homological-algebramodulesprojective-module
I'm able to prove it for finitely generated modules, by appealing to the characterization of a projective module as a summand of a free module, and the fact that finite-rank free modules are isomorphic to their duals.
Is it true for all modules? I have seen seemingly conflicting evidence both ways (mostly against, by observations like the dual of the direct sum of countably many copies of $\mathbb{Z}$ is not free (but could it still be projective?).)
Since $P$ is finitely generated, there is a surjection $\pi\colon R^n\to P$ for some $n$. (Explicitly, if $P$ is generated by $\{a_1,\dots,a_n\}$, and $\{e_1,\dots,e_n\}$ a basis of $R^n$, you can define $\pi$ by $e_i\mapsto a_i$. It is surjective since the $a_i$ are a generating set.) Since $P$ is projective, we find $P\oplus\ker\pi\simeq R^n$, so $P$ is a direct summand of a free $R$-module of finite rank. Standard isomorphisms show, since $R$ is commutative, $$ \operatorname{hom}_R(R^n,R)\simeq\bigoplus_{i=1}^n\hom_R(R,R)\simeq R^n. $$
So we also have $$ R^n\simeq\operatorname{hom}_R(R^n,R)\simeq\operatorname{hom}_R(P\oplus\ker\pi,R)\simeq\operatorname{hom}_R(P,R)\oplus\operatorname{hom}_R(\ker\pi,R). $$
So $\operatorname{hom}_R(P,R)$ is a direct summand of a free $R$-module, hence projective.
Here is the proof I outlined in the comments. I will show a more general fact:
Theorem 1. Let $G$ be a finite group. Let $H$ be a subgroup of $G$. Let $K$ be a commutative ring such that $\left[ G:H\right] $ is invertible in $K$. Let $M$ be a $K\left[ G\right] $-module. Then, $M$ is projective as a $K\left[ G\right] $-module if and only if the restriction $\operatorname{Res}_{H}^{G}M$ is projective as a $K\left[ H\right] $-module.
(Here and in the following, "module" means "left module".)
Our proof of Theorem 1 will rely on the following neat fact:
Theorem 2. Let $G$ be a finite group. Let $H$ be a subgroup of $G$. Let $K$ be a commutative ring such that $\left[ G:H\right] $ is invertible in $K$. Let $M$ be a $K\left[ G\right] $-module. Then, $M$ appears as a direct addend in the $K\left[ G\right] $-module $\operatorname{Ind}_{H} ^{G}\operatorname{Res}_{H}^{G}M$.
Here, if $V$ is any $K\left[ H\right] $-module, then $\operatorname{Ind}_{H}^{G}V$ is defined as the $K\left[ G\right] $-module $K\left[ G\right] \otimes_{K\left[ H\right] }V$, where $K\left[ G\right] $ is considered as a $\left( K\left[ G\right] ,K\left[ H\right] \right) $-bimodule (with both actions given by multiplications inside the group algebra $K\left[ G\right] $).
Our proof of Theorem 2 will rely on the following standard fact:
Lemma 3. Let $A$ be any ring. Let $U$ and $V$ be two $A$-modules. Let $P:U\rightarrow V$ and $Q:V\rightarrow U$ be two $A$-module homomorphisms satisfying $Q\circ P=\operatorname*{id}$. Then, $U$ appears as a direct addend in $V$.
Proof of Lemma 3 (sketched). The map $P$ is injective (since $Q\circ P=\operatorname*{id}$); thus, $P\left( U\right) \cong U$ as $A$-modules. Now argue that $V=P\left( U\right) \oplus\operatorname*{Ker}Q$ (internal direct sum), since each $v\in V$ can be written as $v=\underbrace{P\left( Q\left( v\right) \right) }_{\in P\left( U\right) }+\underbrace{v-P\left( Q\left( v\right) \right) }_{\in\operatorname*{Ker}Q}$ (and since every $v\in P\left( U\right) \cap\operatorname*{Ker}Q$ satisfies $v=0$ due to $Q\circ P=\operatorname*{id}$). Thus, $P\left( U\right) $ is a direct addend of $V$. Hence, $U$ appears as a direct addend in $V$ (since $P\left( U\right) \cong U$). This proves Lemma 3. $\blacksquare$
Proof of Theorem 2. We shall denote the restriction $\operatorname{Res}_{H}^{G}M$ as $M$ when the context makes it clear that we mean a $K\left[ H\right] $-module (and not a $K\left[ G\right] $-module). This is harmless, since $\operatorname{Res}_{H}^{G}M=M$ as $K$-modules.
The definition of $\operatorname{Ind}_{H}^{G}\operatorname{Res}_{H}^{G}M$ yields \begin{equation} \operatorname{Ind}_{H}^{G}\operatorname{Res}_{H} ^{G}M=K\left[ G\right] \otimes_{K\left[ H\right] } \underbrace{\operatorname{Res}_{H}^{G}M}_{=M}=K\left[ G\right] \otimes_{K\left[ H\right] }M. \end{equation}
For each $g\in G$, we define a $K$-linear map \begin{align*} Z_{g}:M & \rightarrow K\left[ G\right] \otimes_{K\left[ H\right] }M,\\ m & \mapsto g\otimes_{K\left[ H\right] }g^{-1}m. \end{align*}
For each $g\in G$, we consider the left coset $gH\in G/H$. It is easy to see that if two elements $u$ and $v$ of $G$ belong to the same left coset $\alpha\in G/H$, then \begin{equation} Z_{u}=Z_{v}. \label{darij1.pf.t2.3} \tag{1} \end{equation}
[Proof of \eqref{darij1.pf.t2.3}: Let $u$ and $v$ be two elements of $G$ that belong to the same left coset $\alpha\in G/H$. Thus, $u\in\alpha$ and $v\in\alpha$. From $v\in\alpha$, we obtain $\alpha=vH$ (since $\alpha$ is a left coset of $H$), so that $u\in\alpha=vH$. In other words, $u=vh$ for some $h\in H$. Now, for each $m\in M$, we have \begin{align*} Z_{u}\left( m\right) & =u\otimes_{K\left[ H\right] }u^{-1}m\qquad\left( \text{by the definition of }Z_{u}\right) \\ & =vh\otimes_{K\left[ H\right] }\underbrace{\left( vh\right) ^{-1} }_{=h^{-1}v^{-1}}m\qquad\left( \text{since }u=vh\right) \\ & =vh\otimes_{K\left[ H\right] }h^{-1}v^{-1}m=v\otimes_{K\left[ H\right] }\underbrace{hh^{-1}}_{=1}v^{-1}m\\ & \qquad\left( \begin{array} [c]{c} \text{since }h\in H\subseteq K\left[ H\right] \text{ and thus }h\text{ can be moved}\\ \text{past the tensor sign} \end{array} \right) \\ & =v\otimes_{K\left[ H\right] }v^{-1}m=Z_{v}\left( m\right) \end{align*} (since the definition of $Z_{v}$ yields $Z_{v}\left( m\right) =v\otimes _{K\left[ H\right] }v^{-1}m$). In other words, $Z_{u}=Z_{v}$. This proves \eqref{darij1.pf.t2.3}.]
Now, for any left coset $\alpha\in G/H$, we can define a $K$-linear map $Z_{\alpha}:M\rightarrow K\left[ G\right] \otimes_{K\left[ H\right] }M$ by picking any representative $u\in\alpha$ and setting $Z_{\alpha}=Z_{u}$. This is well-defined, since $Z_{u}$ does not depend on the choice of representative $u$ (in fact, \eqref{darij1.pf.t2.3} shows that any two representatives $u$ and $v$ yield the same map $Z_{u}=Z_{v}$).
Recall that the set $G/H$ of left cosets is a left $G$-set, with the action being defined by $u\cdot\left( gH\right) =\left( ug\right) H$ for all $u\in G$ and $g\in G$. Next, we claim that any $g\in G$, $\alpha\in G/H$ and $m\in M$ satisfy \begin{equation} Z_{\alpha}\left( gm\right) =g\cdot Z_{g^{-1}\alpha}\left( m\right) . \label{darij1.pf.t2.5} \tag{2} \end{equation}
[Proof of \eqref{darij1.pf.t2.5}: Let $g\in G$, $\alpha\in G/H$ and $m\in M$. Fix any representative $u$ of the left coset $\alpha\in G/H$. Thus, $u\in\alpha$, so that $Z_{\alpha}=Z_{u}$ (by the definition of $Z_{\alpha}$). Also, from $u\in\alpha$, we obtain $\alpha=uH$ (since $\alpha$ is a left coset of $H$) and thus $g^{-1}\alpha=g^{-1}uH$, so that $g^{-1}u\in g^{-1}\alpha$. Hence, the definition of $Z_{g^{-1}\alpha}$ yields $Z_{g^{-1}\alpha} =Z_{g^{-1}u}$.
The definition of $Z_{g^{-1}u}$ yields \begin{equation} Z_{g^{-1}u}\left( m\right) =g^{-1}u\otimes_{K\left[ H\right] }\underbrace{\left( g^{-1}u\right) ^{-1}}_{=u^{-1}g}m=g^{-1}u\otimes _{K\left[ H\right] }u^{-1}gm. \end{equation} Hence, \begin{align*} g\cdot\underbrace{Z_{g^{-1}u}\left( m\right) }_{=g^{-1}u\otimes_{K\left[ H\right] }u^{-1}gm} & =g\cdot\left( g^{-1}u\otimes_{K\left[ H\right] }u^{-1}gm\right) =\underbrace{\left( gg^{-1}\right) }_{=1}u\otimes _{K\left[ H\right] }u^{-1}gm\\ & =u\otimes_{K\left[ H\right] }u^{-1}gm. \end{align*} Comparing this with \begin{equation} Z_{u}\left( gm\right) =u\otimes_{K\left[ H\right] }u^{-1}gm\qquad\left( \text{by the definition of }Z_{u}\right) , \end{equation} we obtain $Z_{u}\left( gm\right) =g\cdot Z_{g^{-1}u}\left( m\right) $. In view of $Z_{\alpha}=Z_{u}$ and $Z_{g^{-1}\alpha}=Z_{g^{-1}u}$, this rewrites as $Z_{\alpha}\left( gm\right) =g\cdot Z_{g^{-1}\alpha}\left( m\right) $. This proves \eqref{darij1.pf.t2.5}.]
Now, define a $K$-linear map \begin{equation} P:M\rightarrow K\left[ G\right] \otimes_{K\left[ H\right] }M \end{equation} by \begin{equation} P=\dfrac{1}{\left[ G:H\right] }\sum_{\alpha\in G/H}Z_{\alpha}. \label{darij1.def.T.2} \tag{3} \end{equation} (This is well-defined, since $\left[ G:H\right] $ is invertible in $K$.)
I claim that this map $P$ is a (left) $K\left[ G\right] $-module homomorphism.
[Proof: Let $g\in G$ and $m\in M$. Recall that $G/H$ is a left $G$-set; thus, the map $G/H\rightarrow G/H,\ \alpha\mapsto g^{-1}\alpha$ is a bijection.
Applying both sides of \eqref{darij1.def.T.2} to $m$, we obtain \begin{equation} P\left( m\right) =\dfrac{1}{\left[ G:H\right] }\sum_{\alpha\in G/H}Z_{\alpha}\left( m\right) . \label{darij1.pf.P-equiv.1} \tag{4} \end{equation}
But applying both sides of \eqref{darij1.def.T.2} to $gm$, we obtain \begin{align*} P\left( gm\right) & =\dfrac{1}{\left[ G:H\right] }\sum_{\alpha\in G/H}\underbrace{Z_{\alpha}\left( gm\right) }_{\substack{=g\cdot Z_{g^{-1}\alpha}\left( m\right) \\\text{(by \eqref{darij1.pf.t2.5})} }}=\dfrac{1}{\left[ G:H\right] }\sum_{\alpha\in G/H}g\cdot Z_{g^{-1}\alpha }\left( m\right) \\ & =\dfrac{1}{\left[ G:H\right] }\sum_{\alpha\in G/H}g\cdot Z_{\alpha }\left( m\right) \\ & \qquad\left( \begin{array} [c]{c} \text{here, we have substituted }\alpha\text{ for }g^{-1}\alpha\text{ in the sum,}\\ \text{since the map }G/H\rightarrow G/H,\ \alpha\mapsto g^{-1}\alpha\text{ is a bijection} \end{array} \right) \\ & =g\cdot\underbrace{\dfrac{1}{\left[ G:H\right] }\sum_{\alpha\in G/H}Z_{\alpha}\left( m\right) }_{\substack{=P\left( m\right) \\\text{(by \eqref{darij1.pf.P-equiv.1})}}}=g\cdot P\left( m\right) . \end{align*}
Now, forget that we fixed $g$ and $m$. We thus have shown that $P\left( gm\right) =g\cdot P\left( m\right) $ for all $g\in G$ and $m\in M$. In other words, the map $P$ is $G$-equivariant. Since $P$ is also $K$-linear, we thus conclude that $P$ is a $K\left[ G\right] $-module homomorphism. Qed.]
On the other hand, we define a $K$-linear map \begin{equation} Q:K\left[ G\right] \otimes_{K\left[ H\right] }M\rightarrow M \end{equation} by \begin{equation} Q\left( u\otimes_{K\left[ H\right] }m\right) =um\qquad\text{for all }u\in K\left[ G\right] \text{ and }m\in M. \end{equation} This is clearly well-defined.
We claim that this map $Q$ is a (left) $K\left[ G\right] $-module homomorphism.
[Proof: Let $v\in K\left[ G\right] $ and $r\in K\left[ G\right] \otimes_{K\left[ H\right] }M$. We shall prove that $Q\left( vr\right) =v\cdot Q\left( r\right) $.
Indeed, both sides of this equality are $K$-linear in $r$; hence, we can WLOG assume that $r$ is a pure tensor (since the $K$-module $K\left[ G\right] \otimes_{K\left[ H\right] }M$ is spanned by pure tensors). Assume this, and write $r$ as $u\otimes_{K\left[ H\right] }m$ for some $u\in K\left[ G\right] $ and $m\in M$. Thus, $r=u\otimes_{K\left[ H\right] }m$, so that $vr=v\left( u\otimes_{K\left[ H\right] }m\right) =vu\otimes_{K\left[ H\right] }m$. Hence, \begin{equation} Q\left( vr\right) =Q\left( vu\otimes_{K\left[ H\right] }m\right) =vum\qquad\left( \text{by the definition of }Q\right) . \end{equation} Comparing this with \begin{equation} v\cdot Q\left( \underbrace{r}_{=u\otimes_{K\left[ H\right] }m}\right) =v\cdot\underbrace{Q\left( u\otimes_{K\left[ H\right] }m\right) }_{\substack{=um\\\text{(by the definition of }Q\text{)}}}=vum, \end{equation} we obtain $Q\left( vr\right) =v\cdot Q\left( r\right) $.
Now, forget that we fixed $v$ and $r$. We thus have shown that $Q\left( vr\right) =v\cdot Q\left( r\right) $ for all $v\in K\left[ G\right] $ and $r\in K\left[ G\right] \otimes_{K\left[ H\right] }M$. In other words, the map $Q$ is a (left) $K\left[ G\right] $-module homomorphism (since $Q$ is $K$-linear). Qed.]
Furthermore, we claim that \begin{equation} Q\circ Z_{\alpha}=\operatorname*{id}\qquad\text{for each }\alpha\in G/H. \label{darij1.pf.t2.9} \tag{5} \end{equation}
[Proof of \eqref{darij1.pf.t2.9}: Let $\alpha\in G/H$. Fix any representative $u$ of the left coset $\alpha\in G/H$. Thus, $u\in\alpha$, so that $Z_{\alpha}=Z_{u}$ (by the definition of $Z_{\alpha}$). Now, for each $m\in M$, we have \begin{align*} Z_{\alpha}\left( m\right) & =Z_{u}\left( m\right) \qquad\left( \text{since }Z_{\alpha}=Z_{u}\right) \\ & =u\otimes_{K\left[ H\right] }u^{-1}m\qquad\left( \text{by the definition of }Z_{u}\right) \end{align*} and thus \begin{align*} \left( Q\circ Z_{\alpha}\right) \left( m\right) & =Q\left( \underbrace{Z_{\alpha}\left( m\right) }_{=u\otimes_{K\left[ H\right] }u^{-1}m}\right) =Q\left( u\otimes_{K\left[ H\right] }u^{-1}m\right) \\ & =\underbrace{uu^{-1}}_{=1}m\qquad\left( \text{by the definition of }Q\right) \\ & =m=\operatorname*{id}\left( m\right) . \end{align*} In other words, $Q\circ Z_{\alpha}=\operatorname*{id}$. This proves \eqref{darij1.pf.t2.9}.]
Now, we can easily show that \begin{equation} Q\circ P=\operatorname*{id}. \label{darij1.pf.t2.10} \tag{6} \end{equation}
[Proof of \eqref{darij1.pf.t2.10}: We have $P=\dfrac{1}{\left[ G:H\right] }\sum_{\alpha\in G/H}Z_{\alpha}$ (by \eqref{darij1.pf.t2.9}) and thus \begin{align*} Q\circ P & =Q\circ\left( \dfrac{1}{\left[ G:H\right] }\sum_{\alpha\in G/H}Z_{\alpha}\right) \\ & =\dfrac{1}{\left[ G:H\right] }\sum_{\alpha\in G/H}\underbrace{Q\circ Z_{\alpha}}_{\substack{=\operatorname*{id}\\\text{(by \eqref{darij1.pf.t2.9})} }}\qquad\left( \text{since the map }Q\text{ is }K\text{-linear}\right) \\ & =\dfrac{1}{\left[ G:H\right] }\underbrace{\sum_{\alpha\in G/H} \operatorname*{id}}_{=\left\vert G/H\right\vert \cdot\operatorname*{id} }=\underbrace{\dfrac{1}{\left[ G:H\right] }\left\vert G/H\right\vert }_{\substack{=1\\\text{(since }\left[ G:H\right] =\left\vert G/H\right\vert \text{)}}}\cdot\operatorname*{id}=\operatorname*{id}. \end{align*} This proves \eqref{darij1.pf.t2.10}.]
Altogether, we thus have found two $K\left[ G\right] $-module homomorphisms $P:M\rightarrow K\left[ G\right] \otimes_{K\left[ H\right] }M$ and $Q:K\left[ G\right] \otimes_{K\left[ H\right] }M\rightarrow M$ that satisfy $Q\circ P=\operatorname*{id}$. Hence, Lemma 3 (applied to $A=K\left[ G\right] $, $U=M$ and $V=K\left[ G\right] \otimes_{K\left[ H\right] }M$) shows that $M$ appears as a direct addend in the $K\left[ G\right] $-module $K\left[ G\right] \otimes_{K\left[ H\right] }M$. In other words, $M$ appears as a direct addend in the $K\left[ G\right] $-module $\operatorname{Ind}_{H}^{G}\operatorname{Res}_{H}^{G}M$ (since $\operatorname{Ind}_{H}^{G}\operatorname{Res}_{H} ^{G}M=K\left[ G\right] \otimes_{K\left[ H\right] }M$). This proves Theorem 2. $\blacksquare$
Proof of Theorem 1. $\Longrightarrow:$ Assume that $M$ is projective as a $K\left[ G\right] $-module. Thus, $M$ is a direct addend of a free $K\left[ G\right] $-module. In other words, $M$ is a direct addend of $\left( K\left[ G\right] \right) ^{\left( I\right) }$ for some set $I$. Consider this $I$.
So we know that $M$ is a direct addend of $\left( K\left[ G\right] \right) ^{\left( I\right) }$. In other words, there exists a $K\left[ G\right] $-module $N$ such that $M\oplus N\cong\left( K\left[ G\right] \right) ^{\left( I\right) }$. Consider this $N$.
The (left) $K\left[ H\right] $-module $\operatorname{Res}_{H} ^{G}\left( K\left[ G\right] \right) $ is precisely the $K\left[ H\right] $-module $K\left[ G\right] $ on which $K\left[ H\right] $ acts by left multiplication. This $K\left[ H\right] $-module $K\left[ G\right] $ is free (indeed, if we pick any system $R$ of representatives of the right $H$-cosets in $G$, so that $G=\bigsqcup\limits_{r\in R}Hr$, then $K\left[ G\right] $ is a free left $K\left[ H\right] $-module with basis $\left( r\right) _{r\in R}$). In other words, there exists a set $R$ such that $K\left[ G\right] \cong\left( K\left[ H\right] \right) ^{\left( R\right) }$ as a $K\left[ H\right] $-module. Consider this $R$.
Now, $K\left[ G\right] \cong\left( K\left[ H\right] \right) ^{\left( R\right) }$ as a $K\left[ H\right] $-module. But restriction of modules respects direct sums; thus, \begin{align*} \left( \operatorname{Res}_{H}^{G}M\right) \oplus\left( \operatorname{Res}_{H}^{G}N\right) & \cong\operatorname{Res}_{H}^{G} \underbrace{\left( M\oplus N\right) }_{\cong\left( K\left[ G\right] \right) ^{\left( I\right) }}\cong \operatorname{Res}_{H}^{G}\left( \left( K\left[ G\right] \right) ^{\left( I\right) }\right) \\ & \cong\left( \underbrace{\operatorname{Res}_{H}^{G}\left( K\left[ G\right] \right) }_{=K\left[ G\right] \cong\left( K\left[ H\right] \right) ^{\left( R\right) }}\right) ^{\left( I\right) } \cong\left( \left( K\left[ H\right] \right) ^{\left( R\right) }\right) ^{\left( I\right) }\\ & \cong\left( K\left[ H\right] \right) ^{\left( R\times I\right) }. \end{align*} Hence, $\operatorname{Res}_{H}^{G}M$ is a direct addend of a free $K\left[ H\right] $-module (namely, of $\left( K\left[ H\right] \right) ^{\left( R\times I\right) }$), and thus is projective as a $K\left[ H\right] $-module. This proves the "$\Longrightarrow$" direction of Theorem 1.
$\Longleftarrow:$ Assume that the restriction $\operatorname{Res}_{H}^{G}M$ is projective as a $K\left[ H\right] $-module. In other words, $\operatorname{Res}_{H}^{G}M$ is a direct addend of a free $K\left[ H\right] $-module. In other words, there exist a set $I$ and a $K\left[ H\right] $-module $N$ such that $\left( \operatorname{Res}_{H}^{G}M\right) \oplus N\cong\left( K\left[ H\right] \right) ^{\left( I\right) }$. Consider these $I$ and $N$.
The definition of induced modules yields $\operatorname{Ind}_{H} ^{G}\left( K\left[ H\right] \right) =K\left[ G\right] \otimes_{K\left[ H\right] }K\left[ H\right] \cong K\left[ G\right] $. But induction of modules respects direct sums; thus, \begin{align*} & \left( \operatorname{Ind}_{H}^{G} \operatorname{Res}_{H}^{G}M\right) \oplus\left( \operatorname{Ind}_{H} ^{G}N\right) \\ & \cong\operatorname{Ind}_{H}^{G}\underbrace{\left( \left( \operatorname{Res}_{H}^{G}M\right) \oplus N\right) }_{\cong\left( K\left[ H\right] \right) ^{\left( I\right) }}\cong \operatorname{Ind}_{H}^{G}\left( \left( K\left[ H\right] \right) ^{\left( I\right) }\right) \cong\left( \underbrace{\operatorname{Ind}_{H}^{G}\left( K\left[ H\right] \right) }_{\cong K\left[ G\right] }\right) ^{\left( I\right) }\\ & \cong\left( K\left[ G\right] \right) ^{\left( I\right) }. \end{align*} Hence, $\operatorname{Ind}_{H}^{G}\operatorname{Res}_{H} ^{G}M$ appears as a direct addend in the $K\left[ G\right] $-module $\left( K\left[ G\right] \right) ^{\left( I\right) }$. Since $M$, in turn, appears as a direct addend in the $K\left[ G\right] $-module $\operatorname{Ind}_{H}^{G}\operatorname{Res}_{H}^{G}M$ (by Theorem 2), we thus conclude that $M$ appears as a direct addend in the $K\left[ G\right] $-module $\left( K\left[ G\right] \right) ^{\left( I\right) }$ (since the binary relation "appears as a direct addend in" is transitive). Thus, $M$ appears as a direct addend in a free $K\left[ G\right] $-module (since $\left( K\left[ G\right] \right) ^{\left( I\right) }$ is clearly a free $K\left[ G\right] $-module). In other words, $M$ is projective as a $K\left[ G\right] $-module. This proves the "$\Longleftarrow$" direction of Theorem 1. $\blacksquare$
Best Answer
Let $P = \bigoplus_{\mathbb{N}}\mathbb{Z}$. Then the dual $\text{Hom}(\bigoplus_{\mathbb{N}}\mathbb{Z},\mathbb{Z})$ is not free.
Assume it is projective,and hence there is some $B$ such that $\text{Hom}(\bigoplus_{\mathbb{N}}\mathbb{Z},\mathbb{Z}) \oplus B$ is free. As Arturo points out subgroups of free Abelian groups are free and so $\text{Hom}(\bigoplus_{\mathbb{N}}\mathbb{Z},\mathbb{Z})$ must be free - which is a contradiction.