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Proving that a ring is non a PID

pantboio

Member
Nov 20, 2012
45
Let \(\displaystyle R=\mathbb{Z}[\sqrt{-13}]\), let \(\displaystyle p\) be a prime in \(\displaystyle \mathbb{N}\), \(\displaystyle p\neq 2,13\). Suppose that \(\displaystyle p\) divides an integer of the form \(\displaystyle a^2+13b^2\), with \(\displaystyle a,b\) integers and coprime. Let \(\displaystyle P=(p,a+b\sqrt{-13})\) be the ideal generated in \(\displaystyle R\) by \(\displaystyle p\) and \(\displaystyle a+b\sqrt{-13}\) and let \(\displaystyle \overline{P}=(p,a-b\sqrt{-13})\).

1)Prove that \(\displaystyle P\cdot\overline{P}=pR\)

2)Prove that if \(\displaystyle P\) is principal then \(\displaystyle p=A^2+13B^2\) for some \(\displaystyle A,B\) integers

3) Prove that if \(\displaystyle p=a^2+13b^2\) then \(\displaystyle P\) is principal

4) Deduce that \(\displaystyle \mathbb{Z}[\sqrt{-13}]\) is not a PID


I have a proof of point 1). Indeed it is easily seen that \(\displaystyle P\cdot\overline{P}\) can be generated by \(\displaystyle p^2,p(a\pm b\sqrt{-13}),a^2+13b^2\), all elements of \(\displaystyle pR\). Conversely, i proved that we may always suppose, without loss of generality, that \(\displaystyle p^2\) does not divide \(\displaystyle a^2+13b^2\) hence \(\displaystyle p=\gcd(p^2,a^2+13b^2)\), and this last fact implies that \(\displaystyle p\) can be expressed as \(\displaystyle \mathbb{Z}\)-linear combination of \(\displaystyle p^2\) and \(\displaystyle a^2+13b^2\), both elements of \(\displaystyle P\cdot\overline{P}\) so that we get the inclusion \(\displaystyle pR\subseteq P\cdot\overline{P}\).

For the other points, any help would be appreciated
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,716
Let \(\displaystyle R=\mathbb{Z}[\sqrt{-13}]\), let \(\displaystyle p\) be a prime in \(\displaystyle \mathbb{N}\), \(\displaystyle p\neq 2,13\). Suppose that \(\displaystyle p\) divides an integer of the form \(\displaystyle a^2+13b^2\), with \(\displaystyle a,b\) integers and coprime. Let \(\displaystyle P=(p,a+b\sqrt{-13})\) be the ideal generated in \(\displaystyle R\) by \(\displaystyle p\) and \(\displaystyle a+b\sqrt{-13}\) and let \(\displaystyle \overline{P}=(p,a-b\sqrt{-13})\).

1)Prove that \(\displaystyle P\cdot\overline{P}=pR\)

2)Prove that if \(\displaystyle P\) is principal then \(\displaystyle p=A^2+13B^2\) for some \(\displaystyle A,B\) integers

3) Prove that if \(\displaystyle p=a^2+13b^2\) then \(\displaystyle P\) is principal

4) Deduce that \(\displaystyle \mathbb{Z}[\sqrt{-13}]\) is not a PID


I have a proof of point 1). Indeed it is easily seen that \(\displaystyle P\cdot\overline{P}\) can be generated by \(\displaystyle p^2,p(a\pm b\sqrt{-13}),a^2+13b^2\), all elements of \(\displaystyle pR\). Conversely, i proved that we may always suppose, without loss of generality, that \(\displaystyle p^2\) does not divide \(\displaystyle a^2+13b^2\) hence \(\displaystyle p=\gcd(p^2,a^2+13b^2)\), and this last fact implies that \(\displaystyle p\) can be expressed as \(\displaystyle \mathbb{Z}\)-linear combination of \(\displaystyle p^2\) and \(\displaystyle a^2+13b^2\), both elements of \(\displaystyle P\cdot\overline{P}\) so that we get the inclusion \(\displaystyle pR\subseteq P\cdot\overline{P}\).

For the other points, any help would be appreciated
For 2), suppose that $P$ is principal, say $P = (A+B\sqrt{-13})$, for some $A,\;B\in\mathbb{Z}$. Then $\overline{P} = (A-B\sqrt{-13})$, and $P\cdot\overline{P} = (A^2+13B^2)$. But $p\in P\cdot\overline{P}$ (by (1)), and so $p$ is a multiple of $A^2+13B^2$. Now use that fact that $p$ is prime in $\mathbb{Z}$ to deduce that $p=A^2+13B^2$.

For 3), if $p=a^2+13b^2 = (a+b\sqrt{-13})(a-b\sqrt{-13})$, then clearly $P = (a+b\sqrt{-13})$, which is principal.

For 4), you just need to find a prime $p$ in $\mathbb{Z}$ which is not of the form $a^2+13b^2$, but which has a multiple of that form (try $p=11$). Then by (1) and (2) $(p,a+b\sqrt{-13})$ is not principal.
 

pantboio

Member
Nov 20, 2012
45
For 2), suppose that $P$ is principal, say $P = (A+B\sqrt{-13})$, for some $A,\;B\in\mathbb{Z}$. Then $\overline{P} = (A-B\sqrt{-13})$, and $P\cdot\overline{P} = (A^2+13B^2)$. But $p\in P\cdot\overline{P}$ (by (1)), and so $p$ is a multiple of $A^2+13B^2$. Now use that fact that $p$ is prime in $\mathbb{Z}$ to deduce that $p=A^2+13B^2$.

For 3), if $p=a^2+13b^2 = (a+b\sqrt{-13})(a-b\sqrt{-13})$, then clearly $P = (a+b\sqrt{-13})$, which is principal.

For 4), you just need to find a prime $p$ in $\mathbb{Z}$ which is not of the form $a^2+13b^2$, but which has a multiple of that form (try $p=11$). Then by (1) and (2) $(p,a+b\sqrt{-13})$ is not principal.
Almost all is clear now for me, thanks. According to your suggestion, i take the prime \(\displaystyle p=11\), which is not of the form \(\displaystyle A^2+13\cdot B^2\), but has a multiple, namely its square \(\displaystyle 121\) such that \(\displaystyle 121=2^2+13\cdot 3^2\) so that the ideal \(\displaystyle (11,2+3\sqrt{-13})\) is not principal, thanks to 2). There are left only few questions I summarize as follows:

1) What is the role played by point 3) in all this?? I mean, i've used only 2) to state that the ideal is non-principal....?

2) Why (in the book where i found this exercise) are we prevented from taking \(\displaystyle p=2,13\)?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,716
1) What is the role played by point 3) in all this?? I mean, i've used only 2) to state that the ideal is non-principal....?
As far as I can see, (3) is not actually needed for the rest of the question. I suppose it is just there for its intrinsic interest. :confused:

2) Why (in the book where i found this exercise) are we prevented from taking \(\displaystyle p=2,13\)?
In your proof of (1), you said "we may always suppose, without loss of generality, that $p^2$ does not divide $a^2+13b^2$".
You may find that supposition is not valid if $p=2$ or $13$.