# Abstract Nonsense

## Using Partial Exactness to Compute Things (Pt. II)

Point of Post: This is a continuation of this post.

This gives a wide array of corollaries, some of them being:

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Corollary: Let $R$ be a PID and let $(a)$ and $(b)$ be two ideals of $R$, then

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$(R/(a))\otimes_R(R/(b))\cong R/(\text{gcd}(a,b))$

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So, for example we have that $\mathbb{Z}/(25)\otimes_\mathbb{Z}\mathbb{Z}/(15)\cong\mathbb{Z}/(5)$ and $\mathbb{C}[x]/((x-i)(x-2i)^2)\otimes_{\mathbb{C}[x]}\mathbb{C}[x](x^6+1)\cong\mathbb{C}[x]/(x-i)\cong\mathbb{C}$.

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While the above may seem slightly laborious consider that if we’re smart we can often get away with computing a lot less. For example, suppose we wanted to compute $\mathbb{Z}_k\otimes_\mathbb{Z}\mathbb{Z}_\ell$ where $k,\ell\in\mathbb{N}$. If $k=p_1^{\alpha_1}\cdots p_m^{\alpha_m}$ and $\ell=p_1^{\beta_1}\cdots p_m^{\beta_m}$ then, using the commuting of tensor and direct products, we have that

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$\displaystyle \mathbb{Z}_k\otimes_\mathbb{Z}\mathbb{Z}_\ell\cong\prod_{i,j=1}^{m}\mathbb{Z}_{p_i^{\alpha_i}}\otimes_\mathbb{Z}\mathbb{Z}_{p_j^{\beta_j}}$

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So, we actually really only have to care about tensors of the form $\mathbb{Z}_{p^m}\otimes_\mathbb{Z}\mathbb{Z}_{q^n}$ where $p,q$ are primes. In fact, we only have to care about the case when $p=q$. Indeed, suppose that $p\ne q$, then since $(p,q)=1$ we know that $p^m+(q^n)=p^m(1+(q^n))$ is invertible and so for any $x+(q^n)$ we have that we can find $y+(q^n)$ such that $p^m(y+(q^n))=1+(q^n)$ thus we see that for

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\begin{aligned}(z+(p^m))\otimes(x+(q^n)) &=(z+(p^m))\otimes p^m(y+(q^n))\\ &=p^m(z+(p^m))\otimes(y+(q^n))\\ &=0\otimes(y+(q^n))\\ &=0\end{aligned}

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so that all simple tensors, and thus everything, in $\mathbb{Z}_{p^m}\otimes_\mathbb{Z}\mathbb{Z}_{q^n}$ is zero. So, to figure out the general case we only actually have to care about $\mathbb{Z}_{p^m}\otimes_\mathbb{Z}\mathbb{Z}_{p^n}$ with $m\geqslant n$, and now we can use partial exactness. Indeed, we have the following exact sequence

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$\mathbb{Z}\xrightarrow{p^m}\mathbb{Z}\xrightarrow{\pi}\mathbb{Z}_{p^m}\to0$

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where the first map is “multiply by $p^m$“. Using right exactness we may then conclude that

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$\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}_{p^n}\overset{p^m\otimes1}{\longrightarrow}\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}_{p^n}\overset{\pi\otimes1}{\longrightarrow}\mathbb{Z}_{p^m}\otimes_\mathbb{Z}\mathbb{Z}_{p^n}\to0$

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We know then that $\mathbb{Z}_{p^m}\otimes_\mathbb{Z}\mathbb{Z}_{p^n}$ is the cokernel of this first map, and so it suffices to find the image of this first map. But, note that for any simple tensor $z\otimes(x+(p^n))$ in the first module we have that the image under $p^m\otimes1$ is $p^mz\otimes(x+(p^n))=z\otimes p^m(x+(p^n))=z\otimes0=0$. Thus, we see that $p^m\otimes1$ has trivial image and so the cokernel is isomorphic to $\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}_{p^n}$ which is isomorphic to $\mathbb{Z}_{p^n}$. That was easy! Thus, we see that

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\displaystyle \begin{aligned}\mathbb{Z}_k\otimes_\mathbb{Z}\mathbb{Z}_\ell &= \prod_{i,j=1}^{m}\mathbb{Z}_{p_i^{\alpha_i}}\otimes_\mathbb{Z}\mathbb{Z}_{p_j^{\beta_j}}\\ &\cong\prod_{i=1}^{m}\mathbb{Z}_{p_i^{\alpha_i}}\otimes_\mathbb{Z}\mathbb{Z}_{p_i^{\beta_i}}\\ &\cong\prod_{i=1}^{m}\mathbb{Z}_{p_i^{\min\{\alpha_i,\beta_i\}}}\\ &\cong\mathbb{Z}_{p_1^{\min\{\alpha_1,\beta_1\}}\cdots p_m^{\min\{\alpha_m,\beta_m\}}}\\ &=\mathbb{Z}_{(k,\ell)}\end{aligned}

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Ok, ok, I can see that this might have seemed like a lot of work, but if you really analyze it you see that everything up until the computation of $\mathbb{Z}_{p^m}\otimes_\mathbb{Z}\mathbb{Z}_{p^n}$ was conceptually easy (you could have guessed the results) and when this computation finally came the actual computation was easy (the image of the pertinent map was trivial)–so overall, not that bad.

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Free Presentations and their Use

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This last computation segues nicely into the the general techinque I’d like to discuss. In particular, it gives a good example of a free presentation. What is a free presentation? Formally, a free presentation for a module $M$ is an exact sequence of left $R$-modules and $R$-maps of the form $F_1\to F_2\to M\to 0$ where $F_1$ and $F_2$ are free modules. Fine, but what does a free presentation really mean? Let’s look at a simple way to construct a free presentation and see if it’ll help our understanding. Suppose that we have that $M$ is generated by some subset $\mathcal{G}$. Note then that if $F[\mathcal{G}]$ denotes a free module on $\mathcal{G}$ the inclusion $\mathcal{G}\hookrightarrow M$ gives way to an $R$-map $F[\mathcal{G}]\to M$ which reduces to the identity map on $\mathcal{G}$ (e.g. $\mathcal{G}\to \mathcal{G}$ is the identity map). Let $K$ denote the kernel of the map $F[\mathcal{G}]\to M$ and note that if $\mathcal R$ denotes a set of generators for $K$ then we can (using the same idea) see that $F[\mathcal{R}]$ surjects onto $K$ by some map $F[\mathcal{R}]\to K$. Thus, we see that $F[\mathcal{R}]\to F[\mathcal{G}]\to M\to0$ is a free resolution. So, what do these $F[\mathcal{R}]$ and $F[\mathcal{G}]$ represent? Well, what $\mathcal{G}$ represents is clearly enough, it’s a set of generators, but what about $\mathcal{R}$? Intuitively an element $\displaystyle \sum_{g\in\mathcal{G}}r_gg$ of $F[\mathcal{G}]$ is in $K$ if and only if when we take the formal sum $\displaystyle \sum_{g\in\mathcal{G}}r_gg$ and pass it into an actual element of $M$ we get zero. In other words, an element of $K$ denotes a “relation” between the generators of $M$ and so $\mathcal{R}$ (a generating set for $K$) roughly denotes some economical (hopefully) and representation choice of these “relations”.

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I don’t think that could have been any more thinly-veiled. Yes, a free presentation is basically the same thing as giving the module in terms of generators and relations. So, how does giving a module in terms of generators and relations lend itself to the computation of tensor products? Well, the basic idea is that in general, giving a module in terms of a cokernel enables us to give that modules tensor with something else in terms of a cokernel, and in the case of the cokernel coming from a free presentation this cokernel is likely to be “nicer looking”. Ok, saying that again without using the word “cokernel” ten times, if we have a free presentation for $M$, say $F_1\xrightarrow{f} F_2\to M\to 0$ then we know that $M$ is equal to $\text{coker}(f)$. Tensoring with some module $N$ then gives us that $M\otimes_R N$ is isomorphic to $\text{coker}(f\otimes 1)$. But, note that $f\otimes1:F_1\otimes_R N\to F_2\otimes_R N$ and since $F_1=R^{\oplus\lambda}$ for some $\lambda$ and $F_2=R^{\oplus\kappa}$ for some $\kappa$ we know from previous discussion that $F_1\otimes_R N\cong N^{\oplus\lambda}$ and $F_2\cong N^{\oplus\kappa}$ with $(r_j)\otimes n)\mapsto (r_j n)$ in both cases. We see then that we have the following diagram

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$\begin{matrix}N^{\oplus\lambda} & \overset{g}{\longrightarrow} & N^{\oplus\kappa}\\ \big\downarrow & & \big\uparrow\\ F_1\otimes_R N & \overset{f\otimes1}{\longrightarrow} & F_2\otimes_R N\end{matrix}$

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where the vertical arrows are the obvious isomorphisms and $g$ is such that the diagram commutes. We see then that $M$ is isomorphic to $\text{coker}(g)$ which is a fairly tame beast. Namely, we see that $g$ basically acts by

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$\displaystyle (n_j)\mapsto \sum_j (\delta_j)\otimes n_j\mapsto \sum_j f((\delta_j))\otimes n_j\mapsto \sum_j f((\delta_j))n_j$

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where $\delta_j$ is the usual Kronecker delta function ($1$ at $j$ and zero elsewhere) and by an unusually bad abuse of notation by $f((\delta_j))n_j$ i mean “write $f((\delta_j))$ in it’s form like $(s_k)$ and then “multiply each term by $n_j$ to get $(s_kn_j)$“. This idea of computing things by generators and relations will be of great use to us in the future.

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.