When does $V/\operatorname{ker}(\phi)\simeq\phi(V)$ imply $V\simeq\operatorname{ker}(\phi)\oplus\phi(V)$?

I wrote it down in an imprecise way on purpose. The notation above is the linear algebra one: we have $V,W$ vector spaces over a field $\mathbb K$ and a $\mathbb K$-linear map $\phi:V\to W$. Then we know that isomorphism theorem says that $$ V/\operatorname{ker}(\phi)\simeq\phi(V) $$ and we also know that $\operatorname{dim_{\mathbb K}}V= \operatorname{dim_{\mathbb K}}(\operatorname{ker}(\phi))+ \operatorname{dim_{\mathbb K}}(\phi(V))$.

But when do we have that from this it follows that $$ V\simeq\operatorname{ker}(\phi)\oplus\phi(V)\;\;? $$

This was to be more precise on the particular case of the vector space, but it would be nice to know how it goes in every situation in which the isomorphism theorem works (rings homomorphism, groups homomorphism and so on).

EDIT: This question came to my mind originally because I was studying the group of units in a number ring $R$ (i.e. $R=\mathbb A\cap K$, where $\mathbb Q\le K\le\mathbb C$, $[K:\mathbb Q]=n$ and $\mathbb A$ is the ring of algebraic integers of $\mathbb C$). There is a theorem which states that in this case the group of units $U$ is a direct product of a finite cyclic group formed by the roots of $1$, call it $V$ and a free abelian group of finite rank $W$ (the rank is $r+s-1$ where $r$ is the number of real embeddings of $K$ and $2s$ is the number of complex embedding of $K$; and clearly $n=r+2s$). In order to show this, we build a moltiplicative-to-additive group homomorphism $U\to\mathbb R^{r+s}$ called $\log$ (it's a special logarithm), whose kernel is exactly $V$ and whose image is $W\simeq\mathbb Z^{r+s-1}$. From this, my book (Daniel Marcus' Number Fields) let follow that $U\simeq V\times W$, and even though it sounds good, I didn't understand exactly why.

SECOND EDIT: if it could help I'll write better the map $\log$. Consider the $r$ real embeddings of $K$: $\sigma_i$, and the $2s$ complex ones $\tau_j, \bar{\tau_j}$. We can define an additive groups monomorphism $$ \varphi:K\to\mathbb R^n $$ defined by $$ \varphi(x)=(\sigma_1(x),\dots,\sigma_r(x),\Re(\tau_1(x)),\Im(\tau_1(x)),\dots,\Re(\tau_s(x)),\Im(\tau_s(x)))\;\;. $$ Then given an integral basis of $R$, say $\{\alpha_i\}_{i=1}^n$ we can consider the $\varphi(\alpha_i)$'s: they form a basis for $\mathbb R^n$, hence we can consider the $n$-dimensional lattice generated by them: $$ \Lambda_{R}:=\operatorname{Span}(\varphi(\alpha_1),\dots,\varphi(\alpha_n{})):=\langle\varphi(\alpha_1),\dots,\varphi(\alpha_n{})\rangle_{\mathbb Z}\;. $$ It's clear that $\varphi$ sends $R$ onto $\Lambda_R$ hence they are isomorphic. Then consider $$ U\subset R\setminus\{0\}\stackrel{\varphi}{\longrightarrow}\Lambda_R\setminus\{0\}\stackrel{\log}{\longrightarrow}\mathbb R^{r+s} $$ where $\log$ is defined as (given $x=(x_1,\dots,x_n)\in\Lambda_R\subset\mathbb R^n$) $$ \log(x):=(\log|x_1|,\dots,\log|x_r|,\log(x_{r+1}^2+x_{r+2}^2),\dots,\log(x_{n-1}^2+x_n^2)). $$

  • 3
    $\begingroup$ For vector spaces, because of the dimension formula, we always have isomorphism $V\cong \ker\phi\oplus \phi(V)$. On the other hand, if $V=W$, we might also ask when will $V=\ker\phi \oplus\phi(V)$. $\endgroup$ – Berci Aug 24 '14 at 22:16
  • $\begingroup$ Thanks! And for group homomorphism? How does it work? $\endgroup$ – Joe Aug 24 '14 at 22:17
  • $\begingroup$ If $\phi:V\to V$ is a linear map such that $\phi^2=\phi$ then $V=\ker\phi\oplus\phi(V)$ $\endgroup$ – user72870 Aug 24 '14 at 22:19
  • $\begingroup$ Similarly, if $\phi:G\to G$ is a group homomorphism such that $\phi^2=\phi$ and $\phi(G)$ is a normal subgroup of $G$, then $G\cong \ker\phi\times \phi(G)$. $\endgroup$ – user72870 Aug 24 '14 at 22:22
  • $\begingroup$ Look to the right of the screen. $\endgroup$ – Troy Woo Aug 24 '14 at 23:05

It's not true in general. (In fact, it's false in most of the common categories: groups, rings, modules, etc; it just happens to hold for finite-dimensional vector spaces because dimension completely classifies them.) To take two examples:

  • Groups: The symmetric group $S_3$ contains a normal subgroup $A_3 = \mathbb{Z}_3$, which is the kernel of the sign map $f:S_3 \to \mathbb{Z}_2$. But $S_3$ is not the direct product of $A_3$ and some other $N\subset S_3$; the group $N$ must then have order $2$ and thus be isomorphic to $\mathbb{Z_2}$, but clearly $S_3\not = \mathbb{Z}_3 \oplus\mathbb{Z}_2$.
  • Modules: The quotient map $f:\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ has kernel $n\mathbb{Z}$, but $n\mathbb{Z}$ is not a direct summand of $\mathbb{Z}$. For if $\mathbb{Z} = n\mathbb{Z} \oplus M$, then $M =\mathbb{Z}_n$, but $\mathbb{Z}$ is torsion-free.

Another way of asking this more generally: When, for any $A,B$, do there exist no nontrivial extensions $0\to A\to E\to B\to 0$?

From homological algebra (at least in the case of an abelian category, perhaps more generally), we can say that this is true when every object in our category is projective.

The relevant fact about vector spaces (modules over a field) is that they all have a basis. So they are all free, hence projective.

The same holds if we consider the category of $R$-modules, where $R$ is a (not necessarily commutative) semisimple artinian ring.


$V\simeq ker(\phi)\oplus \phi(V)$ is always true for a linear map $\phi:V\rightarrow W$, even if $V$ and $W$ are infinite dimensional.

Prop: Let $\phi:V\rightarrow W$ be a linear map. There exists a subspace $R$ of $V$ such that

  1. $R\oplus\ker(\phi)=V$
  2. $\phi:R\rightarrow\phi(V)$ is an isomorphism.

Proof: Choose a basis $\{w_i,\ i\in I\}$ for the subspace $\phi(V)$. Let $v_i\in V$ be such that $\phi(v_i)=w_i$.

Notice that $\{v_i,\ i\in I\}$ is a linear independent set. Let $R$ be the span of $\{v_i,\ i\in I\}$.

Consider $\phi:R\rightarrow\phi(V)$. It is obvious that $\phi$ is surjective.

Now, suppose that $r\in R\cap\ker(\phi)$. Since $r=\sum_{k=1}^m a_kv_{i_k}$ then $0=\phi(r)=\phi(\sum_{k=1}^m a_kv_{i_k})=\sum_{k=1}^m a_kw_{i_k}$. Thus $a_{i_1}=\ldots=a_{i_m}=0$ and $r=0$. Therefore $R\cap\ker(\phi)=\{0\}$ and $\phi:R\rightarrow\phi(V)$ is injective. Thus $\phi:R\rightarrow\phi(V)$ is an isomorphism.

Now, let $v\in V$. Let $\phi(v)=\sum_{s=1}^n b_sw_{i_s}$. Consider $v-\sum_{s=1}^n b_sv_{i_s}$. Notice that $\phi(v-\sum_{s=1}^n b_sv_{i_s})=0$. Notice that $\sum_{s=1}^n b_sv_{i_s}\in R$. So every $v\in V$ is a sum of a vector in $R$ and a vector in $\ker(\phi)$.

  • $\begingroup$ This doesn't answer the question. He knows that it works for vector spaces but wants to know in what more general setting does the decomposition occur. $\endgroup$ – Dan Rust Aug 25 '14 at 1:21
  • $\begingroup$ @DanielRust The vectors spaces don't need to be finite dimensional. He is using the dimension to prove it. $\endgroup$ – Daniel Aug 25 '14 at 1:24
  • $\begingroup$ That's still not answering his question. If you've spotted a mistake in his question, edit it or make a comment. $\endgroup$ – Dan Rust Aug 25 '14 at 1:33
  • $\begingroup$ @DanielRust In the second part of his question, he aks if this theorem can be generalized for rings, fiels, etc. He asked first for vector spaces. $\endgroup$ – Daniel Aug 25 '14 at 1:37
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    $\begingroup$ @Joe You have $log:U\rightarrow W$. $W$ is a free abelian group. Thus $W$ is a projective module over $\mathbb{Z}$. See the lifting property in en.wikipedia.org/wiki/Projective_module. Thus exists $h:W\rightarrow U$ such that $log\circ h=Id$. Now see Daniel Rust's comment. $\endgroup$ – Daniel Aug 25 '14 at 2:35

The reason the map $\log$ has a (noncanonical) splitting is the following:

An algebraic number all of whose Galois conjugates have absolute value $1$ is a root of unity.

It follows immediately from this that the kernel of $\log: R^\times \to \mathbf R^{r+s-1}$ consists precisely of $\text{Tors}(R^\times)$. Moreover, $\log$ is an isomorphism of $R^\times/\text{Tors}(R^\times)$ with a lattice of $\mathbf R^{r+s-1}$. Therefore $R^\times/\text{Tors}(R^\times)$ is finitely generated, and since $\text{Tors}(R^\times)$ is finite, so is $R^\times$; it follows from the structure theorem for finitely generated groups that $R^\times \cong \text{Tors}(R^\times) \oplus R^\times/\text{Tors}(R^\times)$ (noncanonically).


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