- #1
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Given R a ring and M,N two R-modules, we may form their tensor product over Z or R. They can be defined as the group presentations
< A x B | (a + a',b)=(a,b) + (a',b), (a,b+b')=(a,b) + (a,b') >,
< A x B | (a + a',b)=(a,b) + (a',b), (a,b+b')=(a,b) + (a,b'), (ra,b)=(a,rb) >
respectively and the image of (a,b) in the quotient is written [itex]a\otimes b[/itex].
Tensoring over R will generally yield a smaller group since there is the additional relation [itex](ra)\otimes b = a\otimes (rb)[/itex]. But sometimes tensoring over R or Z give the same group.
I have read in a textbook that this happens in the case where R = Z_n or Q. For Z_n, it is quite clear, since in this case, the relation [itex](ra)\otimes b = a\otimes (rb)[/itex] is implied by the relations [itex](a + a') \otimes b = a\otimes b + a'\otimes b[/itex] and [itex]a \otimes (b+b') = a\otimes b + a\otimes b'[/itex].
But for R = Q, I do not quite see why the two tensor products are equal. Thx!
< A x B | (a + a',b)=(a,b) + (a',b), (a,b+b')=(a,b) + (a,b') >,
< A x B | (a + a',b)=(a,b) + (a',b), (a,b+b')=(a,b) + (a,b'), (ra,b)=(a,rb) >
respectively and the image of (a,b) in the quotient is written [itex]a\otimes b[/itex].
Tensoring over R will generally yield a smaller group since there is the additional relation [itex](ra)\otimes b = a\otimes (rb)[/itex]. But sometimes tensoring over R or Z give the same group.
I have read in a textbook that this happens in the case where R = Z_n or Q. For Z_n, it is quite clear, since in this case, the relation [itex](ra)\otimes b = a\otimes (rb)[/itex] is implied by the relations [itex](a + a') \otimes b = a\otimes b + a'\otimes b[/itex] and [itex]a \otimes (b+b') = a\otimes b + a\otimes b'[/itex].
But for R = Q, I do not quite see why the two tensor products are equal. Thx!