Form the free algebra on a set which is in bijection with S A <mo fence="false" stretchy="

Misael Matthews

Misael Matthews

Answered question

2022-06-14

Form the free algebra on a set which is in bijection with S
A i s : s S
and impose the relation that i s is a two-sided inverse of s S for each s S:
S 1 A := A i s : s S s i s 1 , i s s 1 : s S .
Then define ϕ A S 1 A by letting ϕ ( a ) be the image of a in S 1 A. By definition ϕ ( S ) consists of the units of S 1 A.
What I am confused about is what is precisely meant here by
Form the free algebra on a set which is in bijection with A i s : s S .
What is the definition of free algebra over a non-commutative ring A?

Answer & Explanation

nuvolor8

nuvolor8

Beginner2022-06-15Added 32 answers

There's no standard definition of a free A-algebra when A is non-commutative, but my guess is that it can only really mean one thing in this context: the smallest ring that contains A and all of the i s , with no extra relations imposed upon them (in particular, i s a a i s for s S , a A). Elements of A multiply with each other in the usual way, and that's all.
This ring is obviously horrendously large. Even after you quotient out by the relation ‘‘ i s = s 1 "", elements of the resulting ring S 1 A might still look like a 1 ( s 1 s 2 s 3 ) 1 a 2 ( s 4 s 5 ) 1 a 3 a m s n 1 , with no further simplification possible. This is a horrible outcome if what we were hoping for (like the commutative case) was a ring consisting of simple fractions like ‘‘ a / s "".
Luckily, in many 'nice' cases you can in fact simplify everything to a simple left fraction s 1 a, or a simple right fraction a s 1 , or either.
Gaaljh

Gaaljh

Beginner2022-06-16Added 7 answers

good explanation and site

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