Jtdylktuy

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 $h:Irightarrow R^{2}$ such that $gcirc h=id_{I}$, but I failed.

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 $i_tin I, j_tin J$. By the definition of the inverse, all the products $ij, iin I, jin J$ are elements of $R$. It is then 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:
$$
p:R^krightarrow I, (r_1,r_2,ldots,r_k)mapsto sum_{t=1}^k{r_ti_t}.
$$
This is split by the mapping
$$
s:Irightarrow R^k, imapsto (ij_1,ij_2,ldots,ij_k).
$$
As $s$ is clearly a monomorphism $s(I)simeq I$, and we can conclude that $R^ksimeq s(I)oplus mathrm{ker}, p$.

In the specific example case we can use the fact that $I$ is of order two in the class group. More precisely,
$$
I^2=(4,2sqrt{-6},-6)=(2)
$$
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
$$
1=4-3=4cdot1+sqrt{-6}cdotfrac{sqrt{-6}}2
$$
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
$$
begin{eqnarray*} p:&R^2&to I\
& (r_1,r_2)&mapsto 4r_1+sqrt{-6}r_2end{eqnarray*}
$$
that is split by the $R$-homomorphism
$$
begin{eqnarray*} s: &I& to R^2\
&i&mapsto left(i,i,frac{sqrt{-6}}2right).end{eqnarray*}
$$

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 $IJ= (6,2sqrt{-6},3sqrt{-6},-6) = (sqrt{-6})$.

Note that since $-6equiv 2text{ mod }4$ 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.

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 $jin J$ such that $I+jJ^{-1}=R$, hence there exists $iin I$, $jin J$, and $kin J^{-1}$ such that $i+jk=1$.

b) Let $M$ be the matrix $begin{pmatrix}i & j\ -k & 1end{pmatrix}$ and let $N=begin{pmatrix}1 & -j\ k & iend{pmatrix}$. Show that the map $(x,y)mapsto (x,y)M$ maps $Roplus IJ$ to $Ioplus J$, and that $(x,y)mapsto (x,y)N$ is the inverse map. Thus, $Ioplus Jcong Roplus IJ$ 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 $alphain I$ then there exists $betain I$ such that $(alpha,beta)=I$ (in other words, we can choose one of the generators arbitrarily). Thus, note that $alphain IJsubseteq IJ$ and so by prior remark we can find $gammain J$ such that $(alpha,gamma)=J$, we see then that $IJ+(gamma)=(alpha,beta,gamma)$ and since $(alpha,gamma)=J$ and $betain J$ we see that $(alpha,beta,gamma)=J$ as desired.

b) We note that

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

Now, $ixin I$ since $iin I$, and $ykin I$ since $yin IJsubseteq I$, and thus we see that $ix-ykin I$. Similarly, $jxin J$ since $jin J$, and $yin J$ since $yin IJsubseteq J$. Thus, we see that $M$ really does give a map $Roplus IJto Ioplus J$. Now, we see that

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

Now, evidently $x-ykin R$. Note then that $yiin IJ$ since $yin J$ and $iin I$, and $jxin IJ$ since $xin I$ and $jin J$--thus, we see that $yi-jxin IJ$. Thus, we see that $N$ really is a map $Ioplus Jto Roplus IJ$. 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 $Roplus IJxrightarrow{approx}Ioplus J$.

c) We can find an ideal $J$ of $R$ such that $IJ$ is principal, and thus a free $R$-module. We see then that $Ioplus Jcong IJoplus Rcong Roplus R$ and thus $I$ is a direct summand of a free module, and thus projective.