Infinitely many primes in Q[Sqrt(d)]

**Samson** · October 9th 2010, 04:52 PM

Hello all,

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)

**Brimley** · October 12th 2010, 11:56 AM

I am going to 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, wouldn't one write a proof showing that there exists an infinite number of primes of the form $a+b*\sqrt{d}$ ? From my mind, d has 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

If this is wrong, someone feel free to correct me.

**tonio** · October 12th 2010, 12:07 PM

Originally Posted by Brimley

I am going to 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, wouldn't one write a proof showing that there exists an infinite number of primes of the form $a+b*\sqrt{d}$ ? From my mind, d has 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

If this is wrong, someone feel free to correct me.

The original question is flawed: $\mathbb{Q}(\sqrt{d})$ is a field so there's no point to talk about primes...

Tonio

**Brimley** · October 12th 2010, 12:14 PM

Originally Posted by tonio

The original question is flawed: $\mathbb{Q}(\sqrt{d})$ is a field so there's no point to talk about primes...

Tonio

I don't think his original question is flawed. $Q[\sqrt{d}]$ is a field, but surely within that field are primes. I think he is asking how we can prove that there are infinitely many primes within that field.

**Samson** · October 12th 2010, 07:25 PM

Originally Posted by Brimley

I am going to 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 $Q[\sqrt{2}]$ .

In your source you mentioned $d$ not being the square of a rational number, yet you used 4, 16, and 36 to make your point even though they are perfect squares of 2, 4, and 6 respectively.

Can anyone offer any help with this?

**Brimley** · October 12th 2010, 07:52 PM

Originally Posted by Samson

In your source you mentioned $d$ not being the square of a rational number, yet you used 4, 16, and 36 to make your point even though they are perfect squares of 2, 4, and 6 respectively.

Can anyone offer any help with this?

Sorry about that, my mistake. I hope that somebody can help you with this. My idea flew right out the window...

**tonio** · October 12th 2010, 08:42 PM

Originally Posted by Brimley

I don't think his original question is flawed. $Q[\sqrt{d}]$ is a field, but surely within that field are primes. I think he is asking how we can prove that there are infinitely many primes within that field.

There can't be any non-zero primes in field since any non-zero element is a unit...as simple as that.

What the OP probably meant is to show that the RING WHICH IS NOT A FIELD $\mathbb{Z}[\sqrt{d}]$ has infinite

or whatever primes.

Tonio

**Samson** · October 12th 2010, 10:14 PM

Originally Posted by tonio

There can't be any non-zero primes in field since any non-zero element is a unit...as simple as that.

What the OP probably meant is to show that the RING WHICH IS NOT A FIELD $\mathbb{Z}[\sqrt{d}]$ has infinite

or whatever primes.

Tonio

Well I know that describing why there are infinitely many primes in $\mathbb{Q}[\sqrt{d}]$ should be similar to the proof that there are infinitely many primes in $\mathbb{Z}$ . But how can I prove this?

**tonio** · October 13th 2010, 04:36 AM

Originally Posted by Samson

Well I know that describing why there are infinitely many primes in $\mathbb{Q}[\sqrt{d}]$ should be similar to the proof that there are infinitely many primes in $\mathbb{Z}$ . But how can I prove this?

Once again: but for zero there are NO PRIMES in any field since all the elements but zero are units!

Anyway, you can take any rational prime $p=p+0\cdot\sqrt{d}\in\mathbb{Z}[\sqrt{d}]$ : if it remains prime there good, otherwise write it as

a product of different primes there(why?). If $q\neq p$ is another rational prime then either it is prime or its prime

decomposition in the above ring will be different from that of $p$ ...

Tonio

**Samson** · October 13th 2010, 08:52 AM

I see! I got it from here, thank you!

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