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We know that every f.g. algebra is isomorphic to quotient of f.g. free algebra. How can I get the surjective homomorphism map from the finitely generated free algebra A=K to a finitely generated algebra B ? (e.g., take B=K+K (direct sum) with generators (1,0),(0,1) )

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  • $\begingroup$ the finitely generated free algebra A=K<x1,...,xn> $\endgroup$
    – Ramy
    Dec 7 '14 at 20:39
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Let $X=\{x_1,x_2,\ldots,x_n\}$ be a generating set for the finitely generated algebra $B$. The map is given by extending the function $X\to B$ sending $x_i\mapsto x_i$ to a homomorphism from the free algebra generated by $X$. A unique such homomorphism exists by definition of the free algebra, and it is surjective by virtue of the fact that $X$ is a generating set.

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  • $\begingroup$ @Ramy In order for there to be a surjective homomorphism, the free algebra has to have at least as large a generating set as the algebra you're trying to map into. In particular, there's no surjective homomorphism from the field itself onto any algebra over the field that has at least one generator. $\endgroup$ Dec 7 '14 at 21:30
  • $\begingroup$ ٍٍSorry I have Confusion. How can we apply on k<x> to K? Is x ↦ 1? if it is true , does the kernel be <x-1>? thanks for your answer. $\endgroup$
    – Ramy
    Dec 7 '14 at 21:46
  • $\begingroup$ @Ramy that would work. $\endgroup$ Dec 7 '14 at 21:51
  • $\begingroup$ @ Matt S Thanks $\endgroup$
    – Ramy
    Dec 7 '14 at 21:54

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