Let $R=\mathbb Z [\sqrt{-5}]$, and $I=(2,1+\sqrt{-5})$. Prove that the $R$-module $I \bigoplus I$ is free on the generators $(2, 1+\sqrt{-5})$ and $(1-\sqrt{-5},2)$.
Can anyone help me with this question?
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2Can you show us your efforts on this problem? And perhaps at least correct your typos – WhatsUp Dec 16 '19 at 23:51
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@JyrkiLahtonen I think a link might have gotten lost in the mix there? – Badam Baplan Dec 17 '19 at 07:45
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Indeed, @BadamBaplan. I meant to link to this thread. Here is more of the same (generalizing the argument over any Dedekind domain, please read Alex Youcis's answer before mine). – Jyrki Lahtonen Dec 17 '19 at 07:49
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Let $a = (2, 1 + \sqrt{-5}), b = (1 - \sqrt{-5}, 2)$ for convenience.
In order to show that $a,b$ generate $I \oplus I$, let's try to find linear combinations of $a,b$ that give generators of the submodules $I \oplus 0$ or $0 \oplus I$. If you show that $a,b$ span these two submodules then you show they span $I \oplus I$.
For example, let's try to kill the first coordinate... calculate
$$(1 - \sqrt{-5}) a = (2 - 2\sqrt{-5}, 6)$$
$$2b = (2 - 2\sqrt{-5}, 4)$$
Hence $$(1 - \sqrt{-5})a -2b = (0,2)$$
Can you find linear combinations of $a,b$ that yield $(2,0)?$ And use this to finish showing that $a,b$ generate $I \oplus I$?
Well, once you've shown that $a,b$ span $I \oplus I$, you just need to show that they are linearly independent.
To that end, suppose that $r_1a + r_2b = 0$ with $r_1,r_2 \in R$. You aim to show this implies $r_1= r_2 = 0$. Looking at the equation in each coordinate of $I \oplus I$ gives you the two equations:
$$2r_1 + (1 - \sqrt{-5})r_2 = 0$$ $$(1 + \sqrt{-5})r_1 + 2r_2 = 0$$
Can you take it from there?
Badam Baplan
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