Infinitely many primes in Q[Sqrt(d)]

[Brimley]

Hello PhysicsForums people!

Could someone show me how there are inifintely many primes in Q[Sqrt(d)]?

I figure this will be very similar to the proof of there being infinitely many primes in Z.

I cited the above from Wikipedia -Prime Number - The Number of Prime Numbers

There are infinitely many prime numbers. The oldest known proof for this statement, sometimes referred to as Euclid's theorem, is attributed to the Greek mathematician Euclid. Euclid states the result as "there are more than any given [finite] number of primes", and his proof is essentially the following:
Consider any finite set of primes. Multiply all of them together and add 1 (see Euclid number). The resulting number is not divisible by any of the primes in the finite set we considered, because dividing by any of these would give a remainder of 1. Because all non-prime numbers can be decomposed into a product of underlying primes, then either this resultant number is prime itself, or there is a prime number or prime numbers which the resultant number could be decomposed into but are not in the original finite set of primes. Either way, there is at least one more prime that was not in the finite set we started with. This argument applies no matter what finite set we began with. So there are more primes than any given finite number. (Euclid, Elements: Book IX, Proposition 20)

 

[Hurkyl]

The easiest way is to show that each integer prime has algebraic primes lying over it.

 

[Brimley]

The easiest way is to show that each integer prime has algebraic primes lying over it.

Thank you for your response! How can I show this?

 

[Brimley]

Well I have been researching it and I can't find anything on Google like what you had mentioned. Could you possibly break it down further and show how this is done? Thank you! - Brimley

 

[Petek]

You need to study algebraic number theory to understand why Q(sqrt(d)) has infinitely many primes. AFAIK, Euclid's proof doesn't generalize to the case of Q(sqrt(d)). To give you an idea of what's involved, here's an example using Q(i).

We consider what happens to a prime in Z when considered as an element of Q(i). There are three possibilities:

1) The prime number 2 may be written as i(1 - i)^2, where i is a unit in Q(i) and 1-i is a prime in Q(i).

2) The prime number 3 remains prime in Q(i).

3) The prime number 5 may be written as (1 + 2i)(1 - 2i) where both 1 +2i and 1 - 2i are prime in Q(i).

Every other prime in Z will fall under one of the above possibilities. So each prime number in Z corresponds to at least one prime number in Q(i). Since Z contains infinitely many primes, so does Q(i). Of course, all these statements need to be proved and then generalized to the case of Q(sqrt(d)). This would be covered in a book on algebraic number theory.

HTH

Petek

 

[Brimley]

You need to study algebraic number theory to understand why Q(sqrt(d)) has infinitely many primes. AFAIK, Euclid's proof doesn't generalize to the case of Q(sqrt(d)). To give you an idea of what's involved, here's an example using Q(i).

We consider what happens to a prime in Z when considered as an element of Q(i). There are three possibilities:

1) The prime number 2 may be written as i(1 - i)^2, where i is a unit in Q(i) and 1-i is a prime in Q(i).

2) The prime number 3 remains prime in Q(i).

3) The prime number 5 may be written as (1 + 2i)(1 - 2i) where both 1 +2i and 1 - 2i are prime in Q(i).

Every other prime in Z will fall under one of the above possibilities. So each prime number in Z corresponds to at least one prime number in Q(i). Since Z contains infinitely many primes, so does Q(i). Of course, all these statements need to be proved and then generalized to the case of Q(sqrt(d)). This would be covered in a book on algebraic number theory.

HTH

Petek

I will cite the book "An Introduction to Number Theory" (Harold M. Stark) which I recently picked up. In it, they say the following:

Let d be a fixed rational number which is not the square of a rational number. We let Q[Sqrt(d)] denote the set of numbers a+b*Sqrt(d) where a and b are arbitrary rational numbers. We call Q[Sqrt(d)] a quadratic field and if d>0, it is a real quadratic field but if d<0, it is a complex or imaginary quadratic field. As an example 3+4*Sqrt(2) is a member of Q[Sqrt(2)].

So using this, how would I write a proof showing that there exists an infinite number of primes of the form a+b*Sqrt(d) ? From my mind, doesn't d have to be a perfect even square number, b|a and/or b=a=1? That makes sense to me however I'm not sure if its true.

As an example:
1+1*Sqrt(4)=1+2=3 Prime
1+1*Sqrt(9)=1+3=4 Not prime (not an even square)
1+1*Sqrt(16)=1+4=5 Prime
1+1*Sqrt(25)=1+5=6 Not prime (not an even square)
1+1*Sqrt(36)=1+6=7 Prime

Does that make sense or is this completely off? If its off, could someone offer some assistance as to how I would write this?

 

[robert Ihnot]

That is the opposite of the point. Reads: "Which is not the squre of a rational number."

I will cite the book "An Introduction to Number Theory" (Harold M. Stark) which I recently picked up. In it, they say the following:

Let d be a fixed rational number which is not the square of a rational number. We let Q[Sqrt(d)] denote the set of numbers a+b*Sqrt(d) where a and b are arbitrary rational numbers. We call Q[Sqrt(d)] a quadratic field and if d>0, it is a real quadratic field but if d<0, it is a complex or imaginary quadratic field. As an example 3+4*Sqrt(2) is a member of Q[Sqrt(2)].

So using this, how would I write a proof showing that there exists an infinite number of primes of the form a+b*Sqrt(d) ? From my mind, doesn't d have to be a perfect even square number, b|a and/or b=a=1? That makes sense to me however I'm not sure if its true.

As an example:
1+1*Sqrt(4)=1+2=3 Prime
1+1*Sqrt(9)=1+3=4 Not prime (not an even square)
1+1*Sqrt(16)=1+4=5 Prime
1+1*Sqrt(25)=1+5=6 Not prime (not an even square)
1+1*Sqrt(36)=1+6=7 Prime

Does that make sense or is this completely off? If its off, could someone offer some assistance as to how I would write this?

 

[Brimley]

That is the opposite of the point. Reads: "Which is not the squre of a rational number."

I see. Can someone show me how I would prove this then? I've cited a few sources but I can't put all the pieces together.

 

[Petek]

Are you looking for a detailed, rigorous proof or just an outline of one? Here's an outline: Let K = Q([itex]\sqrt {d}[/itex]) be a quadratic number field and let p be a rational prime (that is, a prime number in Z). Then p is an integer in K and hence may be written as a product of primes in K. (There may be more than one way to express p as a product of primes in K, but that doesn't matter.) If q is a rational prime not equal to p, then its representation as a product of primes in K will not have any prime factors in common with the representation of p. Thus, to each rational prime there corresponds at least one distinct prime in K. Since there are infinitely many rational primes, the same is true of primes in K.

If you want a more detailed proof, we have to go into what it means to be an "integer" and a "prime" in K, as well as other notions such as units and norms, for example. Have you taken an undergraduate course in abstract algebra?

 

[Brimley]

If you want a more detailed proof, we have to go into what it means to be an "integer" and a "prime" in K, as well as other notions such as units and norms, for example. Have you taken an undergraduate course in abstract algebra?

That proof makes sense to me! I do know what all of those terms are (Units, Norms, Integers) etc. I had taken a course in Number Theory years ago but I'm brushing up on some things.

Thank you!