Mod 9, every failure is a multiple of 3
Tonight's idea
Work mod 9: after every sum or product, keep only the remainder on dividing by 9. The numbers are 0 to 8, and \(5 \times 7 = 35 \equiv 8\).
To divide by \(a\) you need a partner \(b\) with \(a \times b \equiv 1\). A number with a partner is a unit. Mod 9, \(2 \times 5 = 10 \equiv 1\), so 2 and 5 are units. But \(3 \times b\) only ever lands on 0, 3 or 6, never on 1. You cannot divide by 3.
The non-units mod 9 are 0, 3 and 6. They share two habits. Each is nilpotent: some power of it is 0. And adding two of them gives another, so they form one closed family.
Mod 9, the non-units are exactly the nilpotent numbers, and a sum of non-units is a non-unit. Mod 10, neither statement is true.
The family of nilpotent, "nearly zero" numbers is called a radical. A number system whose non-units are closed under adding is called local: it has just one way to fail.
Watch one: sorting 0 to 8, mod 9
Units. Hunt for partners: \(1 \times 1 = 1\), \(2 \times 5 = 10 \equiv 1\), \(4 \times 7 = 28 \equiv 1\), \(8 \times 8 = 64 \equiv 1\). So 1, 2, 4, 5, 7 and 8 are units.
Non-units. \(0 \times b\) is always 0. As \(b\) runs from 0 to 8, \(3 \times b\) gives 0, 3, 6, 0, 3, 6, 0, 3, 6 and \(6 \times b\) gives 0, 6, 3, 0, 6, 3, 0, 6, 3. Neither list contains 1. So the non-units are 0, 3 and 6, exactly the multiples of 3.
Nilpotent. \(3^2 = 9 \equiv 0\) and \(6^2 = 36 \equiv 0\), and 0 is 0 already. No unit is nilpotent. The powers of 2, for example, go 2, 4, 8, 7, 5, 1 and then repeat, so they never reach 0.
One family. \(3 + 3 = 6\), \(3 + 6 = 9 \equiv 0\), \(6 + 6 = 12 \equiv 3\), and adding 0 changes nothing. Every sum of two non-units is a non-unit.
So mod 9 there is one way to fail: be a multiple of 3. Every failure is a square root of 0.
Your turn
- Work mod 8, where the numbers are 0 to 7. (a) Find a partner for each of 1, 3, 5 and 7. (b) Explain why 0, 2, 4 and 6 have no partner. (c) For each of 2, 4 and 6, find the smallest power that is 0 mod 8.
- Still mod 8. (a) Work out \(2 + 2\), \(2 + 4\), \(2 + 6\), \(4 + 4\), \(4 + 6\) and \(6 + 6\). Is every answer a non-unit? (b) Add the unit 1 to each of the non-units 0, 2, 4 and 6. What kind of number do you get each time?
- Work mod 10. (a) Find the four units and their partners, then list the six non-units. (b) Find two non-units whose sum is a unit. (c) Write out the first few powers of 2, and of 5, mod 10. Is either number nilpotent?
- Mod 12 the non-units are 0, 2, 3, 4, 6, 8, 9 and 10. (a) Write out the powers of each one mod 12 until they repeat. Which are nilpotent? (b) Find two non-units whose sum is a unit. (c) Is the sum of two nilpotent numbers still nilpotent here?
Stretch: every prime power
Let \(p\) be a prime and \(k \ge 1\), and work mod \(p^k\). (Mod 9 is \(p = 3\), \(k = 2\).) (a) Show that the \(k\)th power of any multiple of \(p\) is 0. (b) Explain why a nilpotent number can never be a unit. (c) Harder: let \(a\) be any number that is not a multiple of \(p\). Why are the products \(a \times 0, a \times 1, \dots, a \times (p^k - 1)\) all different mod \(p^k\), and why does that make \(a\) a unit?
Answers
- (a) Each is its own partner: \(1 \times 1 = 1\), \(3 \times 3 = 9 \equiv 1\), \(5 \times 5 = 25 \equiv 1\), \(7 \times 7 = 49 \equiv 1\). (b) An even number times anything is even, and taking away 8s keeps it even. So the product is 0, 2, 4 or 6, never 1. (c) \(2^3 = 8 \equiv 0\) (and \(2^2 = 4\) is not 0), so the third power. \(4^2 = 16 \equiv 0\), the second power. \(6^2 = 36 \equiv 4\) and \(6^3 = 216 = 27 \times 8 \equiv 0\), the third power. Mod 8 the non-units are again exactly the nilpotent numbers.
- (a) 4, 6, 0, 0, 2, 4. All are even, so all are non-units. Sums that use 0 change nothing, so the non-units mod 8 form one family. (b) 1, 3, 5, 7: a unit every time. Odd plus even is odd, and mod 8 the odd numbers are the units.
- (a) \(1 \times 1 = 1\), \(3 \times 7 = 21 \equiv 1\), \(9 \times 9 = 81 \equiv 1\), so the units are 1, 3, 7, 9. The non-units are 0, 2, 4, 5, 6, 8. (b) \(2 + 5 = 7\). Also \(4 + 5 = 9\), \(5 + 6 \equiv 1\) and \(5 + 8 \equiv 3\). So the non-units are not one family: being even and being a multiple of 5 are two different ways to fail. (c) Powers of 2: 2, 4, 8, 6, 2, ... and powers of 5: 5, 5, 5, ... Neither reaches 0, so neither is nilpotent. In fact only 0 is nilpotent mod 10, because a power of \(n\) is a multiple of 10 only when \(n\) itself is a multiple of both 2 and 5.
- (a) 2: 2, 4, 8, 4, 8, ... 3: 3, 9, 3, ... 4: 4, 4, ... 6: 6, 0. 8: 8, 4, 8, ... 9: 9, 9, ... 10: 10, 4, 4, ... Only 0 and 6 are nilpotent. A power of \(n\) is a multiple of 12 only when \(n\) is a multiple of both 2 and 3. (b) \(3 + 4 = 7\) is one; \(2 + 3 = 5\) and \(8 + 9 \equiv 5\) are others. (c) Yes: \(6 + 6 = 12 \equiv 0\). The nilpotent numbers 0 and 6 still form a closed family, but here it is much smaller than the set of non-units.
Stretch. (a) \((pm)^k = p^k m^k\), a multiple of \(p^k\). (b) Suppose \(a^n \equiv 0\) and \(a \times b \equiv 1\). Then \(1 \equiv (ab)^n = a^n b^n \equiv 0\), which is false because \(p^k\) is at least 2. (c) If \(a \times b \equiv a \times c\), then \(p^k\) divides \(a(b - c)\). The prime \(p\) does not divide \(a\), so all \(k\) factors of \(p\) are in \(b - c\). But \(b\) and \(c\) differ by less than \(p^k\), so \(b = c\). That makes \(p^k\) different products among only \(p^k\) remainders, so every remainder appears, including 1. So mod \(p^k\) the non-units, the nilpotent numbers and the multiples of \(p\) are the same family, and it is closed under adding. With two different primes in the modulus, as in 10 and 12, this breaks.