up to speed on primes

wildrabbitt · 2018-11-13, 09:05

Hi,

I'm trying to get a precise definition of 'prime' in sets of numbers beyond the set of natural numbers.

In my open university book it says :

Let $R$ be a commutative ring. A non-zero, non-unit element $p$ of $R$ is prime if whenever $p\mid ab$ then either $p\mid a$ or $p \mid b$.

When I see what this means for the set of integers I get to a conclusion I know must be wrong :

$2 \mid 4\times8$ but 2 divides both 4 and 8. It can't be said that 2 divides either 4 or 8.

In other definitions of prime in rings and fields, there's no either in the wording. This makes more sense but is

Let $R$ be a commutative ring. A non-zero, non-unit element $p$ of $R$ is prime if whenever $p\mid ab$ then $p\mid a$ or $p \mid b$.

the right definition of prime?

Nick · 2018-11-13, 09:29

Yes, you're absolutely right!
The point is that, in the integers for example, 4 divides 6x10 but 4 does not divide 6 or 10.
This is only possible because 4 is not a prime number.

Dr Sardonicus · 2018-11-13, 13:29

Strike out "either." It should be the "inclusive or." If p is prime (a "prime element"), then p|a*b implies p|a or p|b (or both). The condition that p is not a unit, i.e. p does not divide 1, is to avoid trivialities. Of course, 0 is excluded from this definition since division by zero is undefined.

There is however, a special (and important) type of ring called an integral domain, in which a*b = 0 implies a = 0 or b = 0.

The ring of integers Z is an integral domain. The ring of 2x2 matrices with integer entries is not an integral domain. The ring of the integers mod 4 is not an integral domain, since 2*2 = 0.

wildrabbitt · 2018-11-14, 00:33

Thanks to both posters. Very helpful.

Dr Sardonicus · 2018-11-15, 03:44

My example of the ring of 2x2 matrices was inapt. Yes, it has plenty of "divisors of zero," but it's not an integral domain for a much more basic reason.

Its multiplications isn't commutative!

One can take commutative subrings, though. For example, the ring generated by (integer multiples of) the 2x2 identity I2 and M = [0,1;1,0] (which obviously commute) has

M^2 = I2, so that

(M - I2)*(M + I2) = 0.

wildrabbitt · 2018-11-15, 15:19

I guess M = [0,1;1,0], means

0 1
1 0

CRGreathouse · 2018-11-16, 00:29

Quote:

Originally Posted by wildrabbitt

I guess M = [0,1;1,0], means

0 1
1 0

Yes (in PARI/GP notation).

Neutron3529 · 2018-12-24, 14:44

I know a weird thing called "ideal", it might be the definition of 'prime' in sets of numbers beyond the set of natural numbers.

Since it is quite difficult to define a "prime" beyond the set of natural numbers.

For example, in $\mathbb Z[\sqrt{-1}]$ , the defination of "prime" will be what you describe, but in $\mathbb Z[\sqrt{-5}]$ , since $9=3*3=(2+\sqrt{-5})(2-\sqrt{-5})$ , the "prime" you define will be quite different than what you are thinking about.

Actually both $3|(2+\sqrt{-5})$ and $3|(2-\sqrt{-5})$ could not be true, but $3|(2+\sqrt{-5})(2-\sqrt{-5})=9$

Nick · 2018-12-24, 14:54

In our Basic Number Theory course, we look at a more general definition of primes here:
https://www.mersenneforum.org/showthread.php?t=22479

2018-11-13, 09:05	#1
wildrabbitt Jul 2014 3·149 Posts	up to speed on primes Hi, I'm trying to get a precise definition of 'prime' in sets of numbers beyond the set of natural numbers. In my open university book it says : Let \(R\) be a commutative ring. A non-zero, non-unit element \(p\) of \(R\) is prime if whenever \(p\mid ab\) then either \(p\mid a\) or \(p \mid b\). When I see what this means for the set of integers I get to a conclusion I know must be wrong : \(2 \mid 4\times8\) but 2 divides both 4 and 8. It can't be said that 2 divides either 4 or 8. In other definitions of prime in rings and fields, there's no either in the wording. This makes more sense but is Let \(R\) be a commutative ring. A non-zero, non-unit element \(p\) of \(R\) is prime if whenever \(p\mid ab\) then \(p\mid a\) or \(p \mid b\). the right definition of prime? Last fiddled with by wildrabbitt on 2018-11-13 at 09:07

2018-11-13, 13:29	#3
Dr Sardonicus Feb 2017 Nowhere 1001000100011₂ Posts	Strike out "either." It should be the "inclusive or." If p is prime (a "prime element"), then p\|ab implies p\|a or p\|b (or both). The condition that p is not a unit, i.e. p does not divide 1, is to avoid trivialities. Of course, 0 is excluded from this definition since division by zero is undefined. There is however, a special (and important) type of ring called an integral domain, in which ab = 0 implies a = 0 or b = 0. The ring of integers Z is an integral domain. The ring of 2x2 matrices with integer entries is not an integral domain. The ring of the integers mod 4 is not an integral domain, since 22 = 0. Last fiddled with by Dr Sardonicus on 2018-11-13 at 14:06*

2018-12-24, 14:44	#8
Neutron3529 Dec 2018 China 41 Posts	I know a weird thing called "ideal", it might be the definition of 'prime' in sets of numbers beyond the set of natural numbers. Since it is quite difficult to define a "prime" beyond the set of natural numbers. For example, in $\mathbb Z[\sqrt{-1}]$ , the defination of "prime" will be what you describe, but in $\mathbb Z[\sqrt{-5}]$ , since $9=33=(2+\sqrt{-5})(2-\sqrt{-5})$ , the "prime" you define will be quite different than what you are thinking about. Actually both $3\|(2+\sqrt{-5})$ and $3\|(2-\sqrt{-5})$ could not be true, but $3\|(2+\sqrt{-5})(2-\sqrt{-5})=9$ Last fiddled with by Neutron3529 on 2018-12-24 at 14:45*

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2018-11-13, 09:29	#2
Nick Dec 2012 The Netherlands 1702₁₀ Posts	Yes, you're absolutely right! The point is that, in the integers for example, 4 divides 6x10 but 4 does not divide 6 or 10. This is only possible because 4 is not a prime number.

2018-11-14, 00:33	#4
wildrabbitt Jul 2014 3·149 Posts	Thanks to both posters. Very helpful.

2018-11-15, 03:44	#5
Dr Sardonicus Feb 2017 Nowhere 4,643 Posts	My example of the ring of 2x2 matrices was inapt. Yes, it has plenty of "divisors of zero," but it's not an integral domain for a much more basic reason. Its multiplications isn't commutative! One can take commutative subrings, though. For example, the ring generated by (integer multiples of) the 2x2 identity I2 and M = [0,1;1,0] (which obviously commute) has M^2 = I2, so that (M - I2)*(M + I2) = 0.

2018-11-15, 15:19	#6
wildrabbitt Jul 2014 3·149 Posts	I guess M = [0,1;1,0], means 0 1 1 0

2018-12-24, 14:54	#9
Nick Dec 2012 The Netherlands 1702₁₀ Posts	In our Basic Number Theory course, we look at a more general definition of primes here: https://www.mersenneforum.org/showthread.php?t=22479