# Exterior algebra as quotient algebra

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

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

My question is as follows. Suppose we have the element $$$$a\otimes b \otimes c\in V.$$$$ Now I add on the following element in the ideal $$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$$. Adding an element from the ideal should mean I remain in the same equivalence class, so $$$$a\otimes b \otimes c\sim a\otimes b\otimes c+a\otimes (-b+2a)\otimes(-b+2a)=2a\otimes2a\otimes (c-b+2a).$$$$ The term on the right is in the form of an ideal. From this, I conclude that $$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?

• That's not how addition of tensors works. $x\otimes y +z\otimes w \neq (x+z)\otimes (y+w)$ in general. – Matthew Towers Nov 25 '18 at 23:29