Q:

Two-sided ideal $I$ in exterior algebra $T(V)/I$.

I have a confusion regarding two definitions of the two-sided ideal in exterior algebra.
Def 1)
In one definition, the exterior algebra $\Lambda(V)$ is defined as $T(V)/I$, where $I$ is the two-sided ideal generated by the graded commutators $$[a,b]=ab-(-1)^{|a||b|}ba$$ for $a\in T(V)_{|a|}$ and $b\in T(V)_{|b|}$.
Def 2) In another definition, $J_k$ is defined as the vector subspace of $V^{\otimes k}$ spanned by the $k$-fold tensors $$\dots\otimes \alpha\otimes\dots\otimes\alpha\otimes\dots$$ where $\alpha$ appears in the $i$th and $j$th position, $i<j$, $\alpha\in V$. Then $J$ is defined as $\bigoplus_{k=0}^\infty J_k$. And again, $\Lambda(V)=T(V)/J$.
My issue is I can't see how the $I$ and $J$ (in the two definitions respectively) are related. The $I$ and $J$  ought to be the same (aren't they?) but that is not so clear to me.
Thanks for any help.

A:

Since for all $a,b\in V$ we have $$(a+b)\otimes(a+b)=a\otimes a+a\otimes b+b\otimes a+b\otimes b\in J$$ and since also  $(a+b)\otimes(a+b),a\otimes a,b\otimes b\in J$, we deduce that all $a\otimes b+b\otimes a\in J$, so that $I\subset J$ and we have a surjective algebra  map $f:T(V)/I\to T(V)/J=\Lambda V  $.    
Over a field of characteristic $\neq 2$  we actually have $J=I$: indeed  for all $a\in V$ we have  $a\otimes a+a\otimes a=2(a\otimes a)\in I$  and thus also $a\otimes a\in I$ for all $a\in V$, so that $J\subset I$ and since we knew that  $I\subset J$ we have equality.
Thus our map  $f$  is an isomorphism of algebras in characteristic $\neq 2$.   
In characteristic $2$ however the algebra $T(V)/I$ is none other  than the symmetric algebra $Sym(V)$ and $f:Sym (V)\to \Lambda V  $ cannot be an isomorphism because $ Sym(V)$ is infinite dimensional   whereas $\Lambda(V)$ is finite dimensional.

