2
Units in an Unfamiliar Ring
Paul Keeley
The Citadel
Math 495
For this research project I explored unfamiliar rings--their nature, properties, and operations on them. The main object was to consider the units in them and to see how each differed from, for example, the familiar ring of integers as well as from each other.
Section 0. Introduction
A ring is an ordered triple (R, +, *) of elements with two binary operations, called addition and multiplication and denoted by + and *, satisfying the following postulates:
A1, R is closed with respect to addition.
A2, Addition is associative.
A3, Addition is commutative
A4, There exists an element 0 in R such that a + 0 = a for every a in R.
A5, For each a in R there exists an element -a in R such that a + (-a) = 0
M1, R is closed with respect to multiplication
M2, Multiplication is associative but need not be commutative
D, The two distributive laws hold; that is, if a,b, and c are in R, then
a * (b + c) = (a * b) + (a * c)
(a + b) * c = (a * c) + (b * c) [1, 73]
We informally denote any ring (R, +, *) simply by R.
A ring with unity is one with a multiplicative identity, i.e. there exists an element 1 in R not equal to 0 such that a * 1 = 1 * a = a for every a in R [2, p. 172], [1, p. 74].
A multiplicative inverse of an element a in a ring R with unity 1 not equal to 0 is an element a-1 in R such that aa-1 = a-1a = 1 [2, p. 173].
An element u in R is called a unit of R if it has a multiplicative inverse in R. The unity element is of course a unit. Furthermore, it may be helpful to think of a unit as an element which divides every element in the ring.
In the ring (Z, +, *), the integers with the customary arithmetic, 1 is the unity element and {1, -1} are the units. In the Gaussian integers G = { a +bi: a, b Z} the units are {1,-1, i, -i}.
Section 1. An Unfamiliar Ring
The first ring investigated will be denoted T = {a +b: a, b Î Z}. Since the ring operations are the usual ones of arithmetic, a unit in this ring is an element a + bsuch that 1/(a + b ) is an element of the ring as well.
In order to explore which elements of this ring are units, I first analyzed the conditions under which an element would become a unit. By multiplying top and bottom by the conjugate,
we see that a2 - 2b2 must evenly divide both a and b in order for the result to be an element of the ring. For example, 3 + 2is a unit in this ring since 32 - 2(32) = 1 which obviously evenly divides both 3 and 2. In fact, a2 - 2b2 must equal 1 in order for
a + bto be a unit, as shown in the following theorem.
Theorem 1.1: a2 - 2b2 must equal ± 1 in order for a + bto be a unit.
Proof: Suppose a +bis a unit. Let g = gcd (a, b), g ³ 1. Then a = gq1 and b = gq2.
Now a2 - 2b2 = g2[q12 - 2q22]. By the definition of “divides” for integers, g2 divides a2 - 2b2, and, since a2 - 2b2 divides a and b (since a + bis a unit), it follows that g2|a, g2|b. Thus g2 = g since g is an integer and is the greatest common divisor of (a, b), and so g = 1. Therefore gcd (a, b) = 1, q1 = a and q2 = b. And since a2 - 2b2|a, b then a2 - 2b2 = ± 1.
This leads to the idea of a norm in T, which similarly to the Gaussian integers, is defined by N() = |a2 - 2b2|. Thus we may state that if a + bis a unit, then N(a + b) = 1, gcd (a, b) = 1, and a2 - 2b2 = 1. Conversely, when a2 - 2b2 = 1, then gcd (a, b) = 1, N(a + b) = 1, and a + bis a unit.
Theorem 1.2: Let α = a + band β = c + dbe in T. Then N(α*β) = N(α) * N(β).
Proof: (α*β) = (ac + 2bd) + ( ad + bc) , and so
N(α*β) = (ac + 2bd)2 - 2( ad + bc)2
=
Now N(α) = a2 - 2b2, and N(β) = c2 – 2d2, and so
N(α) * N(β) = (a2 - 2b2) * (c2 – 2d2)
= = N(αβ).
This leads to the following corollary about multiplication of units.
Corollary 1.3: The set of units U in T is closed with respect to multiplication.
Proof: As proven above, u is a unit iff N(u) = 1. By the previous theorem, N(αβ) = N(α)N(β) for α, β in T. Thus if α, β are units, then N(αβ) = 1 x 1 = 1, and thus (αβ) is a unit.
I then explored the idea that the set of units formed a field. This was quickly seen to be untrue. For example, they are not closed under addition [(1) + (3 + 2) =
(4 + 2) which is not a unit].
Section 2. Numerical Properties of the Units in T
By numerical experimentation I found 26 units in this ring, all with integers less than 100, viz.
(1), (1), (32), (75), (1712), (2941), (9970).
I then ordered this set of units on the numberline according to the following subsets: (a + b), (-a - b), (-a + b), (a - b).
(a + b)
{1, 2.414, 5.83, 14.07, 33.97, 82.01, 197.995}
(-a - b)
{-1, -2.414, 5.83, -14.07, 33.97, -82.01, 197.995}
(-a + b)
{-1, 0.414, 0.1716, 0.071, 0.029, 0.0122, 0.005051}
(a - b)
{1, -0.414, 0.1716, -0.071, 0.029, -0.0122, 0.005051}
As can be seen from this ordering of units, there exist infinitely many units in (1, ¥) and (-¥, -1) as well as infinitely many units in the intervals (0, 1) and (-1, 0).
Does there exist a unit u such that 1 < u < (1 + )? Obviously, no such unit exists in the set of units u = a + b, a, b > 0, since 1 is the smallest integer greater than zero. The question remains as to whether or not such a unit exists in the sets which have both positive and negative integer coefficients within the same element. If this were true for the set of units {u | u = (-a + b)}, then there exists a unit j such that 1 < a - b < 1 +. Then
12 < (a - b)2 < (1 +)2, and
1 < a2 + 2b2- ab2< 3 + 2.
Now a2 + 2b2 < 3, and a2 - 2b2 = ± 1, since j is a unit. Let a2 - 2b2 = +1. Then
a2 + 2b2 < 3
+a2 - 2b2 = 1
2a2 < 4
Thus a2 < 2, and since a is an integer, a2 must be 1. So 1 – 2b2 = 1, which means that b = 0. Thus j = 1, which contradicts the assumption that it is greater than 1.
Again let a2 - 2b2 = -1. Then
a2 + 2b2 < 3
+a2 - 2b2 = -1
2a2 < 2
Then a2 = 1, so the above results follow. And a similar argument can be made for the set of units {u | u = (a - b)}. Thus there does not exist a unit in the interval (1, 1 + ).
Theorem 2.1: Every unit greater than 1 is a positive power of 1 + .
Proof: Let α be a unit greater than 1. By the previous theorem, certainly 1 + ≤ α. There must be some exponent, say a possibly large n, such that
(1 + )n ≤ α < (1 + )n + 1.
Now since 1 + is a unit, so are its powers. Thus (1 + ) -n is also a unit, and by multiplying it through the previous inequality we obtain
1 ≤ α (1 + ) -n < 1 +. Now since the product of two units is a unit, α (1 + ) -n is obviously a unit. There being no units between 1 and 1 + , we must have
1 = α (1 + ) -n and so α = (1 + ) n. That is, α is a power of 1 + .
This naturally leads to the question as to the existence of a unit between 1 and
(1 + ) -1 = -1 + . Similarly, no such unit exists, and additionally every positive unit less than 1 is a negative power of (1 + ) , as proven below:
Theorem 2.2: No unit exists in the interval (-1 + , 1).
Proof: Let such a unit u exist. Thus -1 + < u < 1. By multiplying through by
1 + , we see that 1< u(1 + ) < 1 +. Since the units are closed under multiplication, u(1 + ) is a unit. But then there is a unit between 1 and 1 + ; which as shown earlier is impossible. Thus there is no unit between -1 + and 1.
Theorem 2.3: Every positive unit less than 1 is a negative power of (1 + ).
Proof: Let b be a unit <1. Now (1 + )-1 = (-1 + ). By the previous theorem, certainly b < (-1 + ). There must be some exponent, say a possibly large positive n, such that (-1 + )n + 1 < b ≤ (-1 + )n. As before, now divide through by (-1 + )n, yielding -1 + < β (-1 + )-n ≤ 1. But since β (-1 + ) n is a unit, and no units exist between (-1 + ) and 1, we must have β (-1 + ) n = 1, and thus
β = (-1 + )n. That is, β = (-1 + )n = (1 + )-n. Thus all units on the interval
(0, -1 + ) are negative powers of 1 + .
Conclusion: By the preceding two theorems, we may conclude that the set of units in T is {± (1 + )n: n = 0, ±1, ±2, …}.
Section 3. Beyond Units
The concept of norms in T led to further analysis of other properties within the ring. The idea of primes was the first to come up. The first problem was to define the concept within the context of the ring. In the integers, a prime is a number such that it can be factored as the product of unity and itself only. How would such a concept be extended to a ring such as T which contains units other than 1?
I first examined the ordinary integer primes (all of which are contained in T). These obviously do not fit the ordinary definition of a prime since, e.g., 7 = (3 + )(3 - ).
What, then, makes an element of T prime? For the answer I looked at the factorization and norms of the elements.
As proven above, an element a Î T is a unit iff N(a) = 1. Since for two elements a, b Î T, N(ab) = N(a)N(b), an element which has a norm equal to a prime number can only be factored by a unit and another element with a prime norm. This situation, analogous to that of integers, leads to the definition of a prime in T as an element which cannot be factored by two elements of lesser norms; i.e., its norm is a prime integer.
Now if p = a2 - 2b2 where p is a prime in Z, then p is not a prime in T. Note that p = (a + b)( a - b). Since a + bhas a prime norm, a + bis a prime in T.
Thus, which has norm 2 (a prime in Z) is prime in T, while 2 is not prime in T, since it has norm 4. Additionally, it can be factored as *.
Now 4 + 3is prime, since N(4+ 3) = 2, and two is a prime integer. But this does not mean, as might be expected from the set of integers, that 4+ 3can only be factored as a product of itself and unity. For example, (1 + )(2 + ) = 4 + 3, with 1 + being a unit and N(2 + ) = 2.