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Let $K$ be an algebraically closed field and let $K\langle t_{1},t_{2},\ldots,t_{n}\rangle$ denote the free associative algebra in $n$ noncommuting indeterminates. Why is this algebra not a local algebra? (i.e the only idempotents are $0$ and $1$, or equivalently the set of non-units is a two sided ideal or equivalently it has a unique left (resp right) maximal ideal)

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up vote 3 down vote accepted

The only units in your algebra are the non-zero scalar polynomials, as you can easily check. In particular $t_1+1$ and $t_1$ are non-units. But their difference is a unit: the set of non-units is therefore not an ideal (it is not even an additive subgroup, in fact)

Alternatively: if $\lambda=(\lambda_1,\dots,\lambda_n)$ is an $n$-tuple of scalars, let $I_\lambda$ be the left ideal generated by $t_1-\lambda_1$, $\dots$, $t_n-\lambda_n$. This is a maximal ideal, and when $\lambda\neq\lambda'$ we have $I_\lambda\neq I_{\lambda'}$. There is then no unique maximal left ideal.

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Nevermind, I was about to type that but wasn't sure. Thank you, you can either delete your answer if you find too trivial this post or I can accept it, please let me know. – user6495 Oct 12 '11 at 6:03
Lots of things become trivial only once you realise they are true :) – Mariano Suárez-Alvarez Oct 12 '11 at 6:11

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