# Abstract Nonsense

## Every Short Exact Sequence of Vector Spaces Splits

Point of Post: In this post we prove that every short exact sequence of vector spaces splits, and why there is a block to proving this in a more general situation.

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Motivation

From our last post it’s not at all clear whether or not a sequence splitting is a happy mistake or a common occurrence. In other words, we now know necessary and sufficient conditions for a short exact sequence of $R$-modules to split, but we don’t know whether or not we should expect it to be so. In fact, it’s not even clear (at least immediately) that short exact sequences don’t either always split or never split (although, if this were true we probably wouldn’t have defined “split”!). Indeed, there are some simple examples of sequences not splitting. For example, consider the  obvious sequence of maps

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$0\to\mathbb{Z}_4\xrightarrow{2}\mathbb{Z}_8\xrightarrow{\text{mod }2}\mathbb{Z}_2\to0$

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where the first map multiplies the representative by $2$ (i.e. $[x]_4\mapsto [2x]_8$) and the second map reduces modulo $2$ (i.e. $[x]_8\mapsto [x]_2$). It is easily verified that this is, in fact, a short exact sequence of $\mathbb{Z}$-modules (i.e. abelian groups) and yet it doesn’t split for the simple reason that $\mathbb{Z}_8\not\cong\mathbb{Z}_4\times\mathbb{Z}_2$. This clearly generalizes to create a class of short exact sequences that doesn’t split

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$0\to\mathbb{Z}_{p^{n-1}}\to\mathbb{Z}_{p^n}\to\mathbb{Z}_p\to0$

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Another example would be that the following sequence

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$0\to 2\mathbb{Z}\hookrightarrow\mathbb{Z}\xrightarrow{\text{mod }2}\mathbb{Z}_2\to0$

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is short exact, yet clearly doesn’t split since $2\mathbb{Z}\times\mathbb{Z}_2$ has torsion and $\mathbb{Z}$ doesn’t. In fact, it should be clear that we are hitting on a general method to construct short exact sequences (of at least $\mathbb{Z}$-modules) that don’t split. Namely, take a $\mathbb{Z}$-module (i.e. abelian group) $A$ and a subgroup $B$ of $A$ such that $A\not\cong A\times A/B$. Then,

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$0\to B\hookrightarrow A\xrightarrow{\pi}A/B\to 0$

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is a short exact sequence that doesn’t split.

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Ok, so fine, it’s clear that it is not shocking that a short exact doesn’t split–we’ve just constructed a huge class of examples where this happens. Well, what about the converse? Is it shocking that a short exact sequence does split? Well, in this post we construct a vast class of examples of short exact sequences that do, in fact, split. In particular, we shall show that every short exact sequence of $k$-modules where $k$ is some field (i.e. vector spaces over $k$ split). There are two things to mention at this point: why we should expect this is the case, and what property allows to actually carry out the proof. The reason why this should not surprise us is that the main obstruction (or at least in the above non-examples) can’t exist in the category of vector spaces. Namely, let’s take finite dimensional vector spaces for a second. Then, as we have discussed if $0\to U\to V\to W\to 0$ is an exact sequence of vector spaces then $V/\text{im }U\cong W$ which tells us that $\dim V-\dim U=\dim W$, or $\dim V=\dim U+\dim W$. But, since dimension is the only invariant of vector spaces this allows us to conclude that $V\cong U\oplus W$. Thus, this is strong evidence for us to believe that every short exact sequence of vector spaces split.

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So, now what is the useful fact that actually allows us to conclude the proof. It’s clear how one would want, in general, to prove that a sequence $0\to A\to B\to C\to 0$ splits. Namely, we’d merely like to define a map $B\to A$ by sending the elements of $\text{im }A$ backwards (in the only way possible, since $A\to B$ is a monomorphism) and send $B-A$ to, whatever. The problem is what “whatever” should mean. It’s not at all obvious how to do this, or even clear that it can be done in a way to make a morphism (in fact, we know that it can’t always). But, with vector spaces we have the ability to do this precisely because vector spaces have the property that they’re easy to define morphisms on–just define a map on a basis and extend by linearity. So, roughly we send $\text{im }A\to A$ in the way we want, and then we “form a basis” for $B-A$ and map this to $0\in A$.

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Thus, the operative thing about vector spaces is that they always admit a basis. It’s clear that if more general objects are to have bases, then the same argument should apply (this shall be the case, when we define free modules).

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The Result

Now that we have an idea of how the proof should go, it’s extremely easy to write it out:

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Theorem: Let $U,V,W$ be vector spaces over the field $k$ and $C:0\to U\xrightarrow{f}V\xrightarrow{g}W\to0$ a short exact sequence. Then, $C$ splits.

Proof: Since $\text{im }f$ is a subspace of $V$ we may form a basis $\mathscr{B}$ for $\text{im }f$ in $V$. We may then extend $\mathscr{B}$ to a basis $\mathscr{B}\sqcup\mathscr{C}$ for $V$. Define a map $f^{\leftarrow}:V\to U$ by mapping every element of $\mathscr{B}$ back to the unique element of $U$ which maps to it (we can do this because $f$ is injective) and map $\mathscr{C}$ to $0$, and then extending this basis map to a morphism by linearity. Clearly this is a morphism and $f^{\leftarrow}\circ f=\text{id}_U$. The conclusion follows from the splitting lemma. $\blacksquare$

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Note that as a consequence of this we immediately get the following, famous, “rank-nullity theorem”

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Theorem(Rank-Nullity): Let $V,W$ be vector spaces over some field $k$ and $T:V\to W$ a linear transformation. Then,  we know that$\dim_k V=\dim_k\text{im }T+\dim_k \ker T$.

Proof: Evidently $0\to \ker T\hookrightarrow V\xrightarrow{T}\text{im }T\to 0$ is a short exact sequence of vector spaces. By the above theorem we know that this chain splits, and this tells us that $V\cong \ker T\oplus \text{im }T$, or said differently $\dim_k V=\dim_K\text{im }T+\dim_k \ker T$. $\blacksquare$

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By considering the natural projection $\pi:V\to V/W$ we get a somewhat inefficient way of proving the dimension of a quotient space

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Corollary: Let $V$ be an $n$-dimensional $k$-vector space, and $W\leqslant V$ with $\dim W=m$. Then, $\dim V/W=n-m$.

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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.