1. ## 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 ?

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

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