If I is an ideal of the ring R, show how to make the quotient ring...

If I is an ideal of the ring R, show how to make the quotient ring R/I into a left R-module, and also show how to make R/I into a right R-module.

Solutions

Expert Solution

R/I is the quotient ring of R by the ideal I. More explicitly, R/I = { I+r : r is in R } with the operations +,x defined by, (I+r) + (I+s) = I+(r+s) and (I+r) x (I+s) = I+(rs) for all r,s in R.

R/I being a ring, is an abelian group under +.

Define the map . : R x R/I R/I given by, r.(I+a) = I+(ra) for all a in R and for all r in R.

Observe that, if I+a = I+b for some a,b in R, then a-b is in I. Thus, for any r in R, r.(a-b) is in I as I is an ideal in R. Thus, r.a-r.b is in I. Hence, I+(ra) = I+(rb). Hence, r.(I+a)=r.(I+b).

Hence, . is well-defined. Further, r.(I+a + I+b) = r.(I+(a+b)) = I+r.(a+b) = I+ra + I+rb = r.(I+a) + r.(I+b)

(r+s).(I+a) = I+(r+s)a = I+ra + I+sa = r.(I+a) + s.(I+a)

r.(s.(I+a)) = r.(I+sa) = I+rsa = (rs).(I+a)

1.(I+a) = I+1a = I+a for all a,b,r in R and r,s in R.

This verifies all the module axioms. This shows that R/I can be made into a left R-module.

Similarly, R/I can also be made into a right R-module using the scalar multiplication . : R/I x R R/I given by, (I+a).r = I+(ar) for all a in R, r in R.

Colby Messinger answered 8 months ago

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