fundamental theorem of coalgebras

Fundamental Theorem of Coalgebras. Let $(C,\mathrm{\Delta },\epsilon )$ be a coalgebra over a field $k$ and $x\in C$. Then there exists subcoalgebra $D\subseteq C$ such that $x\in D$ and .

Proof. Let

$$\mathrm{\Delta }(x)=\sum _i{b}_i\otimes c_i.$$

Consider the element

$${\mathrm{\Delta }}_2(x)=\sum _i\mathrm{\Delta }(b_i)\otimes c_i=\sum _{i,j}{a}_j\otimes b_{ij}\otimes c_i.$$

Note that we may assume that $(a_j)$ are linearly independent and so are $(c_i)$. Let $D$ be a subspace spanned by $(b_{ij})$. Of course . Furthermore $x\in D$, because

$$x=\sum _{i,j}\epsilon (a_j)\epsilon (c_i)b_{ij}.$$

We will show that $D$ is a subcoalgebra, i.e. $\mathrm{\Delta }(D)\subseteq D\otimes D$. Indeed, note that

$$\sum _{i,j}\mathrm{\Delta }(a_j)\otimes b_{ij}\otimes c_i=\sum _{i,j}{a}_j\otimes \mathrm{\Delta }(b_{ij})\otimes c_i$$

and since $c_i$ are linearly independent we obtain that

$$\sum _j\mathrm{\Delta }(a_j)\otimes b_{ij}=\sum _j{a}_j\otimes \mathrm{\Delta }(b_{ij})$$

for all $i$. Thus

$$\sum _j{a}_j\otimes \mathrm{\Delta }(b_{ij})\in C\otimes C\otimes D$$

and since $a_j$ are linearly independent, we obtain that $\mathrm{\Delta }(b_{ij})\in C\otimes D$ for all $i,j$. Analogously we show that $\mathrm{\Delta }(b_{ij})\in D\otimes C$, thus

$$\mathrm{\Delta }(b_{ij})\in C\otimes D\cap D\otimes C=D\otimes D,$$

(please, see this entry (http://planetmath.org/TensorProductOfSubspacesOfVectorSpaces) for last equality) which completes the proof. $\mathrm{\square }$

Remark. The category of finite dimensional coalgebras is dual to the category of finite dimensional algebras (via dual space functor), so one could think that generally they are similar. Unfortunetly Fundamental Theorem of Coalgebras is major diffrence between algebras and coalgebras. For example consider a field $k$ and its polynomial algebra $k[X]$. Then whenever $f\in k[X]$ is such that $\mathrm{deg}(f)>0$, then a subalgebra generated by $f$ is always infinite dimensional (if $\mathrm{deg}(f)=0$ then subalgebra generated by $f$ is $k$). This can never occur in coalgebras.