Let $K\rightarrow M\rightarrow N\rightarrow 0$ be a short exact sequence of modules over a commutative ring $R$, then prove that $T(N)\otimes K\otimes T(N)\rightarrow T(N)\rightarrow T(M)\rightarrow 0$ is a short exact sequence, where $T(-)$ is the tensor algebra of the given module.

I'm reading through Eisenbud's Commutative Algebra with a View Twoards Algebraic Geometry, and in in one of the appendices, it seems to assume this sequence is exact to prove that another sequence is exact. I understand how the sequence is exact at $T(M)$ since $\otimes$ is an exact functor, but I'm not understanding how to define the maps induced but $T(-)$. If I understood the maps, I could possibly deduce the exactness at $T(N)$ on my own.

  • $\begingroup$ Are you sure the order of the modules in both sequences is correct? I thought about what the natural maps in the second sequence would be, but it seems that if we had $K=M=R=\mathbb{Z}$, $N=0$, the second sequence would be $\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow T(\mathbb{Z})$, which I think would be weird? My modules are a bit rusty so I expect I might be wrong though (: $\endgroup$
    – mcwiggler
    Jun 3, 2018 at 0:19

1 Answer 1


(The given exact sequence is not short, at least for me, it can maybe be called a "left incomplete short exact sequence.) The given exact sequence is not the one in loc. cit, page 571, Proposition A2.2, (d). (There is no map $TN\to TM$ in the given setting.) So i have to restate:

Given: $K\to N\to M\to 0$ is an exact sequence of $R$-modules. We show that $$ TN\otimes K\otimes TN\to TN\to TM\to 0$$ is exact.

For this, let us start with two exact sequences in the category of $R$-modules, $$ \begin{aligned} &K_1\to N_1\to M_1\to 0\ ,\\ &K_2\to N_2\to M_2\to 0\ , \end{aligned} $$ and show $$ \color{blue}{N_1\otimes K_2\ \oplus\ K_1\otimes N_2} \qquad\to\qquad \color{red}{N_1\otimes N_2} \qquad\to\qquad \color{red}{M_1\otimes M_2} \qquad\to\qquad 0 $$ is exact.

Consider for the last the diagram with exact rows and columns ($\otimes$ being right exact): $\require{AMScd}$ \begin{CD} 0 @<<< M_1\otimes K_2@<<< \color{blue}{N_1\otimes K_2}@<<< K_1\otimes K_2\\ @. @VVV @VVV @VVV \\ 0 @<<< M_1\otimes N_2@<<< \color{red}{N_1\otimes N_2}@<<< \color{blue}{K_1\otimes N_2}\\ @. @VVV (*) @VVbV @VVV \\ 0 @<<< \color{red}{M_1\otimes M_2}@<<a< N_1\otimes M_2@<<< K_1\otimes M_2\\ @. @VVV @VVV @VVV \\ @. 0 @. 0 @. 0 \end{CD} We are searching for the kernel of the surjective diagonal map $M_1\otimes M_2\leftarrow N_1\otimes N_2$, $ab$, between the red entries. We will show it is the sum of the blue entries. This follows from the exactness in $$ 0 \leftarrow \operatorname{Coker} a \leftarrow \operatorname{Coker} ab \leftarrow \operatorname{Coker} b $$ or from the corresponding diagram chase, here explicitly as follows.

Start with $\color{red}{\sum n_1\otimes n_2}$ (in slightly abusive notation) mapped to zero diagonally via $ab$.

We want to show that this sum is of the shape $\color{red}{\sum n_1\otimes n_2}\sim \color{blue}{\sum n'_1\otimes k_2} \color{red}{+} \color{blue}{\sum k_1\otimes n'_2} $ where we use "lower case notations" for elements in capital case objects. The $\sim$ means that the blue entries have to be mapped first in the red world, where the plus finally makes sense.

Then $b$ maps this starting element in an element, denoted in the same spirit $\sum n_1\otimes m_2$. It is mapped to zero horizontally via $a$, so it comes from an element $\sum k_1\otimes m'_2$. We lift this in the blue entry above, get an element $\sum k_1\otimes n'_2$. And map this element back in the middle, getting an element $\sum n'_1\otimes n'_2$, say. Now consider the difference $\sum n_1\otimes n_2-\sum n'_1\otimes n'_2$. It is vertically mapped via $b$ to zero. So it comes from the second blue entry above, say from $\sum n''_1\otimes k_2$. Then $$ \begin{aligned} &\color{blue}{\sum n''_1\otimes k_2} \oplus \color{blue}{\sum k_1\otimes n'_2} \\ &\qquad\qquad\qquad \to \left( \sum n_1\otimes n_2-\sum n'_1\otimes n'_2 \right) + \left( \sum n'_1\otimes n'_2 \right) \\ &\qquad\qquad\qquad =\sum n_1\otimes n_2\ . \end{aligned} $$

This show the special form for the element "in the kernel", as an element "in the image", i.e. the needed exactity.

The above generalizes inductively. Given: $$ \begin{aligned} &K_1\to N_1\to M_1\to 0\ ,\\ &K_2\to N_2\to M_2\to 0\ ,\\ &\vdots\qquad\vdots\qquad\vdots\qquad\vdots\qquad\vdots\qquad\\ &K_r\to N_r\to M_r\to 0\ ,\\ \end{aligned} $$ we have the exact sequence: $$ \color{blue}{\bigoplus \dots N_{j-1}\otimes K_j\otimes N_{j+1}\dots} \to \color{red}{N_1\otimes N_2\otimes\dots\otimes N_r} \to \color{red}{M_1\otimes M_2\otimes\dots\otimes M_r} \to 0\ . $$ We need the above for equal values of the $K$'s, of the $N$'s, and of the $M$'s, explicitly $$ \color{blue}{\bigoplus_{p+q+1=r} T^p N\otimes K\otimes T^qN} \to \color{red}{T^r N} \to \color{red}{T^r M} \to 0\ . $$

  • $\begingroup$ Okay, I'm following along with the diagram chase all the way up to the end but I don't understand why the kernel must equal the direct sum of the two blue modules because of the chase we did. Could you elaborate on this a little bit? $\endgroup$ Jun 5, 2018 at 21:14
  • 1
    $\begingroup$ I inserted some more lines, hope it makes the things clear now. $\endgroup$
    – dan_fulea
    Jun 6, 2018 at 0:49
  • $\begingroup$ That make sense now! Thank you for the clarification! $\endgroup$ Jun 6, 2018 at 0:54
  • $\begingroup$ (Small typo corrected now...) Thanks also for the comment, it is important for me to always have clear arguments, this makes the own work transparent! $\endgroup$
    – dan_fulea
    Jun 6, 2018 at 1:00

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