Rings and Fields

• Feb 24th 2009, 10:05 AM
vincisonfire
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 ?
• Mar 22nd 2009, 05:44 PM
NonCommAlg
Quote:

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 $0 \neq x \in R.$ every element of $R$ is algebraic over $F$ because $\dim_F R < \infty.$ so there exists a natural number $n$ and $a_j \in F$ such that $x^n + a_1x^{n-1} + \cdots + a_n = 0.$ now since $R$ is an integral

domain and $x \neq 0,$ we may assume that $a_n \neq 0.$ thus $a_n^{-1}(-x^{n-1} - a_1x^{n-2} - \cdots - a_{n-1})x=1,$ which proves that $x$ is invertible. Q.E.D.