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I was reading about Algebra over a field, and I see the definition of $\mathbb{K}\langle X \rangle$ as follows:

Let $X \neq \emptyset$ a set, a word $w$ is an expression in the following way:

$w=x_{1}^{m_{1}}x_{2}^{m_{2}} \cdots x_{k}^{m_{k}}, \space \forall x_{i} \in X$ and $\forall m_{i} \in \mathbb{N}$, $\forall i \in \{1,2, \cdots, k\}$ where $k \in \mathbb{N^{*}}$.

Then the algebra $\mathbb{K}\langle X \rangle$ is based on each word $w$ defined above. The bilinear product is the concatenation of the corresponding words.

Does anyone know a pdf that has a more formal setting (including showing why $\mathbb{K}\langle X \rangle$ is a vector space, defining its operations, etc ...) or could answer me with a more formal definition?

Thanks!


If in addition also show how $\mathbb{K}\langle X \rangle$ generalizes the polynomials ring $\mathbb{K}\left[ X \right]$ would be great !!


Jim, I didn't know the definition of free vector space , so I searched on W. Greub, Linear Algebra, Springer-Verlag, Fourth edition, 1975. Let's see if I understand:

$\mathbb{K}\langle X \rangle := \{f: \langle X \rangle \longrightarrow \mathbb{K}/ \space f^{-1}(\mathbb{K^{*}}) \mbox{ is finite}\}$, where $\langle X \rangle$ is the set of words with letters in $X$. This is a vector space if we consider the usual sum of functions and the usual scalar multiplication.

So one basis for $\mathbb{K}\langle X \rangle$ is:

$B=\{f_{w}/ \space w \in \langle X \rangle\}$, where $f_{w}:\langle X \rangle \longrightarrow \mathbb{K}$ is a function, such that:

$f_{w}(x)=\begin{cases} 1_{\mathbb{K}} & \mbox{if } x=w \\ 0_{\mathbb{K}} & \mbox{if } x\neq w\\ \end{cases}$.

This is basis because:

  1. $B$ generate $\mathbb{K}\langle X \rangle$:

    $\forall f \in \mathbb{K}\langle X \rangle$ we have $f^{-1}(\mathbb{K^{*}})=\{ w_{1}, w_{2}, \cdots, w_{n} \}$, for $n \in \mathbb{N}$ and $w_{i}\in\langle X \rangle$, $1 \leq i \leq n$. In this way:

    $f=\sum_{i=1}^{n}f(w_{i})f_{w_{i}}\in [B] \Longrightarrow \mathbb{K}\langle X \rangle=[B]$.

  2. $B$ is L.I.:

    Let $\{\lambda_{w}\}_{w \in B}$ be a family of scalars in $\mathbb{K}$ such that:

    $g=\sum_{w \in B}\lambda_{w}f_{w}=0_{\mathbb{K}\langle X \rangle}$.

    So $\forall w \in B, \space g(w)=0_{\mathbb{K}}\Longrightarrow \forall w \in B, \space \lambda_{w}=0_{\mathbb{K}}$.

I agree with you, to define a bilinear map (multiplication) one only needs to define what happens to basis vectors. In this case, how can I match the base $B$ with the words in $\langle X \rangle$? (I'm not able to see as a function $f_{w} \in B \subset \mathbb{K}\langle X \rangle$ can be a word in $\langle X \rangle$).

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The reason $\mathbb K\langle X\rangle$ is a vector space is because it's defined that way! $\mathbb K\langle X\rangle$ is the free vector space with basis the words with letters in $X$. Then to define a bilinear map $\mathbb K\langle X\rangle \otimes \mathbb K\langle X\rangle \to \mathbb K\langle X\rangle$ (multiplication) one only needs to define what happens to basis vectors. Well, if $\mathbb K\langle X\rangle$ has basis corresponding to words in $X$, then $\mathbb K\langle X\rangle \otimes \mathbb K\langle X\rangle$ has basis corresponding to pairs of words in $X$. So a map $\mathbb K\langle X\rangle \otimes \mathbb K\langle X\rangle \to \mathbb K\langle X\rangle$ can be defined by saying how to create a word in $X$ from a pair of words in $X$. And for multiplication this is done using concatenation.

One reason $\mathbb K\langle X\rangle$ generalizes $\mathbb K[X]$ is simply because one is a quotient of the other. You just mod $\mathbb K\langle X\rangle$ by the ideal generated by all commutators of letters. Another reason is that $\mathbb K[X]$ satisfies a universal property with respect to commutative $\mathbb K$-algebras, whereas $\mathbb K\langle X\rangle$ satisfies essentially the same universal property with respect to all (not necessarily commutative) $\mathbb K$-algebras.

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  • $\begingroup$ Jim see the edit I've made please. $\endgroup$
    – kurtzdoni
    Commented Dec 20, 2014 at 23:52
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    $\begingroup$ One way to define a free vector space is as you've done it, another is to say it's the set of formal linear combinations of the elements in $\langle X\rangle$. If you do it that last way then your basis elements are actually elements of $\langle X\rangle$, if you do your way then your basis elements aren't actually elements of $\langle X\rangle$, they just correspond with elements of $\langle X\rangle$. Either way is fine. $\endgroup$
    – Jim
    Commented Dec 21, 2014 at 0:36

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