# Defining an ideal in the tensor algebra

In the wikipedia article about exterior algebra:

The exterior algebra $Λ(V)$ over a vector space $V$ over a field $K$ is defined as the Quotient algebra of the tensor algebra by the two-sided Ideal $I$ generated by all elements of the form $x \otimes x$ such that $x \in V$.

I want to show that $I$ is an ideal of $T(V)$ but i was confused about the elements of $I$, i mean what is exactly $I$ as a set and more particularly as a subset of $T(V)$.

• You don't have to show $I$ is an ideal: by definition it is the ideal generated by elements of the form $x\otimes x$, for $x\in V$. Glad egreg's explanation clears up the rest: it's a pretty good explanation! – rschwieb Oct 15 '13 at 12:34

If $S$ is a subset of a (non commutative) ring $R$, the ideal generated by $S$ consists of all elements of the form $$\sum_{i=1}^{n} a_i s_i b_i$$ where $n$ is an arbitrary integer, $a_i,b_i\in R$ and $s_i\in S$.
This set is obviously closed under addition (by construction) and contains $0$; it's also closed by left and right multiplication by elements of $R$, so it's an ideal. Any ideal containing $S$ as a subset must contain these elements, so this set is indeed the ideal generated by $S$.
In the particular case, you can think to $I$ as the set of elements in $T(V)$ of the form $$\sum_{i=1}^n A_i(x_i\otimes x_i)B_i$$ where $x_i\in V$ and $A_i,B_i\in T(V)$. I'm afraid there's not much more that can be said, even in the finite dimensional case. What's important is that the relation $$x\wedge x=0$$ is satisfied in $E(V)=T(V)/I$ (where $\wedge$ denotes the induced operation on the quotient ring) for all $x\in V$ because $x\otimes x\in I$ by definition.
This is just one particular construction of the exterior algebra: others are possible, but they will always give an isomorphic ring, because the exterior algebra satisfies a universal property. Therefore it's not really useful knowing what the elements of $I$ look like.
• @BenjaminThoburn In the case of the exterior algebra over a finite dimensional vector space it's much more important to know what a basis looks like, in terms of a basis of the vector space: the products $e_{i_1}\wedge e_{i_2}\wedge\dots\wedge e_{i_k}$, with $i_1<i_2\dots<i_k$, for instance. This is not proved through the definition $\Lambda(V)=T(V)/I$. This definition can be handy in other cases, though. – egreg Jul 31 at 19:49