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For some of you, this question is going to seem extremely basic.

I think I understand what an ideal is. Such as $\langle 2, 1 + \sqrt{-5} \rangle$, it consists of all numbers in this ring of the form $2a + (1 + \sqrt{-5})b$.

But then what is $\langle 2, 1 + \sqrt{-5} \rangle^2$? My first thought was $\langle 4, -4 + 2 \sqrt{-5} \rangle$, but that seems wrong somehow. I also had something in the back of my mind saying $\langle 2 \rangle$, but I'm not sure about that one either.

Then I thought about trying to figure out $\langle 2, 1 + \sqrt{-5} \rangle \langle 2, 1 + \sqrt{-5} \rangle$ when I realized I don't actually understand how to multiply ideals to begin with.

I'm only using $\langle 2, 1 + \sqrt{-5} \rangle$ as an example (though that does draw in one question identified as similar that looks much more relevant than all the questions identified as "Questions that may already have your answer"). In a principal ideal domain, would it be correct to think that $\langle a \rangle \langle b \rangle = \langle ab \rangle$?

Any help would be much appreciated.

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    $\begingroup$ The product of two ideals $IJ$ is (in a commutative ring) the collection of all sums of products $ij$ where $i \in I, j \in J$. $\endgroup$ – Qiaochu Yuan Nov 9 '15 at 22:26
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Both other answers are right but fail to give the answer in its simplest form. Note that $$-4+(2 + 2 \sqrt{-5}) - (-4+2 \sqrt{-5}) = 2.$$ So $2$ is in the ideal $\langle 4, (2 + 2 \sqrt{-5}), (-4+2 \sqrt{-5}) \rangle$ and, conversely, it is easy to see that each of $4$, $2 + 2 \sqrt{-5}$ and $-4+2 \sqrt{-5}$ is divisible by $2$. So $$\langle 2, 1 + \sqrt{-5} \rangle^2 = \langle 4, (2 + 2 \sqrt{-5}), (-4+2 \sqrt{-5}) \rangle = \langle 2 \rangle.$$

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  • $\begingroup$ @DavidESpeyer Doesn't it matter what the parent ring is (from where we take $a$ and $b$.)? Will we get the same answer for $\mathbb{Z}$ and $\mathbb{Q}$? $\endgroup$ – SARTHAK GUPTA Dec 10 '20 at 17:24
  • $\begingroup$ I was assuming $\mathbb{Z}$ although I think technically every thing I wrote is true, but vacuously so, if you consider $\mathbb{Q}$. (Vacuously because $\mathbb{Q}[\sqrt{-5}]$ is a field, so it only has one nonzero ideal.) $\endgroup$ – David E Speyer Dec 10 '20 at 17:37
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When you square an ideal you're really looking at products of two things coming from your ideal. In you case $\langle 2, 1+\sqrt{-5}\rangle^2$ means take something in $\langle 2, 1+\sqrt{-5}\rangle$ and multiply it by something else in the ideal; the set of all the things you get by doing this is the ideal $\langle 2, 1+\sqrt{-5}\rangle^2$.

In order to work with this you can just multiply your generators together, so $\langle 2, 1+\sqrt{-5}\rangle^2=\langle 4,2+2\sqrt{-5},-4+2\sqrt{-5}\rangle$. Then you might look for relations between these generators to reduce this set. In this case it does not appear that there are any relations between them.

You are correct that in a P.I.D., $\langle a \rangle \langle b\rangle=\langle ab\rangle$ provided $a,b$ aren't units (in which case $\langle a \rangle$ or $\langle b \rangle$ would be the entire ring).

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  • $\begingroup$ As long as neither is a unit ideal, right? $\endgroup$ – Bob Happ Nov 9 '15 at 22:33
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    $\begingroup$ @BobHapp yes sorry, let me sort out the other glaring errors first.. $\endgroup$ – Sam Weatherhog Nov 9 '15 at 22:33
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For a P. I. D., yes. For the general case, the product of two ideals $I=(a_1,\dots, a_n)$ and $J=(b_1,\dots, b_p)$ is the ideal generated by all possible products $a_ib_j$.

Hence the square of $I=(2, 1+\sqrt{-5})$ is the ideal $I^2=(4,2+2\sqrt{-5},-4+2\sqrt{-5})$.

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  • $\begingroup$ I think the third generator of $I^2$ should be $-4+2\sqrt{-5}$? And I don't think $I=\mathbb{Z}[1+\sqrt{-5}]$. $\endgroup$ – Sam Weatherhog Nov 9 '15 at 22:46
  • $\begingroup$ I don't think we can get $-6+2\sqrt{-5}$ from these generators either. I don't think $2\in \langle 2, 1+\sqrt{-5} \rangle^2$. $\endgroup$ – Sam Weatherhog Nov 9 '15 at 22:50
  • $\begingroup$ Yes I've just checked, as it seemed unlikely. It's corrected $\endgroup$ – Bernard Nov 9 '15 at 22:51
  • $\begingroup$ Telescoping errors. Thanks for pointing it. $\endgroup$ – Bernard Nov 9 '15 at 22:52
  • $\begingroup$ No worries. I had similar issues trying to find relations between generators and also forgot to square 2 :) $\endgroup$ – Sam Weatherhog Nov 9 '15 at 22:53

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