Show that an element of a free algebra is not generated by some relations

Let $$A$$ be the free associative algebra over a field $$k$$ generated by countably many indeterminates $$x_1, x_2, \ldots$$.

I want to show that for any $$n$$, $$x_1 \ldots x_n$$ is not in the ideal $$I$$ generated by $$S=\{x_i^2, x_ix_j+x_jx_i : i,j \geq 0\}$$.

My attempt: Suppose for a contradiction that it is. Then it would be a sum of terms of the form $$xsy$$ where $$x,y \in A$$ and $$s \in S$$. I would like to argue that each such term $$xsy$$ must be $$0$$ since $$s$$ has the form $$x_i^2$$ (so it has degree $$2$$ in $$x_i$$, whereas $$x_1 \ldots x_n$$ does not) or $$x_ix_j+x_jx_i$$ (and this is 'bad' because it contributes two terms where the orders in which $$x_i$$ and $$x_j$$ occur are flipped). A problem with this argument (other than it being too informal), is that it might be that (the 'bad' part of) such a term $$xsy$$ cancels with (the 'bad' part of) another term $$xs'y$$, neither term being $$0$$. How do I correct and formalize my argument?

Footnote: I am only interested in the case where $$k$$ has characteristic not equal to $$2$$; in this case the ideal $$I$$ can be generated just by the relations $$\{ x_ix_j+x_jx_i : i,j \geq 0\}$$, since $$x_i^2 = \frac{1}{2} (x_ix_i+x_ix_i)$$. Feel free to assume this, but if there is a proof that works in full generality even in characteristic $$2$$ I would love to see it.

• Typically, making an argument like this explicitly in terms of elements of the ideal the way you're trying is very messy. Instead, it is better to use the universal property of the quotient and construct an explicit algebra $B$ with a homomorphism $A/I\to B$. You can then conclude certain elements of $A/I$ must be nonzero because they map to nonzero elements of $B$. – Eric Wofsey Mar 13 at 21:41
• In other words, it suffices to just construct some associative algebra $B$ with elements $x_1,x_2,\dots\in B$ such that $x_i^2=0$ and $x_ix_j+x_jx_i=0$, but such that $x_1\dots x_n\neq 0$. – Eric Wofsey Mar 13 at 21:45
• This is helpful. I'll probably want $B$ some matrix algebra. – SSF Mar 13 at 22:00