When a prime isn’t a prime 

It's too bad that the definitions of irreducible and prime elements at the end are wrong. (the definition of prime is almost right - it just needs the proviso that _p_ is non-zero and does not have a (multiplicative) inverse. the definition of irreducible also needs this proviso, but the correct condition is that the only divisors of _p_ are invertible elements _u_ and invertible elements times _p_ , _up_ .)

The positive (or non-negative) integers under multiplication aren't a ring even if we consider the usual definition of addition of integers.

@@schweinmachtbree1013 Zero can't be a prime in the integers (resp., non-negative integers) under multiplication since 1 (resp, +-1) is (resp., are) the only unit (resp, units), and Zero = (Zero )(any integer). i.e., Zero is divisible by any integer.

@@someperson188 The positive/non-negative integers are what's called a "semiring". If a semiring has a multiplicative identity then one can speak of units. Zero is excluded from being prime/composite or irreducible/reducible exactly because of the behaviour you describe - in terms of the "for all a and b, p | ab => p | a or p | b" and "the only divisors of p are units and the associates of p" parts of the definitions however, the former is satisfied by 0 (in any domain) while the latter is not (except in fields). That is, 0 would be prime but not irreducible if it weren't prohibited by the definitions, and this in itself is further reason to prohibit 0 from being prime/composite or irreducible/reducible: if 0 were permitted then it would mess with the implication prime=>irreducible which holds in any domain.

@@schweinmachtbree1013 prime means irreducible and not a unit (i.e., a divisor of the identity). If there's a zero element (it doesn't exist in the positive integers under the usual multiplication, I assume you mean the commutative monoid with identity has at least 2 elements and zero = (zero)(anything). Hence, zero cant be a unit. Since zero = (zero)(zero), the product of two non-units, zero isn't a prime.

Well yes, but the three summand way is much more intuitive as an introduction to the idea.

--every element of a field would be prime by Michael's definition-- , but the standard definition of primes and irreducibles is that they are non-zero non-units by definition, so that a field has no primes nor irreducibles. (Edit: deleted the first part of my comment - I was talking nonsense, time for my coffee)

@@schweinmachtbree1013 Why would that be? His definition demands p e1, so 1 wouldn't be prime. And any other element also wouldn't be prime by his definition because -1 is also a divisor. Well, disregarding characteristic 2.

@@artey6671 Yes you're right ignore me lol. But now that you mention -1, a better (but still wrong in general) definition for the natural number case would have been "p ϵ *N* is prime if p ≠ 1 and its only divisors are ±1 and ±p" (where "divisor" is now being understood as "positive or negative divisor"). This would then be adapted to the integers as "p ϵ *Z* is prime if p ≠ 0,±1 and its only divisors are ±1 and ±p" (and this would be the correct definition of an irreducible element in any ring whose group of units has order 1 or 2)

Lord of the Ring(theorie)s

yeah he should have written ac=2 ⇒ a=2,c=1 or a=-2,c=-1 (or vice versa swapping a and c)

To be picky, primality and irreducibility are the same "for algebraic numbers" because the set of all algebraic numbers forms a field and primality and irreducibility are equivalent in any field (because there are no primes or irreducibles). Likewise, primality and irreducibility are the same "for algebraic integers" because the ring of all algebraic integers is a GCD domain which means irreducible⇒prime and hence irreducible⇔prime. The (pedantically) correct statement would be that primality and irreducibility are not the same in all algebraic number rings (rings consisting of algebraic numbers).

in fact if d is a squarefree integer>=3, Z[sqrt(-d)] has nonunique factorization

what exactly do you mean squarefree? like a non-perfect square integer?

@@Ninja20704 An integer is squarefree if 1 is the only perfect square that divides that integer

@@darkmask4767 ok thank you for the clarification. I've never heard this term before.

Not sure if this is number theory class or algebraic number theory class. If the latter, then it's quite easy: We can take the minimal polynomial of sqrt(-5) to be x^2+5, the discriminant of this polynomial is -5. In particular this is squarefree so Z[sqrt(-5)] is the entire ring of integers, looking at 29, it'll be prime if (29) is inert in the ring of integers as an ideal. Reduction mod n

13 letters?

@@chimkinovania5237 nope. 18

@@yahoo5726 you high on god

Spaces and ( and : aren’t letters

@@angelmendez-rivera351 Can you give me an example of where it matters?

@@angelmendez-rivera351 What's the difference between [(1, 1)] and [(2, 2)]?

mu * mu cant be written as a+b*mu, so multiplication doesnt stay inside your set and primes arent really meaningful to talk about

Either that or he just bought his shirt in a charity shop.

time traveling prime unraveling math man

Surely ZF is far more abstract than this?

Artin’s algebra (freely available on the internet) is good for this

In fact, it’s the same exact distribution as normal primes, x/ln(x) (if by absolute value you mean norm, which is more natural to talk about in a general sense anyway). The idea is that the special primes that break apart completely make up about half the primes, but they each break up into 2 distinct primes. And the others have norm p^2, so their norms occur sparsely enough that they don’t contribute to the asymptotic. So in the Eisenstein setting, it’s like instead of the list 2, 3, 5, 7, 11, 13…, the list is 3, 4, 7, 7, 13, 13, 19, 19, 25, … The 1 mod 3 primes form half the original list, but they appear twice in the new list to compensate.

I was thinking the same. When you allow factors that aren't part of the natural numbers, then obviously the whole concept of only having 1 and itself as possible factors breaks.

What is it written as?

You’re right, but 2 is unique in the Gaussian integers because it’s the only prime that “ramifies”, or factors into a unit times a square. I also noticed he didn’t make a similar comment regarding 3 and the Eisenstein primes, even though 3 plays the same role there as 2 does in the Gaussian integers

interesting point. could it be because binary parity is the simplest form of variation? as in, if all things are not necessarily “this,” then some things must be “that.”

@@RandyKing314 I'm not sure that holds. Divisibility by three sorts things into "this" (is divisible) and "that" (isn't). I suppose there's a minor argument favouring 2 in that divisibility by 2 shorts things into equal sets

@@RandyKing314 well, then that's simply the fact that 2 is the first prime. SOME prime has to be divisible into this exclusive group.

We don’t, though. The natural numbers are what’s known as a Unique Factorization Domain, which has the property that all irreducibles are prime. So we’ve recovered the traditional definition of prime.

is induction with bounded quantifiers enough to show that prime(x) implies irred(x)?

@@schweinmachtbree1013 I don't think induction is needed at all. My logic is a little rusty. But here goes. I'm using Kaye's Models of Peano Arithmetic as my reference. In this book, we are given as an axiom that ∀xyz [ (0 < z ∧ x < y) → xz < yz ] - essentially these axioms for

@@Chalisque I thought you didn't need induction in the other direction irred(x) ⇒ prime(x) either, but thinking about things all the way back to first principals you do: irred(x) ⇒ prime(x) follows from Euclid's lemma. Euclid's lemma can be proved by induction, but we're trying to avoid that so we can instead say that it follows by Bezout's identity. Bezout's identity can be proved using the Euclidean algorithm but that requires induction (one uses the fact that any strictly decreasing sequence of natural numbers will reach zero to prove that it always terminates, which is equivalent to induction), so let's instead say that Bezout's identity follows from the fact that *Z* is a principal ideal domain, however here we finally get cornered because the proof that *Z* is a PID makes a lot of use of induction: it not only uses division-with-remainder (which relies on the fact that any strictly decreasing sequence of integers will become negative, which is again equivalent to induction) but also the Well Ordering Principle that any subset of natural numbers has a least element, which is also equivalent to (unbounded) induction. To be honest it is extremely rare for a statement about the natural numbers or integers to *not* depend on induction, as the natural numbers are defined by induction (recursion), but I find it interesting that sometimes statements about *N* or *Z* depend only on bounded induction. Thanks for bringing this curiosity to my attention! :)

@@schweinmachtbree1013 I remember this factlet as an exercise from my grad student days. In Kaye's Models of Peano Arithmetic chapter 5. The proof of Euclidean division uses induction on ∃rs [ ... ], which is not Δ0, but Σ1.

He isn't squaring both sides, he's multiplying both sides by their complex conjugates.

a^2 + 5b^2 = (a + b sqrt(-5))(a - b sqrt(-5))

It means There's more Than 2. For example, For 5, There is 1, 5, (2+i), (2-i) That is 4 Distinct factors. Or, if You prefer, (1+0i),(5+0i), (2+i),(2-i)

No you don't get the same problem in ℤ[ω] as you do in ℤ[sqrt(-5)]. Michael showed that in ℤ[sqrt(-5)] it is not the case that irreducible ⇒ prime. In any "domain" (a kind of ring), prime ⇒ irreducible. In ℤ[ω] it turns out that irreducible _does_ imply prime, so that irreducible ⇔ prime, and this is due to ℤ[ω] having unique factorization like ℤ (the same goes for ℤ[i] as it is also a unique factorization domain). As michael demonstrated, on the other hand ℤ[sqrt(-5)] does not have unique factorization.

@@schweinmachtbree1013 gotcha

@@schweinmachtbree1013 What happens if you do stop right before the end?

@@Anonymous-df8it Well I don't see exactly what Ethan was proposing. Michael showed that 7 equals (2 + 3ω)(2 + 3ω^2) and hence isn't prime in ℤ[ω] - I don't see how multiplying out the brackets would lead to a different factorization (and indeed because ℤ[ω] is a unique factorization domain, it can't lead to an essentially different factorization). If you or Ethan can clarify what he's suggesting though then please fill me in

@@schweinmachtbree1013 Oh, ok!

Yeah michael got the definition wrong - what he should have said at the end is that _p_ ≠ 0 in _R_ is irreducible if _p_ does not have a multiplicative inverse and its only divisors are of the form _u_ and _up_ where _u_ has a multiplicative inverse (we say _u_ is a "unit"). Likewise _p_ ≠ 0 in _R_ is prime if _p_ does not have a multiplicative inverse (i.e. _p_ is not a unit) and _p_ | _ab_ ⇒ _p_ | _a_ or _p_ | _b_ for all _a_ and _b_

i is a unit of the Gaussian integers (i.e., i divides 1). That's like saying that 3 isn't a prime of the integers because 3 = (-1)(-3). See my comment above.

(1-6i)*(1+6i) =37, which is a prime, but not the prime 7.

For some reason it is standard practice for mathematicians to write "if" when they really mean "iff" for definitions (I for one think it is better to write "iff" and would encourage doing so)

@@schweinmachtbree1013 I agree. Writing "iff" instead of "if" makes the statement less ambiguous. It can make a drastic difference when a statement is a one-way implication versus when it is a two-way implication.

It makes no sense to use 'iff' in a definition since it does not state a logical equivalence.

@@pedroteran5885 definitions certainly _do_ state logical equivalences. for example a number _n_ is even by definition if and only if _n_ = 2 _k_ for some integer _k_