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Math Help - Rings and Fields

  1. #1
    Senior Member vincisonfire's Avatar
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    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 ?
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    Quote Originally Posted by vincisonfire View Post
    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.
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