Thread: Rings and Fields

If F is a field and R is a ring containing F as a subring, we can just as
easily regard R as a vector space over F as we can in the case that R itself
is field. Suppose that R is an integral domain containing the field F , and
as a vector space over F , it has finite dimension. Show that R itself must
be a field.
As I understand it, we only need to show that every elements of R have a multiplicative inverse. I don't see how to show that it is true independantly of the definition of multiplication.
Can you help me ?

Originally Posted by vincisonfire

If F is a field and R is a ring containing F as a subring, we can just as
easily regard R as a vector space over F as we can in the case that R itself
is field. Suppose that R is an integral domain containing the field F , and
as a vector space over F , it has finite dimension. Show that R itself must
be a field.
As I understand it, we only need to show that every elements of R have a multiplicative inverse. I don't see how to show that it is true independantly of the definition of multiplication.
Can you help me ?

i was looking at the questions posted during my short absence and i came across this one, which is, although simple, but very important! here's a proof:

let every element of is algebraic over because so there exists a natural number and such that now since is an integral

domain and we may assume that thus which proves that is invertible. Q.E.D.