Today I Learned

June 16th

Today I learned finish to the proof that any $G$-module can be embedded into an injective $G$-module. This turns out to be somewhat involved. We recall from yesterday that any abelian group can be embedded into a divisible abelian group, which is the same thing as an injective $\langle e\rangle$-module. Here is the main attraction for today.

The key construction is to use the above lemma to embed the abelian group $M$ into a divisible group, which we will call $N,$ and then fix $\op{Hom}_\ZZ(\ZZ[G],N),$ which consist of the group homomorphisms $\ZZ[G]\to N.$ The $\ZZ$ in the subscript means that we require $\varphi:\ZZ[G]\to N$ to be $\ZZ$-linear, meaning that\[\varphi(ng)=n\varphi(g)\]for $n\in\ZZ$ and $g\in\ZZ[G].$ However, this follows by, say, induction on the fact $\varphi$ is a group homomorphism already, so the $\ZZ$ is mostly redundant to point out that we are not trying to be $G$-linear.

Anyways, to turn $\op{Hom}_\ZZ(\ZZ[G],N)$ into a right (!) $G$-module, we first note that $\ZZ[G]$ is a $G$-module, where $g$ acts on $\sum_{h\in G}a_hh$ by taking it to $\sum_{h\in G}a_h(hg).$ Indeed, $\times g$ is a group homomorphism because\[\left(\sum_{h\in G}(a_h+b_h)h\right)g=\sum_{h\in G}(a_h+b_h)(hg)=\sum_{h\in G}a_h(hg)+\sum_{h\in G}b_h(hg).\]This action is $G$-compatible because\[\left(\sum_{h\in G}a_h\right)(gg')=\sum_{h\in G}a_hh(gg')=\left(\left(\sum_{h\in H}a_hh\right)g\right)g'.\]Now, given $\varphi:\ZZ[G]\to N$ and $g\in G,$ we define the action (note the multiplication on the right!)\[(g\varphi)(a):=\varphi(ag),\]which works because $\times g$ is a group homomorphism, and the composition of group homomorphisms is still a group homomorphism. The inversion here makes this $G$-compatible: $(gh)\varphi(a)=\varphi\left(a(gh)\right)=\varphi\left((ag)h\right)=g(h\varphi)(a).$ The simplicity of this (not requiring any inversion) is why we made $\ZZ[G]$ a right $G$-module, though of course it is also a left $G$-module.

Anyways, to finish showing that $\op{Hom}_\ZZ(\ZZ[G],N)$ is a $G$-module, we need to know the action is linear, but we can fix $\varphi,\varphi':\ZZ[G]\to N$ and $g\in G$ so that\[(g(\varphi+\varphi'))(r)=(\varphi+\varphi')(rg)=\varphi(rg)+\varphi'(rg)=(g\varphi+g\varphi')(r).\]So indeed, $\op{Hom}_\ZZ(\ZZ[G],N)$ is a $G$-module. Note that the proof of this only required knowing that $\ZZ[G]$ was a right $G$-module.

We divide the rest of the proof of the proposition into steps. The first part will focus on $M$ because we need to embed $M$ into $\op{Hom}_\ZZ(\ZZ[G],N).$

We begin by describing our embedding $M\to\op{Hom}_\ZZ(\ZZ[G],N).$ Write $\iota_N:M\to N$ for the embedding of $M$ in $N,$ and then we send $m\in M$ to\[\varphi_m\left(\sum_{g\in G}a_gg\right):=\sum_{g\in G}a_g\cdot\iota_N(gm),\]where the final sum is evaluate in $N.$ Of note $gm$ is the group action of $G$ on $M.$ This $\varphi_m$ is a group homorphism because\[\varphi_m\left(\sum_{g\in G}(a_g+b_g)g\right)=\sum_{g\in G}(a_g+b_g)\cdot\iota_N(gm)=\sum_{g\in G}a_g\cdot\iota_N(gm)+\sum_{g\in G}b_g\cdot\iota_N(gm),\]which collapses back into $\varphi_m\left(\sum_{g\in G}a_gg\right)+\varphi_m\left(\sum_{g\in G}b_gg\right).$

We have left to show that the map $m\mapsto\varphi_m$ is an injective $G$-module homomorphism. It is a group homomorphism because\[\varphi_{m+n}\left(\sum_{g\in G}a_gg\right)=\sum_{g\in G}a_g\cdot\iota_N(g(m+n))=\sum_{g\in G}a_g\cdot\iota_N(gm)+\sum_{g\in G}a_g\cdot\iota_N(gn)\]by using the fact $G$ is an action and $\iota_N$ a group homomorphism. This collapses to $\varphi_m+\varphi_n,$ so we are safe. To finish showing that the embedding is a $G$-module homomorphism, we have to show that it preserves the group action, which means we need $\varphi_{gm}=g\varphi_m,$ for which we note\[(g\varphi_m)\left(\sum_{h\in G}a_hh\right)=\sum_{h\in G}a_h\iota_N((hg)m)=\varphi_{gm}\left(\sum_{h\in G}a_hh\right).\]So we have established that our map $M\to\op{Hom}_\ZZ(\ZZ[G],N)$ is a $G$-module homomorphism. Our final promise is that it is an embedding, but note\[\varphi_m(1e)=1\iota_N(em)=\iota_N(m),\]so the input $1e$ detects $m.$ Explicitly, if $\varphi_m=\varphi_n,$ then plugging in $1e$ means $\iota_N(m)=\iota_N(n),$ from which $m=n$ follows because $\iota_N$ is an embedding. $\blacksquare$

At this point we can throw out the $G$-module $M.$ Indeed, the second part of the proof will be showing that $\op{Hom}_\ZZ(\ZZ[G],N)$ is an injective $G$-module.

The main idea in the proof is currying. Indeed, fix $G$-modules $A\subseteq B$ with a map $\varphi_A:A\to\op{Hom}_\ZZ(\ZZ[G],N)$ so that we need to extend $\varphi_A$ out to $\varphi_B:B\to\op{Hom}_\ZZ(\ZZ[G],N).$ Because $N$ is divisible, we know group homomorphisms into $N$ can be extended, so we try to message such a group homomorphism.

If we ignore all the structure present, then the way to turn a map of the form $A\to\op{Hom}_\ZZ(\ZZ[G],N)$ into a map into $N$ is to think about this like $A\to(\ZZ[G]\to N)$ and then un-curry into $A\times\ZZ[G]\to N.$ However, this will not let us preserve our group structure, for\[\begin{cases} \varphi_A(a_1+a_2)(r_1)=\varphi_A(a_1)(r_1)+\varphi_A(a_2)(r_1), \\ \varphi_A(a_1)(r_1+r_2)=\varphi_A(a_1)(r_1)+\varphi_A(a_1)(r_2),\end{cases}\]for any $a_1,a_2\in A$ and $r_1,r_2\in\ZZ[G].$ The point is that our map $A\times\ZZ[G]\to N$ needs to be linear in both arguments.

To keep track of this algebraic structure, the key observation is to push it into the domain! Namely, we need the tensor product. Simply being linear in both arguments gives $A\otimes_\ZZ\ZZ[G],$ but we note that we also have\[\varphi_A(ga)(r)=(g\varphi_A(a))(r)=\varphi_A(a)(rg)\]by definition of the action on $\op{Hom}_\ZZ(\ZZ[G],N),$ so we also want linearity in $G.$

Thus, we define the following object. Define $A\otimes_G\ZZ[G]$ as\[A\otimes_G\ZZ[G]=(A\times\ZZ[G])/\sim,\]which is the group generated $\ZZ$-linearly by pairs in $A\times\ZZ[G]$ modulo $\sim$ defined to give all of the following relations: given $a,a_1,a_2\in A$ and $r,r_1,r_2\in\ZZ[G]$ with $g\in G,$

1. $(a,r_1+r_2)\sim(a,r_1)+(a,r_2),$

2. $(a_1+a_2,r)\sim(a_1,r)+(a_2,r),$

3. $(ga,r)\sim(a,rg).$

We will denote the equivalence classes of the pair $(a,r)$ under $\sim$ by $a\otimes r$ for $a\in A$ and $r\in\ZZ[G].$ The definition we gave technically defines this group as elements of the form\[\sum_{a\in A,r\in\ZZ[G]}k_{a,r}(a\otimes r)\]where all but finitely many of the $k_{a,r}$ are zero. Defining addition pointwise on the coordinates $a\otimes r,$ we need to check that the addition is well-defined and that it creates a group.

Checking that this forms a group is done in the same way as checking $\ZZ[G]$ is a group. Checking that addition is well-defined is annoying, but we will say that if two $\sum_{a,r}k_{a,r}(a\otimes r)$ and $\sum_{a,r}k_{a,r}'(a\otimes r)$ are equal, then there is a finite sequence of steps (a), (b), and (c) proving their equality. So if $\sum_{a,r}\ell_{a,r}(a\otimes r)=\sum_{a,r}\ell_{a,r}'(a\otimes r)$ are also equal by some sequence of steps,\[\sum_{a,r}(k_{a,r}+\ell_{a,r})(a\otimes r)=\sum_{a,r}(k_{a,r}'+\ell_{a,r}')(a\otimes r)\]can be established by doing the sequence for $\sum k=\sum k'$ and $\sum\ell=\sum\ell'$ one after the other. We won't dwell on this point.

Anyways, with this construction in hand, we have the following lemma.

For the forwards direction, suppose we have a $G$-module homomorphism $\varphi_A:A\to\op{Hom}_\ZZ(\ZZ[G],N)$ and want to induce a group homomorphism $\psi_A:A\otimes_G\ZZ[G]\to N.$ We claim that extending\[\psi_A(a\otimes r):=\varphi_A(a)(r)\]linearly will work. Most of our time is going to be spent showing that this is a well-defined map. It suffices to show that this definition $\psi_A$ respects the equivalence relation $\sim.$ (If two elements in $A\otimes_G\ZZ[G]$ are equal with $\sim,$ then there should be a finite number of moves of the type given above showing the equality. So respecting the individual moves is enough by, say, induction.) Fix $a,a_1,a_2\in A$ and $r,r_1,r_2\in\ZZ[G]$ with $g\in G,$

1. Note $\varphi_A(a)(r_1+r_2)=\varphi_A(a)(r_1)+\varphi_A(a)(r_2)$ because $\varphi_A(a)$ is a group homomorphism. This gives $\psi_A(a\otimes(r_1+r_2))=\psi_A(a\otimes r_2)+\psi_A(a\otimes r_2).$

2. Note $\varphi_A(a_1+a_2)(r)=\varphi_A(a_1)(r)+\varphi_A(a_2)(r)$ because $\varphi_A$ is itself a group homomorphism. This gives $\psi_A((a_1+a_2)\otimes r)=\psi_A(a_1\otimes r)+\psi_A(a_2\otimes r).$

3. Note $\varphi_A(ga)(r)=\big(g\varphi_A(a)\big)(r)=\varphi_A(a)(rg)$ because $\varphi_A$ is a $G$-module homomorphism. This gives $\psi_A(ga\otimes r)=\psi_A(a\otimes rg).$

Having shown all the individual components of $\sim,$ we can be comforted that $\psi_A$ is well-defined. It remains to show that $\psi_A$ is a group homomorphism. Well, we defined $\psi_A$ to extend linearly, which gives this for free. (I am growing tired of being rigorous.) This finishes this direction.

For the backwards direction, suppose we have a $\psi_A:A\otimes_G\ZZ[G]\to N$ and want to induce $\varphi_A:A\to\op{Hom}_\ZZ(\ZZ[G],N).$ Then we simply set\[\varphi_A(a)(r):=\psi_A(a\otimes r).\]This time around this is a perfectly good function. We might be afraid of collisions, but at least this function is surely well-defined. To start, this $\varphi_A$ does indeed give homomorphisms $\op{Hom}_\ZZ(\ZZ[G],N)$ because, fixing $a\in A$ and picking up $r_1,r_2\in\ZZ[G],$ we see\[\varphi_A(a)(r_1+r_2)=\psi_A(a\otimes(r_1+r_2))=\psi_A(a\otimes r_1+a\otimes r_2)=\psi_A(a\otimes r_1)+\psi_A(a\otimes r_2),\]which is indeed $\varphi_A(a)(r_1)+\varphi(a)(r_2).$ Secondly, $\varphi_A$ is itself a group homomorphism because, picking up $a_1,a_2\in A$ and arbitrary $r\in\ZZ[G],$\[\varphi_A(a_1+a_2)(r)=\psi_A((a_1+a_2)\otimes r)=\psi_A(a_1\otimes r+a_2\otimes r)=\psi_A(a_1\otimes r)+\psi_A(a_2\otimes r),\]which again correctly comes out to $\varphi_A(a_1)(r)+\varphi(a_2)(r).$ Lastly, $\varphi_A$ is a $G$-module homomorphism because, fixing $a\in A,$ $r\in\ZZ[G]$ and now $g\in G,$ we have\[\big(g\varphi_A(a)\big)(r)=\varphi_A(a)(rg)=\psi_A(a\otimes rg)=\psi_A(ga\otimes r)=\varphi_A(ga)(r).\]This finishes this direction and so the proof of the lemma. $\blacksquare$

With the above lemma in hand, we can quickly get the result. Note that the give $\varphi_A$ induces a group homomorphism $\psi_A:A\otimes_G\ZZ[G]\to N.$ Now, the divisibility of $N$ (we finally use the divisibility!) implies that we can pull this $\psi_A$ back to a $\psi_B:B\otimes_G\ZZ[G]$ because\[A\otimes_G\ZZ[G]\subseteq B\otimes_G\ZZ[G].\](There are literally more pairs in the second.) But then the $\psi_B$ induces a homomorphism of $G$-modules $\varphi_B:B\to\op{Hom}(\ZZ[G],N).$ This is what we wanted. $\blacksquare$

As some final commentary, I am amused by the trick of encoding algebraic information about maps into the structures themselves. From the outset, it is quite reasonable to expect that $G$-module homomorphisms $A\to\op{Hom}_\ZZ(\ZZ[G],N)$ to be harder to talk about than group homomorphisms, but if we make the groups sufficient complicated, it turns out to be the same data!

Anyways, we remark that we have somewhat explicitly given a way to get an injective resolution, with the exception of the use of the Zorn's lemma to retrieve the injectivity maps for a divisible abelian group. Regardless, we do know that we can embed an arbitrary $G$-module $M$ into\[M\hookrightarrow\op{Hom}_\ZZ(\ZZ[G],\QQ[M]).\]Now we can write the injective resolution from yesterday as having $I^{-2}=0$ and $I^{-1}=M$ as base cases and then\[I^{k+2}=\op{Hom}_\ZZ\left(\ZZ[G],\QQ[I^{k+1}/\del^k(I^k)]\right),\]where $\del^{-2}:0\to M$ and the rest are defined recursively by the composition of\[I^{k+1}\to I^{k+1}/\del^k\left(I^k\right)\hookrightarrow I^{k+2}.\]Granted, this construction isn't pretty, but it is still a construction, and we can be guaranteed that it always works.