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I want to ask about Problem 7 on the page 96 in this.

  1. (a) If $0\to M_1\xrightarrow{f}M\xrightarrow{g}M_2\to 0$ is a short exact sequence of $R-$modules and $R-$module homomorphisms, prove that $$0\to\text{Hom}_R(R^{(n)},M_1)\xrightarrow{f_{\ast}}\text{Hom}_R(R^{(n)},M)\xrightarrow{g_{\ast}}\text{Hom}_R(R^{(n)},M_2)\to0$$ is exact for any integer $n\geq 1.$

(b) Does the conclusion of (a) remain valid if “for any integer $n\geq1$” is replaced by “for a set $\Delta$” in the statement of Part (a) and $R^{(n)}$ is replaced by $R^{\Delta}$?

I proved that $f_{\ast}$ is injective and $\ker g_{\ast}=\text{Im}f_{\ast}.$ However, I don't find how to proof $g_{\ast}$ is surjective. Morever, I think that (b) isn't right. But I don't find counterexample. Please help me!

Chris Eagle
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Tran Nam Son
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2 Answers2

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You've already been given a method for surjectivity, so I'll try to explain a different general method for this problem which should help explain part (b).

As the other answer explained, a map out of $R^n$ is the same as defining where the basis vectors $e_i$ go. In other words, there is an isomorphism $Hom_R(R^n, M) \cong M^n$ sending $f \mapsto (f(e_i))_i$. This is, in fact, "natural". In other words, the following diagram commutes for any map $f: M \rightarrow N$.

enter image description here

where the vertical maps are the isomorphisms I described before. Sorry for the poor typesetting, I'll try to improve it when I can.

Anyway, this tells us that the sequence you gave is isomorphic to

$$ 0 \rightarrow M_1^n \xrightarrow{f^n} M_2^n \xrightarrow{g^n} M_3^n \rightarrow 0. $$

Exactness of this sequence is more easily checked.

For part (b), one would hope to do the same thing. However, it's not in general true that $Hom_R(R^\Delta, M) \cong M^\Delta$ naturally for $\Delta$ infinite. We'd have to replace $R^\Delta$ with $\bigoplus_{d \in \Delta} R$, the free $R$-module on $\Delta$. For ease of notation I'll call this $R^{\oplus \Delta}$. This is defined as $\{a \in R^\Delta : \text{ all but finitely many } a_d = 0\}$. Observe that this is just $R^\Delta$ for $\Delta$ finite. It turns out $Hom_R(R^{\oplus \Delta}, M) \cong M^\Delta$ naturally. You can then apply similar reasoning to prove this.

To get to the question you asked, the answer is that the sequence need not be exact. It will always be exact in the places you've already checked, but surjectivity can fail. The key notion here is projective modules. An $R$ module $P$ is said to be projective if $Hom_R(P, -)$ preserves exact sequences. It turns out that all modules preserve the left part of the exact sequence, so projectivity is equivalent to preserving the surjectivity part. The discussion above, by the way, leads to the result that arbitrary direct sums of $R$ are projective. However, it is not the case that all products of $R$ are projective. In fact, $\mathbb Z^{\mathbb N}$ is not projective. I'm not sure of an easy way to prove this, but as $\mathbb Z$ is a PID, it suffices to show that $\mathbb Z^{\mathbb N}$ is not free (i.e. isomorphic to a direct sum of copies of $\mathbb Z$). This isn't the easiest result, and is proven here.

paul blart math cop
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The key point is a map (of $R$-modules) out of $R^n$ is determined by its image on the basis vectors $e_i$, and vice versa. Here are some more details:

Let $e_i=(0, 0, \ldots, 1, 0, \ldots 0) \in R^n$ Given a map $Y:R^n \to M_2$, let $y_i=Y(e_i) \in M_2$ be the image of the basis vector $e_i \in R^n$, for each $1 \leq i\leq n$. Let $x_i \in M$ be a lift of $y_i$, i.e. $g(x_i)=y_i$. Then define $X:R^n \to M$ by $X(e_i)=x_i$; this defines $X$ on all of $R^n$ via $x(\sum_i r_ie_i) = \sum_i r_ix_i$. Then $g_\ast(X)=Y$.

For $R^\Delta$ I originally said the same argument applies, but as pointed out in the comments that’s not the case for $\Delta$ infinite... I don’t know the answer in this case

usr0192
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