This is my first time posting so I apologise for the stupid question. I am trying to understand the idea of exterior algebra as a quotient algebra. Let $T(V)$ be the tensor algebra of a vector space $V$. The exterior algebra is defined as

\begin{equation} \Lambda(V)=T(V)/I, \end{equation} where $I$ is the two-sided ideal generated by elements of the form $v\otimes v$ with $v\in V$. This means the ideal is the set of elements \begin{equation} I=\left\{\sum_i r_i\otimes v\otimes v\otimes s_i\bigg|v\in V,r_i,s_i\in T(V) \right\}. \end{equation} My understand of taking the quotient is that elements of the ideal should be identified as the additive identity of the algebra, $0_R$, under projection.

My question is as follows. Suppose we have the element \begin{equation} a\otimes b \otimes c\in V. \end{equation} Now I add on the following element in the ideal $a\otimes (-b+2a)\otimes (-b+2a)\in I$ -- this is in the ideal because it is of the form $a\otimes v\otimes v$. Adding an element from the ideal should mean I remain in the same equivalence class, so \begin{equation} a\otimes b \otimes c\sim a\otimes b\otimes c+a\otimes (-b+2a)\otimes(-b+2a)=2a\otimes2a\otimes (c-b+2a). \end{equation} The term on the right is in the form of an ideal. From this, I conclude that $a\otimes b\otimes c$ is also in the ideal! This is blatantly wrong. It appears as if I can always add elements of the ideal to an element in $T(V)$ that is not in the ideal to obtain one that is in the ideal.

Where did I go wrong?

Thank you in advance!

  • $\begingroup$ That's not how addition of tensors works. $x\otimes y +z\otimes w \neq (x+z)\otimes (y+w)$ in general. $\endgroup$ – Matthew Towers Nov 25 '18 at 23:29
  • 1
    $\begingroup$ Oh, of course. Thank you! Can't believe I missed that! $\endgroup$ – KinLong Nov 25 '18 at 23:56

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