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let $X=\{X_0, X_1,\ldots \}$ be a finite or countable set, $K$ a field, and $K\langle X\rangle$ the free associative algebra generated by $X$.

It is known that all associative algebras, generated by a finite or countable number of elements, can be represented as the factor ring $K\langle X\rangle/I$, where $I$ is an ideal of $K\langle X\rangle$.

If we considere the matrix algebra $M_n(K)$, over the field $K$, what is the representation of this algebra in the form $K\langle X\rangle/I$?

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"The" is not an appropriate word here; presentations are very far from being unique. –  Qiaochu Yuan Aug 26 '12 at 18:34

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up vote 1 down vote accepted

You need to be careful: there are two fields floating around here, $K$ and $\mathbb{F}$. The statement should be that all associative $K$-algebras are factor rings of some free $K$-algebra $K\langle X \rangle$; it is not true as you stated it.

Example: $\mathbb{F}_p$ is not a factor algebra of any free $\mathbb{Q}$-algebra

If $K=\mathbb{F}$, you may take the generating set $X$ to be $\{E_{ij}: 1 \leq i,j \leq n\}$ and the ideal $I$ to be generated by the elements of this free algebra coming from the usual relations amongst elementary matrices, namely $E_{ij}E_{kl} - \delta_{jk}E_{il}$ together with $1 - \sum_i E_{ii}$.

Here $\delta_{jk} = 0$ if $j \neq k$, otherwise it is $1$.

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Thre is an error. I have edited the question. Thank you –  zacarias Aug 26 '12 at 18:32

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