There is the definition of exterior algebra over $R$-module $M$ as quotient algebra of tensor algebra: $$\Lambda(M):= \frac{T(M)}{I}$$ Where $I$ is the ideal generated by $X=\{x_i\otimes x_i |x_i\in M\}$.
First: By definition, ideal $I$ of algebra $A$ must be a subalgebra of the algebra $A$. And when we say that ideal $I$ is generated by $X$, then also $X \subseteq A $.
Since tensor algebra $T(M)$ is a direct sum: $T(M)=\bigoplus\limits_{k=0}^{\infty}T^k_0M$, elements of $T(M)$ must be a sequences (by definition of direct sum): $T(M)\ni v = (v_0,v_1,v_2, \dots)$
Where $v_0 \in R$, $v_1 \in M$,$v_2 \in M \otimes M$, $\dots$ and $v_i=0$ for all but for finite number.
Question: How can we say about ideal $I$ generated by $X$ (as defined above) if $X$ is not even a subset of $T(M)$ since it's elements are not sequences? I agree that $X \subset M\otimes M = T^2_0M$,
but $T^2_0M \not\subset T(M)$ since they have different cartesian products behind them.