Can the Arithmetic Derivative be Defined on a Non-Unique Factorization Domain?

Pentti Haukkanen, Mika Mattila, and Jorma K. Merikoski School of Information Sciences FI-33014 University of Tampere Finland

                pentti.haukkanen@uta.fi
                  mika.mattila@uta.fi
                jorma.merikoski@uta.fi

Timo Tossavainen

School of Applied Educational Science and Teacher Education

University of Eastern Finland P.O. Box 86 FI-57101 Savonlinna Finland

                timo.tossavainen@uef.fi

Abstract

Given n ∈ ℤ, its arithmetic derivative n′ is defined as follows: (i) 0′ = 1′ = (−1)′ = 0. (ii) If n = up_{1} ⋯ p_{k}, where u = ±1 and p_{1}, …, p_{k} are primes (some of them possibly equal), then

n′ = n ∑_{j=1}^{k} \frac{1}{p_{j}} = u ∑_{j=1}^{k} p_{1} ⋯ p_{j−1}p_{j+1} ⋯ p_{k}.

An analogous definition can be given in any unique factorization domain. What about the converse? Can the arithmetic derivative be (well-)defined on a non-unique factorization domain? In the general case, this remains to be seen, but we answer the question negatively for the integers of certain quadratic fields. We also give a sufficient condition under which the answer is negative.

1 The arithmetic derivative

Let n ∈ ℤ. Its arithmetic derivative n′ (\underline{A003415} in [4]) is defined [1, 6] as follows:

(i) 0′ = 1′ = (−1)′ = 0. (ii) If n = up_{1} ⋯ p_{k}, where u = ±1 and p_{1}, …, p_{k} ∈ ℙ, the set of primes, (some of them possibly equal), then

n′ = n ∑_{j=1}^{k} \frac{1}{p_{j}} = u ∑_{j=1}^{k} p_{1} ⋯ p_{j−1}p_{j+1} ⋯ p_{k}.  (1)

If k = 1, we set p_{1} ⋯ p_{k−1}p_{k+1} ⋯ p_{k} = 1 in the last expression.

A few basic properties of n′ follow:

∀p ∈ ℙ : p′ = 1,
∀n ∈ ℤ : (−n)′ = −n′,
∀m, n ∈ ℤ : (mn)′ = m′n + mn′.

The third equality is called the Leibniz rule. Moreover, f(n) = n′ is the only mapping ℤ → ℤ having these properties. For details, see [6, Theorems 1 and 13].

Kovič [3, Proposition 1] studied how to extend f to ℚ(i) = {a+bi | a, b ∈ ℚ}. Ufnarovski and Åhlander [6, Section 10] outlined how to define the arithmetic derivative on a unique factorization domain (UFD from now on). We begin by performing this task in detail. To that end, we follow the terminology of [2, Section 6.5].

Let D be a UFD. First, we must decide what atoms (irreducible elements) are “positive”. Write P for a set of atoms of D such that every atom of D is associated with one and only one element of P. Call P the set of positive atoms. Further, denote by U the set of units of D.

Given a ∈ D, we define its arithmetic derivative a′ as follows: If a = 0 or a ∈ U, then a′ = 0. Otherwise, there are unique (up to the ordering) p_{1}, …, p_{k} ∈ P (some of them possibly equal) and u ∈ U such that

a = up_{1} ⋯ p_{k}.

Then

a′ = u ∑_{j=1}^{k} p_{1} ⋯ p_{j−1}p_{j+1} ⋯ p_{k}.  (2)

To be precise, we should actually write a′_{P} (or something like that) for the derivative of a, since a′ depends on P. However, for the simplicity of notation, we will omit this practice if there is no need to emphasize P.

The given definition implies the analogous equalities as above:

∀p ∈ P : p′ = 1,
∀v ∈ U, a ∈ D : (va)′ = va′,
∀a, b ∈ D : (ab)′ = a′b + ab′.

Again, f(x) = x′ is the only mapping D → D with these properties.

Example 1. Let D = ℤ. If P_{1} = ℙ, we obtain the ordinary arithmetic derivative defined above. For example, because 30 = 2 · 3 · 5, we have 30′_{P_{1}} = 30′ = 3 · 5 + 2 · 5 + 2 · 3 = 31. Obviously, another selection of positive atoms results in a different derivative function. For instance, if P_{2} = {2, −3, 5, −7, 11, … }, then 30 = (−1) · 2 · (−3) · 5 and 30′_{P_{2}} = (−1) · [(−3) · 5 + 2 · 5 + 2 · (−3)] = 11.

Example 2. Let D be an arbitrary field F. Since all nonzero elements of F are units, then P = ∅ and, hence, a′ = 0 for all a ∈ F.

Example 3. To give an example of a nontrivial derivative on the field ℚ, we define [6, Theorem 14]

(\frac{m}{n})′ = \frac{m′n − mn′}{n^{2}} .  (3)

Here m, n ∈ ℤ, n ≠ 0, and m′ and n′ are ordinary arithmetic derivatives on ℤ. An analogous definition can be given in the division field of any UFD.

Let us summarize the above discussion.

Proposition 4.

(i) Let D be a UFD. The mapping f(a) = a′_{P} defined on D by (2) is an arithmetic derivative. It depends on the chosen set P of positive atoms.

(ii) The mapping g(a) = a′ defined on ℚ by (3) is an extension of the mapping f(a) = a′ defined on ℤ by (1). Similarly, (2) can be extended to the division field of D.

2 A problem and its partial answers

If a factorization domain (FD in the sequel) is not a UFD, we call it a non-unique factorization domain (NUFD in the sequel). We saw above that the arithmetic derivative can be defined on any UFD. What about the converse?

Problem 5. Is it possible to define an arithmetic derivative on some NUFD?

The next theorem gives a partial answer which is negative. In the following, we mostly apply the same terminology and notation as in [5, Chapter 4].

Theorem 6. Let D_{m} be the integral domain of integers of ℚ(\sqrt{m}), where m ∈ ℤ\\{1\} is squarefree (\underline{A005117} in [4]). If m satisfies

m ≢ 1 (mod 4) and m < −2,  (4)

then either 1 − m or 4 − m does not have a well-defined derivative as an element of D_{m}.

Proof. Clearly, D_{m} is an FD, yet it is not a UFD, see [5, p. 93] or [7, Theorem (actually, in Finnish: Lause) 4.23]. We modify and enhance the argument used in the latter reference.

Case 1. m ≡ 3 (mod 4). Then m is odd and m ≤ −5. Since 1 − m is even and greater than five, it is a composite number (when considered as a positive integer) and expressible as

1 − m = p_{1} ⋯ p_{k},  (5)

where k ≥ 2 and p_{1}, …, p_{k} ∈ ℙ with 2 = p_{1} ≤ ⋯ ≤ p_{k}. On the other hand,

1 − m = (1 − \sqrt{m})(1 + \sqrt{m}).  (6)

We will see later that the factors of the right-hand sides of both (5) and (6) are atoms in D_{m}.

Next, if some of the p_{j}’s in (5) are equal, we omit their repetition; let {p_{j_{1}}, …, p_{j_{h}}} be the set obtained so. Since the only units of D_{m} are ±1, see [5, Proposition 4.2] or [7, Lause 4.8], the atoms p_{j_{1}}, …, p_{j_{h}}, 1 − \sqrt{m}, 1 + \sqrt{m} are pairwise non-associated.

Assume first that P is such that

p_{j_{1}}, …, p_{j_{h}}, 1 − \sqrt{m}, 1 + \sqrt{m} ∈ P.  (7)

If (1 − m)′ exists, then, by (6),

(1 − m)′ = 1 + \sqrt{m} + 1 − \sqrt{m} = 2.

On the other hand, (5) implies that

(1 − m)′ = p_{2} ⋯ p_{k} + ⋯ + p_{1} ⋯ p_{k−1} ≥ p_{2} + p_{1} ≥ 2 + 2 = 4

which contradicts the previous conclusion. So, (1 − m)′ is not well-defined under (7). Second, if (7) does not hold, we anyway have

±p_{j_{1}}, …, ±p_{j_{h}}, ±(1 − \sqrt{m}), ±(1 + \sqrt{m}) ∈ P

with an appropriate selection of signs. Hence a simple modification of the above argument is sufficient to show that the derivative of 1 − m is not well-definable.

Case 2. m ≡ 2 (mod 4). Now m is even and m ≤ −6. Thus, 4 − m is also an even composite number such that 4 − m ≥ 10. So, again

4 − m = p_{1} ⋯ p_{k},  (8)

where k and p_{1}, …, p_{k} are as described above. On the other hand,

4 − m = (2 − \sqrt{m})(2 + \sqrt{m}).  (9)

The factors of the right-hand sides of (8) and (9) are atoms in D_{m}, see below.

We continue similarly as in Case 1 only replacing 1 ± \sqrt{m} with 2 ± \sqrt{m}. So, assume first that

p_{j_{1}}, …, p_{j_{h}}, 2 − \sqrt{m}, 2 + \sqrt{m} ∈ P.  (10)

If (4 − m)′ exists, then (4 − m)′ = 2 + \sqrt{m} + 2 − \sqrt{m} = 4 by (9). However, we have k > 2 or p_{k} > 2, since otherwise 4 − m = 2 · 2 = 4 contradicting the assumption m ≤ −6. By (8), we encounter a dilemma in both cases; if k > 2, then

(4 − m)′ ≥ p_{2}p_{3} + p_{1}p_{3} + p_{1}p_{2} ≥ 4 + 4 + 4 = 12,

and

(4 − m)′ ≥ p_{k−1} + p_{k} ≥ 2 + 3 = 5

if p_{k} > 2. Consequently, (1 − m)′ is not well-defined under (10). If P does not satisfy this condition, an analogous argument as at the end of Case 1 applies again.

To complete the proof, we still have to verify that the factors of the right-hand sides of (5), (6), (8) and (9) are atoms. We do so by using the norm function. We begin by noticing that D_{m} = ℤ(\sqrt{m}) = \{x + y\sqrt{m} | x, y ∈ ℤ\}, see [5, Theorem 3.2] or [7, Theorem 4.2]. An element a = x + y\sqrt{m} ∈ D_{m} is rational if y = 0 and irrational if y ≠ 0. If a is irrational, then, recalling that m is negative, we have

N(a) = x^{2} − my^{2} = x^{2} + |m|y^{2} ≥ |m|.

If also b ∈ D_{m} is irrational, then

N(ab) = N(a)N(b) ≥ m^{2}.  (11)

Let c ∈ D_{m} so that c ≠ 0, ±1. If c = ab where a and b are irrational, then N(c) ≥ m^{2} by (11). Therefore, c is an atom if the following two conditions are satisfied: (i) c has no rational atom divisor (except possibly ±c) and (ii) N(c) < m^{2}.

Now choose any p_{j} from (5) or (8). It clearly satisfies the condition (i). Since |m| ≥ 5, we have

p_{j} ≤ p_{k} = \frac{1}{2} · 2p_{k} ≤ \frac{1}{2} p_{1} ⋯ p_{k} ≤ \frac{1}{2}(4 − m) < \frac{1}{2}(5 + |m|) ≤ \frac{1}{2} · 2|m| = |m|.

Hence N(p_{j}) = p_{j}^{2} < m^{2} and, consequently, also (ii) is satisfied. In other words, p_{j} is an atom in D_{m}.

Next, consider the factors of the right-hand sides of (6) and (9), i.e., q = t ± \sqrt{m}, where t ∈ \{1, 2\}. If r, x, y ∈ ℤ so that t ± \sqrt{m} = r(x + y\sqrt{m}), then ry = ±1 implying also that r = ±1. Consequently, q satisfies (i). Also the condition (ii) is now satisfied because

N(1 ± \sqrt{m}) < N(2 ± \sqrt{m}) = 4 − m < 5 + |m| ≤ 2|m| < m^{2}.

So, these numbers are atoms also.

We conclude this section by stating a sufficient condition under which the answer to Problem 5 is negative.

Theorem 7. Let D be an NUFD. Assume that a ∈ D can be factorized so that

a = p_{1}p_{2} = q_{1}q_{2},

where p_{1}, p_{2}, q_{1}, q_{2} are atoms with {p_{1}, p_{2}} ≠ {q_{1}, q_{2}}. Then a does not have a well-defined derivative.

Proof. If a′ exists, then, by the Leibniz rule,

a′ = p_{1} + p_{2} = q_{1} + q_{2}.

But two elements are uniquely defined by their sum and product. To show this, simply note that

{p_{1}, p_{2}} = {q_{1}, q_{2}} ⟺ ∀x ∈ D : (x − p_{1})(x − p_{2}) = (x − q_{1})(x − q_{2})
⟺ ∀x ∈ D : x^{2} − (p_{1} + p_{2})x + p_{1}p_{2} = x^{2} − (q_{1} + q_{2})x + q_{1}q_{2}
⟺ p_{1} + p_{2} = q_{1} + q_{2} ∧ p_{1}p_{2} = q_{1}q_{2}.

(To show the forward implication in the last equivalence, substitute x = 0 and x = 1.) Therefore p_{1} + p_{2} ≠ q_{1} + q_{2} and our claim follows.

Now we encounter another problem arising from Theorem 7.

Problem 8. Does every NUFD contain an element a satisfying the assumption of Theorem 7?

If the answer to this question is positive, then the answer to Problem 5 is negative.

3 Concluding remarks

We defined an arithmetic derivative on a UFD and asked in Problem 5 about the possibility to do so also in some NUFD. If m ≢ 1 (mod 4) and m < −2, the FD of integers of ℚ(\sqrt{m}) is an NUFD. We proved in Theorem 6 that the answer is negative in this case. If m ≡ 1 (mod 4) and m < 0, then this FD is an NUFD if and only if m ≠ −3, −7, −11, −19, −43, −67, −163, see [5, p. 93].

Surveying these m’s might be the next step. However, it is an essentially more laborious task. Namely, assuming m ≡ 1 (mod 4) implies ([5, Theorem 3.2] or [7, Theorem 4.2]) that D_{m} = \{x + y(1 + \sqrt{m})/2 | x, y ∈ ℤ\} which already complicates the situation compared to that in the proof of Theorem 6. More difficulties arise as the simple condition m < −2 is replaced with the condition excluding the mentioned values of m. Studying the integers of ℚ(\sqrt{m}) for m > 1 may be even more difficult since, according to our knowledge, it is not completely understood which m’s yield a UFD and which do not.

Obviously, an alternative way to try to advance is to study NUFD’s different from those described above.

4 Acknowledgment

We thank the referee for valuable remarks, in particular for those that led us to formulate Theorem 7 and Problem 8.

References

[1] E. J. Barbeau, Remark on an arithmetic derivative, Canad. Math. Bull. 4 (1961), 117– 122.

[2] P. M. Cohn, Algebra, Volume 1, John Wiley, 1974.

[3] J. Kovič, The arithmetic derivative and antiderivative, J. Integer Seq. 15 (2012), Article 12.3.8.

[4] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, http://oeis.org.

[5] I. N. Stewart and D. O. Tall, Algebraic Number Theory, Second Edition, Chapman and Hall, 1987.

[6] V. Ufnarovski and B. Åhlander, How to differentiate a number, J. Integer Seq. 6 (2003), Article 03.3.4.

[7] K. Väisälä, Lukuteorian ja korkeamman algebran alkeet [in Finnish], Otava, 1950.

2010 Mathematics Subject Classification: Primary 11A25; Secondary 11A51, 11R27. Keywords: arithmetic derivative, unique factorization, non-unique factorization, quadratic field.

(Concerned with sequences \underline{A000040}, \underline{A003415} and \underline{A005117}.)

Received October 30 2012; revised version received January 1 2013. Published in Journal of Integer Sequences, January 1 2013.

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