# Proving that a ring is non a PID

#### pantboio

##### Member
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
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
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
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.

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 $13$.