Thread: Not Principal Ideal

How would you prove it's not a P.I.D. by using I?

Let be an ideal of . Prove that I is not a principal ideal of .

I've been trying this for a long time... and I'm just saying okay it has this form, and then I multiply it by (a+bsqrt(-5)) and this happens... but I keep getting so many terms, that I can't find a contradiction. Please help!

Originally Posted by gummy_ratz

How would you prove it's not a P.I.D. by using I?

Let be an ideal of . Prove that I is not a principal ideal of .

I've been trying this for a long time... and I'm just saying okay it has this form, and then I multiply it by (a+bsqrt(-5)) and this happens... but I keep getting so many terms, that I can't find a contradiction. Please help!

Think about norms. $\displaystyle N(2)=4$, $\displaystyle N(1+\sqrt{-5})=1+25 = 26$ (in general, $\displaystyle N(a+b\sqrt{-5})=a^2+25b^2$).

You should therefore notice that if I is a principle ideal, the principle element (or what ever it is called - lets denote it by x) has norm 2 (why?). You should also notice that there exists some a and b such that $\displaystyle x(a+b\sqrt{-5}) = 1+\sqrt{-5}$ and $\displaystyle N(a+b\sqrt{-5})=13$. So...can you find a contradiction in what I have just written?

hmm okay so, 2 is in I. So x*(a + b*sqrt(-5)) = 2. So N(x*a)*N(b*sqrt(-5)) = N(2)
So x^2*a^2 + x^2*25b^2 = 4. But 1 is not in I. So b=0, a=1, x=2.

But then 1 + sqrt(-5) is in I. So x^2*a^2 + x^2*25b^2 = 26. But we know x=2, so 4a^2 + 100b^2 =26, which is not possible.

Is that what you mean? Also, could you create a norm like this for any ideal you were considering in Z[sqrt(-5)]?

Originally Posted by gummy_ratz

hmm okay so, 2 is in I. So x*(a + b*sqrt(-5)) = 2. So N(x*a)*N(b*sqrt(-5)) = N(2)
So x^2*a^2 + x^2*25b^2 = 4. But 1 is not in I. So b=0, a=1, x=2.

But then 1 + sqrt(-5) is in I. So x^2*a^2 + x^2*25b^2 = 26. But we know x=2, so 4a^2 + 100b^2 =26, which is not possible.

Is that what you mean? Also, could you create a norm like this for any ideal you were considering in Z[sqrt(-5)]?

Not quite...have you come across norms before? The point is that $\displaystyle N(xy)=N(x)N(y)$ for all x, y in your ring. So, specifically, $\displaystyle N(x(a+b\sqrt{-5})=N(x)N(a+b\sqrt{-5})$. Use this.

Why would the norm be ?

Wouldn't it be a^2 + 5b^2?

Since (a+bsqrt(-5))*(a-bsqrt(-5)) = a^2 + 5b^2?

Originally Posted by gummy_ratz

How would you prove it's not a P.I.D. by using I?

Let be an ideal of . Prove that I is not a principal ideal of .

I've been trying this for a long time... and I'm just saying okay it has this form, and then I multiply it by (a+bsqrt(-5)) and this happens... but I keep getting so many terms, that I can't find a contradiction. Please help!

Using, perhaps, the idea of norm Swlabr wrote (and you're right, it must be 5 and not 25. A

typo, surely) , prove that $\displaystyle 6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$ are two fundamentally

different factorizations of 6 in that ring and, thus, there is not unique factorization there, so

it cannot be a PID.

Tonio

Originally Posted by tonio

Using, perhaps, the idea of norm Swlabr wrote (and you're right, it must be 5 and not 25. A

typo, surely) , prove that $\displaystyle 6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$ are two fundamentally

different factorizations of 6 in that ring and, thus, there is not unique factorization there, so

it cannot be a PID.

Tonio

Yes, well, more of a mental-o (I was just squaring everything in sight!)