Today I Learned

June 22nd

Today I learned some stuff about short exact sequences. I am under the impression that a short exact sequences more or less acts like a size limitation on the middle term: the left term lower-bounds the size, and the right-term upper-bounds the size. Here is an example of this.

For concreteness, we let $f:M\to N$ and $g:N\to P$ be the maps from the short exact sequence. The intuitive idea here is that we can "pull back'' any element from $P$ to a coset of $N,$ and the exact element of the coset can be chosen from a particular push forward from $M.$

Let $M=(m_1,\ldots,m_a)$ and $P=(p_1,\ldots,p_c)$ generate $M$ and $P$ respectively. As promised, we pull back each generator $p_k$ to some $n_k\in N$ such that $g(n_k)=p_k$; this is legal because $g$ is surjective. We also $n'_k=f(m_k)$ to be our push forward of $M,$ and we claim that\[N\stackrel?=(n'_1,\ldots,n'_a,n_1,\ldots,n_c).\]This will finish because then $N$ is generated by the above $a+c$ elements. On one hand, $n'_\bullet,n_\bullet\in N$ implies that $(n'_1,\ldots,n'_a,n_1,\ldots,n_c)\subseteq N$ by taking $R$-linear combinations.

On the other hand, we fix any $n\in N,$ and we attempt to find its representation. We begin by finding which coset we are in with respect to $P.$ Well, $g(n)\in P,$ so we can write\[g(n)=\sum_{k=1}^cr_kp_k\]for some sequence $\{r_k\}_{k=1}^a\subseteq R$ because $P=(p_1,\ldots,p_k).$ Because $g(n_k)=p_k,$ the above rearranges into\[0=g(n)-\sum_{k=1}^cr_kg(n_k)=g\left(n-\sum_{k=1}^cr_kn_k\right)\]using the $R$-linearity of $g.$ The point is that we have found an element of $\ker g=\im f$ by exactness. So we can find some $m\in M$ for which\[f(m)=n-\sum_{k=1}^cr_kn_k.\]Now, $m\in M=(m_1,\ldots,m_a),$ so we can write $m=\sum_{k=1}^ar'_km_k$ for some sequence $\{r'_k\}_{k=1}^a\subseteq R.$ Pushing this through $f,$ we see\[n-\sum_{k=1}^cr_kn_k=f\left(\sum_{k=1}^ar'_km_k\right)=\sum_{k=1}^ar'_kf(m_k),\]where we have again used the $R$-linearity of $f.$ To finish, we recall $f(m_k)=n'_k,$ so the above rearranges into\[n=\sum_{k=1}^ar'_kn'_k+\sum_{k=1}^cr_kn_k\in(n'_1,\ldots,n'_c,n_1,\ldots,n_c).\]Thus, $N\subseteq(n'_1,\ldots,n'_a,n_1,\ldots,n_c),$ finishing the proof. $\blacksquare$

It is not exactly true that the left-hand element $M$ in the above short exact sequence is lower-bounding the size of $N$ insofar as everything is finitely generated, but we do see that the number of generators we needed for $N$ is lower-bounded by those needed for $M.$ This is in fact a common case.

The requirement that $\varphi$ is well-behaved with respect to $f$ and $g$ is important to this definition, albeit for reasons which aren't clear to me. I am told that for Noetherian rings this behavior is free, but for non-Noetherian rings counterexamples can be given where these do not behave. I am not thinking too hard about this.

It feels like the intuition from a short exact sequence is that, given $m\in M$ and $p\in P,$ we "should'' be able to determine an element in the middle: pull $p$ back up to $N$ to determine our coset of $\ker g,$ and then push forward from $m$ to determine the precise element in the coset of $\im f=\ker g.$ Split exact sequences give the best way for this to happen.

Anyways, it turns out that we can describe split exact sequences in a nicer way: the functions $f$ and $g$ have reasonable semi-inverses. Here is one of the directions.

As a quick remark, we note that $\gamma$ is not a full-power inverse of $g$ because $g$ is not injective. But $g$ is surjective, so we can hope that after pulling back along $\gamma,$ maybe $g$ will take us where we started.

The forward direction is fairly easy. Fix $\varphi:N\to M\oplus P$ the isomorphism promised by the split short exact sequence, and we construct $\gamma:P\to N$ by\[\gamma(p):=\varphi^{-1}((0,p)).\]This is an $R$-module homomorphism because \[\varphi^{-1}((0,a_1p_1+a_2p_2))=\varphi^{-1}((0,a_1p_1))+\varphi^{-1}((0,a_2p_2))=a_1\varphi^{-1}((0,p_1))+a_2\varphi^{-1}((0,p_2))\]for any $p_1,p_2\in P$ and $a_1,a_2\in R.$ And we see\[(g\circ\gamma)(p)=g\left(\varphi^{-1}((0,p))\right)=\left(g\circ\varphi^{-1}\right)((0,p))=p\]by construction of $\varphi.$ This finishes this direction.

In the other direction, suppose we have $\gamma:P\to N$ such that $g\circ\gamma=\op{id}_P,$ and we need to construct our $\varphi^{-1}:M\oplus P\to N.$ We work backwards: pick up $n\in N.$ We would like to make the $N$-coordinate $g(n),$ but we need a way to afterwards retrieve an element of $M.$ The key observation is that, pulling back along $\gamma,$\[g\big(n-(\gamma\circ g)(n)\big)=g(n)-(g\circ\gamma)(g(n))=g(n)-g(n)=0.\]Thus, $n-(\gamma\circ g)(n)\in\ker g=\im f$ by exactness, allowing us to find $m\in M$ such that\[n=f(m)+\gamma(g(n)).\]This tells us that we want $\varphi^{-1}$ to take the pair $(m,p)\in M\oplus P$ to $f(m)+\gamma(p).$ The above argument gives surjectivity: we just constructed $m$ and $p:=g(n)$ such that $(m,p)$ maps to $f(m)+\gamma(p)=n.$

This $\varphi^{-1}$ is an $R$-module homomorphism because $(a_1m_1+a_1m_2,a_1p_1+a_2p_2)$ goes to\[f(a_1m_1+a_2m_2)+\gamma(a_1p_1+a_2p_2)=a_1(f(m_1)+\gamma(p_1))+a_2(f(m_2)+\gamma(p_2)).\]Lastly, we need to check the kernel of $\varphi^{-1}.$ Well, if $(m,p)$ has $f(m)+\gamma(p)=0,$ then applying $g$ gives\[(g\circ f)(m)+(g\circ\gamma)(p)=g(0)=0.\]The first summand vanishes because $\im f=\ker g,$ and the second term is just $\op{id}_P(p)=p.$ Thus, $p=0.$ As for $m,$ we see $\gamma(p)=0$ implies $f(m)=0,$ which requires $m=0$ by the injectivity of $f.$ It follows that $(m,p)=(0,0),$ making our kernel trivial.

Thus, we have constructed an $A$-module isomorphism $\varphi^{-1}:M\oplus P\to N.$ It remains to check that $(\varphi\circ f)(m)=(m,0)$ and $(g\circ\varphi^{-1})((m,p))=p.$ For the first condition, we have to check that $f(m)=\varphi^{-1}((m,0)),$ but\[\varphi^{-1}((m,0))=f(m)+\gamma(0)=f(m)\]is exactly what we need. For the second condition, we simply note\[(g\circ\varphi^{-1})((m,p))=g\big(f(m)+\gamma(p)\big)=(g\circ f)(m)+(g\circ\gamma)(p).\]Now, $g\circ f$ is the zero map by exactness, and $g\circ\gamma=\op{id}_P$ by definition of $\gamma.$ Thus, this is indeed $p.$ This finishes the proof. $\blacksquare$

Of course, the same thing happens if we try to invert $f.$

Again, we quickly remark that $\gamma$ is not a full-power inverse of $f$ because $f$ is not necessarily surjective. However, we can hope that $\gamma$ is an inverse on $\im f.$

Again, the forward direction is not that bad. Fix $\varphi:N\to M\oplus P$ the isomorphism promised by the split short exact sequence, and we construct $\gamma:N\to M$ by\[\gamma(n):=\op{pr}_1(\varphi(n)),\]the left coordinate of $\varphi(n).$ This is an $R$-module homomorphism because $\varphi$ and $\op{pr}_1$ are, so they're composition is as well. And we see\[(\gamma\circ f)(m)=\op{pr}_1\big((\varphi\circ f)(m)\big)=\op{pr}_1((m,0))=m\]by construction of $\varphi.$ This finishes this direction.

In the other direction, suppose we have $\gamma:N\to M$ such that $\gamma\circ f=\op{id}_M,$ and we need to construct our $\varphi:N\to M\oplus P.$ Looking around, we already have functions $\gamma:N\to M$ and $g:N\to P,$ so we hope that\[\varphi(n):=(\gamma(n),g(n))\]will do the trick. This is an $R$-module homomorphism because, for any $a_1,a_2\in R$ and $n_1,n_2\in N,$ we see $\varphi(a_1n_1+a_2n_2)$ is\[(\gamma(a_1n_1+a_2n_2),g(a_1n_1+a_2n_2))=(a_1\gamma(n_1)+a_2\gamma(n_2),a_1g(n_1)+a_2g(n_2)),\]which does indeed split into $a_1\varphi(n_1)+a_2\varphi(n_2).$

Next we show surjectivity and injectivity of $\varphi.$ For injectivity, we check the kernel: $\varphi(n)=(0,0)$ would require\[\gamma(n)=0\qquad\text{and}\qquad g(n)=0.\]The fact that $g(n)=0$ implies $n\in\ker g=\im f$ by exactness. Then $n\in\im f$ promises $m\in M$ such that $n=f(m),$ but now we see $\gamma(n)=(\gamma\circ f)(m)=m$ by definition of $\gamma.$ Thus, $\gamma(n)=0$ requires $m=0,$ and $n=f(m)=0$ as well. Thus, our kernel is trivial.

As for surjectivity, fix a pair $(m,p)\in M\oplus P$ so that we need to find $n\in N$ such that $\gamma(n)=m$ and $g(n)=p.$ Here we use the intuition we built a while ago: $g(n)=p$ is equivalent to $n\in n_0+\ker g$ for any $n_0\in N$ such that $g(n_0)=p.$ (Such an $n_0$ exists because $g$ is surjective.) Indeed,\[g(n-n_0)=g(n)-g(n_0)=p-p=0,\]so $n-n_0\in\ker g=\im f.$ Now that we have an element in $\im f,$ we can get a bearing on the behavior of $\gamma.$ Pick up $m_0$ such that $f(m_0)=n-n_0,$ and it follows\[m_0=(\gamma\circ f)(m_0)=\gamma(n)-\gamma(n_0).\]Because we want $\gamma(n)=m,$ we had better have $m_0=m-\gamma(n_0).$ So in total, we have\[n:=n_0+f(m-\gamma(n_0))\]for any $n_0$ such that $g(n_0)=p.$ We can quickly verify that applying $g$ to this gives $g(n)=g(n_0)=p$ because $f(m-\gamma(n_0))\in\im f=\ker g.$ Further, applying $\gamma$ to this gives $\gamma(n_0)+m-\gamma(n_0)=m$ because $\gamma\circ f=\id_M.$

Thus, we have verified that $\varphi:N\to M\oplus P$ is an $R$-module isomorphism. It remains to check the conditions on $\varphi.$ Well, for any $m\in M,$\[(\varphi\circ f)(m)=\big(\gamma(f(m)),g(f(m))\big)=(m,0),\]where $\gamma\circ f=\id_M$ by construction of $\gamma$ and $g\circ f=0$ by exactness. As for $g\circ\varphi^{-1},$ we remark that we have constructed our $\varphi^{-1}$ above in the proof of surjectivity. So fix some $(m,p)$ and pick up the $n$ for which $\varphi(n)=(m,p)$ so that $\varphi^{-1}((m,p))=n.$ Then\[g(n)=\op{pr}_2\big(\gamma(n),g(n)\big)=\op{pr}_2(\varphi(n))=p.\]This is the last condition on $\varphi$ we had to check, so this finishes the proof. $\blacksquare$

In essence, the above two propositions say that a short exact sequence where there is hope of reversing the maps turn out to have pretty strong ability to reversetheir maps because they're just $0\to M\to M\oplus P\to P\to 0.$ Perhaps this is disappointing as this short exact sequence is somewhat boring, but it is also quite surprising that we have this power-up.

Anyways, we close out our narrative that short exact sequences give size conditions on the middle term by proving the short five lemma.

We can do this as a direct application of the Snake lemma, and we will do so momentarily, but for now, it feels significantly more concrete to do this directly. I feel fuzzy about the maps in the Snake lemma anyways, so we can view this proof as "evidence'' that the snake lemma is really correct.

We merely have to prove that $\varphi_1$ has trivial kernel and is surjective. This comes down to a diagram chase. Suppose we have $m_1\in\ker\varphi_1.$ Then, using the commutativity of the diagram,\[\left(\varphi_2\circ\del_M^1\right)(m_1)=\left(\del_N^1\circ\varphi_1\right)(m_1)=0,\]so $\del_M^1(m_1)\in\ker\varphi_2.$ However, $\varphi_2$ has trivial kernel, so $\del_M^1(m_1)=0,$ so $m_1\in\ker\del_M^1=\im\del_M^0$ by exactness, which promises $m_0\in M_0$ such that $m_1=\del_M^0(m_0).$ Using the commutativity of the diagram again,\[\left(\del_N^0\circ\varphi_0\right)(m_0)=\left(\varphi_1\circ\del_M^0\right)(m_0)=\varphi_1(m_1)=0.\]However, $\del_N^0$ is injective, so we must have $\varphi_0(m_0)=0,$ and $\varphi_0$ is injective as well, so $m_0=0.$ It follows $m_1=\del_M^0(m_0)=0,$ so indeed, $\varphi_1$ has trivial kernel. Note that we have used the injectivity of $\del_N^0$ here but not the surjectivity of $\del_M^1.$

Now we show the surjectivity of $\varphi_1$ and hope to see the surjectivity of $\del_M^1$ appear. Fix any $n_1\in N_1$ which we need to be in the image of $\varphi_1.$ Well, note $\varphi_2$ and $\del_M^1$ are both surjective, so we can pull $\del_N^1(n_1)\in N_2$ back to $M_2$ through $\varphi_2$ and then back to $M_1$ through $\del_M^1.$ So we have $m_1\in M_1$ such that\[\left(\del_N^1\circ\varphi_1\right)(m_1)=\left(\varphi_2\circ\del_M^1\right)(m_1)=\del_N^1(n_1).\]This $m_1$ is not the $m_1$ we want, however. After all, $m_1$ is not even canonically chosen because the pull back along $\del_M^1$ only gives a canonical coset $m_1+\ker\del_M^1.$ Anyways, what we do know is that\[\del_N^1\big(n_1-\varphi_1(m_1)\big)=0,\]so $n_1-\varphi_1(m_1)\in\ker\del_N^1=\im\del_N^0$ by exactness. Thus, we can pull $n_1-\varphi_1(m_1)$ backwards along $\del_N^0$ to $N_0$ (we no longer care that this is injective, but it is), and then the surjectivity of $\varphi_0$ lets us pull back to $M_0.$ So we get $m_0\in M_0$ such that\[\left(\varphi_1\circ\del_M^0\right)(m_0)=\left(\del_N^0\circ\varphi_0\right)(m_0)=n_1-\varphi_1(m_1).\]The above rearranges to $n_1=\varphi_1\left(\del_N^0(m_0)+m_1\right)\in\im\varphi_1.$ This completes the proof of surjectivity and hence the proof. $\blacksquare$

As promised, we now prove the short five lemma again by the Snake lemma. The Snake lemma provides us a long exact sequence\[\ker\varphi_0\to\ker\varphi_1\to\ker\varphi_2\to\coker\varphi_0\to\coker\varphi_1\to\coker\varphi_2.\]We remark that $\varphi:M\to N$ is an isomorphism if and only if it has trivial kernel so that $\ker\varphi=0$ and full image so that $\coker\varphi=N/\varphi(M)=0.$ So the fact that $\varphi_0$ and $\varphi_2$ are isomorphisms lets us fill this in as\[0\to\ker\varphi_1\to0\to0\to\coker\varphi_1\to0.\]Now, we state the following "fill in the blank'' lemma.

The image of $0\to M$ must be equal to the kernel of $M\to 0$ for this to be exact. Well, $0\to M$ has image $0\in M$ and $M\to 0$ takes everything to $0$ and hence has kernel $M.$ It follows that $M=\{0\}.$ This is what we wanted. $\blacksquare$

In light of the above, $0\to\ker\varphi_1\to0$ forces $\ker\varphi_1=0,$ and $0\to\coker\varphi_1\to0$ forces $\coker\varphi_1=0.$ Thus, $\varphi_1$ is indeed an isomorphism, and we are done here. $\blacksquare$

We could be more careful about exactly what conditions are needed in the above proof. For example, injectivity of $\varphi_1$ only required $\varphi_0,\varphi_2,\del_N^0$ to be injective, and surjectivity of $\varphi_1$ only required $\varphi_0,\varphi_2,\del_M^1$ to be surjective. This doesn't really matter but should be remarked.

Anyways, the point of bringing up the short five lemma at all is that our exact sequences essentially forced an isomorphism in the middle term. The rows are not necessarily short exact sequences, but even still, we can feel the lower-bound of injectivity and the upper-bound of surjectivity along the left and right forcing the middle column to behave in a particular way.