore, the nilpotent elements of a ring R form a subring of R.
[Note from Laura: It is important to show this set N is non-empty (otherwise, it certainly isn’t a
subring). This is easy though as 0 ∈ N.]
4. Let R = {(a, b, c)|a ∈ Z2 , b ∈ Z3 , c ∈ Z}. Define addition and multiplication coordinate-wise. What is
the characteristic of R?
Proof (by Chris, Rob, Jayna, Julia). Let R = {(a, b, c)| a ∈ Z2 , b ∈ Z3 , c ∈ Z}. We note that R has
unity (1, 1, 1) since (a, b, c) · (1, 1, 1) = (a · 1, b · 1, c · 1) = (a, b, c). Also, the unity of R has infinite order
under addition since there is no n ∈ Z satisfying n · (1, 1, 1) = (0, 0, 0). By theorem, this implies char
R = 0.
5. Give an example of an integral domain which has an infinite number of elements yet is of nonzero
characteristic.
Proof (by Rob, Anne, Kirsten, Nathan S). Consider the integral domain Zp [x] = a0 + a1 x + · · · +
ap−1 xp−1 where ai ∈ Zp for p a prime. This has an infinite number of elements, but characteristic
p 6= 0.
6. Let R be a commutative ring with no nonzero zero-divisors.
(a) Let a and b be nonzero elements of R. Prove that for any integer k, ka = 0 if and only if kb = 0.
(b) Let a, b be nonzero elements of R, and let ◦(a) denote the smallest positive integer n such that
na = 0. Prove that ◦(a) = ◦(b).
Proof (by Sarah, Zach, Tim, Lisa).
a) Let a, b ∈ R with a 6= 0, b 6= 0. Let k ∈ Z. Then,
ka = 0 ⇐⇒ a + a + ... + a = 0
|
{z
}
k
⇐⇒ (a + a + ... + a)b = 0 · b
{z
}
|
k
⇐⇒ ab + ab + ... + ab = 0
|
{z
}
k
⇐⇒ a(b + b + ... + b) = 0
{z
}
|
k
⇐⇒ b + b + ... + b = 0
|
{z
}
since there are no nonzero zero divisors
k
⇐⇒ kb = 0
b) Let a, b ∈ R with a 6= 0, b 6= 0. Let o(a) denote the smallest positive integer n such that na = 0.
Then, by part a), nb = 0. Suppose o(b) = k ∈ N with k < n. Then, kb = 0 which implies ka = 0
by part a). So, o(a) 6= n which is a contradiction. Therefore, o(b) = n.