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Let $I=(2,1+\sqrt{-5})$ be an ideal of the ring of integers of $\mathbb Q(\sqrt{-5})$. What is its norm $N(I)$? And is $I^2$ principal?

My notes say:

$1$, $\sqrt{-5}$ is a $\mathbb Z$-basis for $\mathcal O_K=\mathbb Z[\sqrt{-5}]$, and especially $1$, $1+\sqrt{-5}$, hence N$(I)=2$. Also, $I^2=(4,2(1+\sqrt{-5}),2(-2+\sqrt{-5}))\subseteq(2)$ and $N(I^2)=N(I)^2=4=N((2))$. So $I^2=(2)$ is principal.

Well, now I am trying to figure out why the ideal $I$ has norm $2$. We know that $N(2)=4$ and $N(1+\sqrt{-5})=6$. So $N(I)$ divides $4$ and $6$, i.e. it is $1$ or $2$. It cannot be $1$, because $I$ is a proper ideal. So it is $2$. But how is this related to the $\mathbb Z$-basis of $\mathcal O_K$?

And I am really clueless as to how I can compute $I^2$. I understand the norm argument, but I just don't know how to compute $I^2$.

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    $\begingroup$ $I$ has index $2$ in $\mathcal{O}_K$ (as seen by looking at the basis you give) $\endgroup$
    – yoyo
    May 1, 2013 at 21:56

3 Answers 3

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Re your first question....

By choosing a $\mathbb{Z}$-basis $\{ 1, 1 + \sqrt{-5} \}$ for $\mathcal{O}_K$, we can write algebraic integers as coordinate "vectors".

The generators of the ideal $I$ are, relative to this basis, $(2,0)$ and $(0, 1)$. This turns out to be a $\mathbb{Z}$-basis for $I$, although you might want to double check by throwing in all four elements of $\{ 2, 1 + \sqrt{-5} \} \cdot \{ 1, \sqrt{-5} \}$ into a matrix and doing (integer) row reduction to simplify. (using all four of these elements ensures that their span is closed under multiplication by elements of $\mathcal{O}_K$).

One definition of the norm of an ideal is that it is the size of the quotient ring $\mathcal{O}_K / I$. Knowing that the basis for $I$ (relative to the basis for $\mathcal{O}_K$) has coordinate matrix

$$ \left( \begin{matrix}2 & 0 \\ 0 & 1 \end{matrix} \right)$$

makes it easy to see that the quotient group has two elements.

Actually, when we do things this way, we can obtain the norm of $I$ as the determinant of its basis matrix.

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  • $\begingroup$ So if we had the ideal $(k,1+\sqrt{-5})$ for $k\in\mathbb Z$, then its norm would always be $|k|$? But what if our ideal had three generators? $\endgroup$
    – Phil-ZXX
    May 2, 2013 at 12:38
  • $\begingroup$ Go through the process I mentioned of doing row reduction on four elements to find the basis. The abelian group generated by $k$ and $1+\sqrt{-5}$ is usually not actually an ideal, so you have to use other generators I mentioned. If I've not made any mistakes, you'll find that the end result is that you have two nonzero rows: $[x,0]$ and $[0,1]$, and that $x$ is usually something that is not $|k|$. If you wrote your ideal with three generators, the whole procedure I described still applies. $\endgroup$
    – user14972
    May 3, 2013 at 1:16
  • $\begingroup$ But when we have the ideal $J=(k,1+\sqrt{-5})$, then the generators (relative to our basis) are $(k,0)$ and $(0,1)$, so the matrix is $$ \left( \begin{matrix}k & 0 \\ 0 & 1 \end{matrix} \right)$$ and its determinant is $k$, which is the norm of the ideal $J$? $\endgroup$
    – Phil-ZXX
    May 3, 2013 at 1:20
  • $\begingroup$ @Thomas: Those are its generators as an ideal, and as a $\mathcal{O}_K$-module. But to do the calculation I described using integer linear algebra, you need its generators as a $\mathbb{Z}$-module -- i.e. its generators as an abelian group. $\endgroup$
    – user14972
    May 3, 2013 at 8:36
  • $\begingroup$ @Hurkyl:I am confused at your last comment. I am in the same position at Tom, so in the case $J=(3,1+u)$, I get the matrix [3,0;0,1]. What can I conclude after this? $\endgroup$
    – lifin
    Feb 18, 2015 at 14:46
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As has already been said in the comments by yoyo, $N(I)=[\mathcal{O}_K:I]=2$. One way of thinking about this is that if you multiply any element of $\mathcal{O}_K$ by $2$, then it will always land in $I$, and you can see this by looking at the $\mathbb{Z}$-basis and the generators of $i$.

To compute $I^2$, you just simply multiply the generators together. So here:

  • $2\cdot 2=4$
  • $2\cdot(1+\sqrt{-5})=2+2\sqrt{-5}$
  • $(1+\sqrt{-5})\cdot 2=2+2\sqrt{-5}$ again, so we only need include this in our list of generators once
  • $(1+\sqrt{-5})\cdot(1+\sqrt{-5})=-4+2\sqrt{-5}$

This gives us all the generators of the ideal, and

$$ I^2=(2,1+\sqrt{-5})(2,1+\sqrt{-5})=(4,2+2\sqrt{-5},-4+2\sqrt{-5}) $$

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About the second point:
In fact, the class number of $\mathbb Q(\sqrt{-5})$ is $2$, so that $I^2$ is principal, whatever the ideal $I$ is. Moreover, by Kummer theory, or Dedekind(?), we know that prime $p$ splits in $\mathbb Q(\theta)$ according to how $f(x)$ factors modulo $p$, where $f$ is the minimal polynomial of $\theta$ over $\mathbb Q$. Now, the minimal polynomial of $\sqrt{-5}$ is $x^2+5\equiv (x+1)^2\pmod2$, so $(2,1+\sqrt{-5})^2=(2)$ by the above-mentioned theory.
Please tell me if some errors occur. Thanks.

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