Proving Grassmann's Algebra Using Free Vector Spaces

[ilia1987]

I currently self study from the book "A Course in Modern Mathematical Physics" by Peter Szekeres, and I'm currently reading the chapter on tensors, which he defines using the concept of Free Vector Spaces.
He gives a re-definition of Grassmann's algebras introduced in the previous section by using the concept of free algebra. And I had the following problem while reading the text:
(This is a Citation:) "
Let \mathcal{F}(V) be the free associative algebra over a real vector space V, and let \mathcal{S} be the ideal generated by all elements of \mathcal{F}(V) of the form u \otimes T \otimes v + v \otimes T \otimes u where u,v\in V and T\in \mathcal{F}(V). The general element of \mathcal{S} is S \otimes u \otimes T \otimes v \otimes U + S \otimes v \otimes T \otimes u \otimes U where u,v\in V and S,T,U\in \mathcal{F}(V)
"
My question is: Why can every element be expressed in this way?
An Ideal is first of all a Vector Subspace, right? So the sum of any two such "general elements" is supposed to be a general element too.
In other words, how do I prove that :
S_1 \otimes u_1 \otimes T_1 \otimes v_1 \otimes U_1 + S_1 \otimes v_1 \otimes T_1 \otimes u_1 \otimes U_1 \ \ \ \ + \ \ \ \ <br /> S_2 \otimes u_2 \otimes T_2 \otimes v_2 \otimes U_2 + S_2 \otimes v_2 \otimes T_2 \otimes u_2 \otimes U_2 \ \ \ \ = \ \ \ \ <br /> S_3 \otimes u_3 \otimes T_1 \otimes v_3 \otimes U_3 + S_3 \otimes v_3 \otimes T_3 \otimes u_3 \otimes U_3
?

Thank you for your time.

 

[HallsofIvy]

I currently self study from the book "A Course in Modern Mathematical Physics" by Peter Szekeres, and I'm currently reading the chapter on tensors, which he defines using the concept of Free Vector Spaces.
He gives a re-definition of Grassmann's algebras introduced in the previous section by using the concept of free algebra. And I had the following problem while reading the text:
(This is a Citation:) "
Let \mathcal{F}(V) be the free associative algebra over a real vector space V, and let \mathcal{S} be the ideal generated by all elements of \mathcal{F}(V) of the form u \otimes T \otimes v + v \otimes T \otimes u where u,v\in V and T\in \mathcal{F}(V). The general element of \mathcal{S} is S \otimes u \otimes T \otimes v \otimes U + S \otimes v \otimes T \otimes u \otimes U where u,v\in V and S,T,U\in \mathcal{F}(V)
"
My question is: Why can every element be expressed in this way?

Because that is what "generated by" means!

An Ideal is first of all a Vector Subspace, right So the sum of any two such "general elements" is supposed to be a general element too.
In other words, how do I prove that :
S_1 \otimes u_1 \otimes T_1 \otimes v_1 \otimes U_1 + S_1 \otimes v_1 \otimes T_1 \otimes u_1 \otimes U_1 \ \ \ \ + \ \ \ \ <br /> S_2 \otimes u_2 \otimes T_2 \otimes v_2 \otimes U_2 + S_2 \otimes v_2 \otimes T_2 \otimes u_2 \otimes U_2 \ \ \ \ = \ \ \ \ <br /> S_3 \otimes u_3 \otimes T_1 \otimes v_3 \otimes U_3 + S_3 \otimes v_3 \otimes T_3 \otimes u_3 \otimes U_3
?

Thank you for your time.

 

[ilia1987]

I'm sorry, but if that's what generated means, the general element should be of the form:

\forall \ x\in \mathcal{S} , \ x=\displaystyle\sum_r (S_r \otimes v_r \otimes T_r<br /> \otimes u_r \otimes U_r \ + \ S_r \otimes u_r \otimes T_r<br /> \otimes v_r \otimes U_r)

where S_r,T_r,U_r \in \mathcal{F}(V) and u_r,v_r\in V.
The question is whether any sum of this form can be expressed as a single element of the same form.

 

[Goldbeetle]

Are you still reading that book? I am, and I'm just a few lines before the part you were discussing in the first message of this thread. Maybe we can exchange some comments.