Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $R=\mathbb{Z}[\sqrt{-6}]$ and $I=(2,\sqrt{-6})$ the ideal generated by $2$ and $\sqrt{-6}$.

I want to show that $I$ is a projective $R$-module by producing a short exact sequence that splits, has a free $R$-module as the middle term, and with $I$ as one of the other terms.

I first tried to produce an $R$-module homomorphism $g:R^{2}\rightarrow I$ by $(r,s)\mapsto 2r+\sqrt{-6}s$, then find another $R$-module homomorphism $h:I\rightarrow R^{2}$ such that $g\circ h=id_{I}$, but I failed.

share|improve this question
1  
Why do you think it is not projective? –  user27126 Feb 9 '13 at 8:19
    
@Sanchez: Indeed it was a typo. –  user44532 Feb 9 '13 at 16:26

3 Answers 3

up vote 1 down vote accepted

Alex Youcis gave the general result. A mechanism for expressing an ideal $I$ of a Dedekind domain as a direct summand of free module is to use the inverse in the class group. If $J$ is a fractional ideal representing the inverse, then we can (by scaling) arrange things such that $IJ=R$. This means that we can represent $1$ $$ 1=\sum_{t=1}^k i_tj_t $$ as a finite sum with some elements $i_t\in I, j_t\in J$. Observe that by the definition of the inverse, al the products $ij, i\in I, j\in J$ are elements of $R$. It is easy to show that the elements $i_t$ (resp. $j_t$) generate $I$ (resp. $J$) as an $R$-module. Therefore we have the following covering of $I$ by a free $R$-module: $$ p:R^k\rightarrow I, (r_1,r_2,\ldots,r_k)\mapsto \sum_{t=1}^k{r_ti_t}. $$ This is split by the mapping $$ s:I\rightarrow R^k, i\mapsto (ij_1,ij_2,\ldots,ij_k), $$ so we get $R^k\simeq s(i)\oplus \mathrm{ker}\, p$. As $s$ is clearly a monomorphism $s(I)\simeq I$.

In the present specific example case we can spot that $I$ is of order two in the class group, because $$ I^2=(4,2\sqrt{-6},-6)=(2) $$ is principal. Thus we can use $J=\frac12 I$ as the inverse fractional ideal. So $J$ is the $R$-module generated by $1$ and $\sqrt{-6}/2$. We can write $$ 1=4-3=4\cdot1+\sqrt{-6}\cdot\frac{\sqrt{-6}}2 $$ in the prescribed way, where in all the terms the former factor is from $I$ and the latter from $J$. The general recipe gives us a cover mapping $$ \begin{eqnarray*} p:&R^2&\to I\\ & (r_1,r_2)&\mapsto 4r_1+\sqrt{-6}r_2\end{eqnarray*} $$ that is split by the $R$-homomorphism $$ \begin{eqnarray*} s: &I& \to R^2\\ &i&\mapsto \left(i,i\,\frac{\sqrt{-6}}2\right).\end{eqnarray*} $$

share|improve this answer
1  
I edited your answer above (some of the formatting); please tell me if it looks ok. –  user38268 Feb 9 '13 at 10:32

I'm pretty sure that this is incorrect. Indeed, note that since since $-6\equiv 2\text{ mod }4$ we have that $\mathcal{O}_{\mathbb{Q}(\sqrt{-6})}=\mathbb{Z}[\sqrt{-6}]$ and so $\mathbb{Z}[\sqrt{-6}]$ is a Dedekind domain. It is then a common fact that an integral domain $R$ is a Dedekind domain if and only if every ideal is projective.

EDIT: Assuming YACP is correct, here's how one might proceed:

Theorem: Every ideal of a Dedekind domain $R$ is projective.

Let's prove this.

Let $R$ be a Dedekind domain and let $I$ and $J$ be nonzero ideals of $R$. We proceed as follows:

a) Show that there exists $j\in J$ such that $I+jJ^{-1}=R$, hence there exists $i\in I$, $j\in J$, and $k\in J^{-1}$ such that $i+jk=1$.

b) Let $M$ be the matrix $\begin{pmatrix}i & j\\ -k & 1\end{pmatrix}$ and let $N=\begin{pmatrix}1 & -j\\ k & i\end{pmatrix}$. Show that the map $(x,y)\mapsto (x,y)M$ maps $R\oplus IJ$ to $I\oplus J$, and that $(x,y)\mapsto (x,y)N$ is the inverse map. Thus, $I\oplus J\cong R\oplus IJ$ as $R$-modules.

c) Show that $I$ is a projective $R$-module.

PROOF:

a) We rewrite this equation as $IJ+(j)=J$. Now, since $R$ is a Dedekind domain we know that $IJ$ is $2$-generated, say $IJ=(\alpha,\beta)$. Now, looking at the proof that every ideal in a Dedekind domain is $2$-generated one can see that we, in fact, can prove that if $I$ is an ideal and $\alpha\in I$ then there exists $\beta\in I$ such that $(\alpha,\beta)=I$ (in other words, we can choose one of the generators arbitrarily). Thus, note that $\alpha\in IJ\subseteq IJ$ and so by prior remark we can find $\gamma\in J$ such that $(\alpha,\gamma)=J$, we see then that $IJ+(\gamma)=(\alpha,\beta,\gamma)$ and since $(\alpha,\gamma)=J$ and $\beta\in J$ we see that $(\alpha,\beta,\gamma)=J$ as desired.

b) We note that

$$(x,y)M=(ix-yk,jx+y)$$

Now, $ix\in I$ since $i\in I$, and $yk\in I$ since $y\in IJ\subseteq I$, and thus we see that $ix-yk\in I$. Similarly, $jx\in J$ since $j\in J$, and $y\in J$ since $y\in IJ\subseteq J$. Thus, we see that $M$ really does give a map $R\oplus IJ\to I\oplus J$. Now, we see that

$$(x,y)N=(x-yk,yi-jx)$$

Now, evidently $x-yk\in R$. Note then that $yi\in IJ$ since $y\in J$ and $i\in I$, and $jx\in IJ$ since $x\in I$ and $j\in J$--thus, we see that $yi-jx\in IJ$. Thus, we see that $N$ really is a map $I\oplus J\to R\oplus IJ$. Now, one just computes that

$$MN=NM=(i+jk)I=I$$

and thus $M$ and $N$ are inverses of each other as desired. Since $M$ is a linear bijective map it follows that it is an isomorphism $R\oplus IJ\xrightarrow{\approx}I\oplus J$.

c) We can find an ideal $J$ of $R$ such that $IJ$ is principle, and thus a free $R$-module. We see then that $I\oplus J\cong IJ\oplus R\cong R\oplus R$ and thus $I$ is a direct summand of a free module, and thus projective.

share|improve this answer
3  
I have the feeling that the OP wants to prove that $I$ is projective by showing that $I$ is a direct summand of $R^2$. –  user26857 Feb 9 '13 at 8:54
    
@AlexYoucis +1 For your answer. However what puzzles me is whether we can always find an ideal $J$ so that 1) $I + J =R$ and 2) $IJ$ is principal. If this is true, my answer below then shows that every ideal in a Dedekind domain is a projective $R$ - module. –  user38268 Feb 9 '13 at 10:34

First we note that your ring $R$ is the ring of integers of $\Bbb{Q}(\sqrt{-6})$. Now suppose we can find an ideal $J$ so that

  1. $I + J = R$
  2. $IJ$ is principal equal to $(x)$, for some $x \in R$.

Then we claim that $I \oplus J \cong R \oplus IJ$ and thus $I$ is a direct summand of a free $R$ - module, hence projective. Indeed to see this, first we establish the existence of an ses

$$0 \longrightarrow I \cap J \stackrel{(x,-x)}{\longrightarrow} I \oplus J \stackrel{x+y}{\longrightarrow} R \longrightarrow 0.$$

The surjectivity of the map $I \oplus J \to R$ is established because $I + J = R$, while the injectivity of $x \mapsto (x,-x)$ is clear. Exactness at $I \oplus J$ is clear, so this shows that the sequence above is indeed an ses. Because $R$ is free, by the splitting lemma the ses above is a split ses and so

$$I \oplus J \cong IJ \oplus R.$$

In your example above we can simply take $J = (3,\sqrt{-6})$. This is because $1 = 3-2 \in I+J$ and $IJ= (6,2\sqrt{-6},3\sqrt{-6},-6) = (\sqrt{-6})$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.