Theorem: If L is algebraic over M, and M is algebraic over K, then L is algebraic over K
Proof: Pick x in L. Then since L is algebraic over M, there exist m_0, m_1,...,m_n in M such that
m_n*x^n+m_(n-1)*x^(n-1)+
...+m_1*x+m_0=0.
Since this is a polynomial in K(m_0, m_1,...,m_n), immediately we have that x is algebraic over K(m_0, m_1,...,m_n). Thus
[K(m_0, m_1,...,m_n,x):K(m_0, m_1,...,m_n)]
[K(m_0, m_1,...,m_n):K]
Hence
[K(m_0, m_1,...,m_n,x):K]=
[K(m_0, m_1,...,m_n,x):K(m_0, m_1,...,m_n)]
*[K(m_0, m_1,...,m_n):K]
So K(m_0, m_1,...,m_n,x) is a finite extension of K, hence K(m_0, m_1,...,m_n,x) is algebraic over K. This implies x is algebraic over K. Since x was an arbitrary element of L, it follows that L is algebraic over K.
2007-07-10
03:57:38
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2 answers
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Anonymous
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Mathematics