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I want to show the following: "a field L containing $\mathbb{C}$ with $[L:\mathbb{C}]< \infty\ $ then $L=\mathbb{C}$"

What I know:

1.$\mathbb{C}$ is algebraically closed, so every polynomial in $\mathbb{C}[x]$ has at least one root in $\mathbb{C}$.

2.If $[L:\mathbb{C}]=\infty$, then there are infinity elements of L that are lin. indep on $\mathbb{C}$.

Can anyone help?

3 Answers 3

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If $[L:\mathbf C]<\infty]$, every element of $L$ is algebraic over $\mathbf C$. As $\mathbf C$ is algebraically closed, this means every element of $L$ lies in $\mathbf C$, hence $L\subset\mathbf C$. As the opposite inclusion is true by hypothesis, we conclude that $L=\mathbf C$.

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    There are a few steps in between "algebraically closed" and "every element of $L$ lies in $\mathbb C$." It's not part of the usual definition of algebraically closed. – 2017-02-20
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    If the minimal polynomial of an element of $L$ already has all its roots in $\mathbf C$, which is contained in $L$, it is obvious for me. – 2017-02-20
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Let $a\in L$. Since $L$ is finite dimensional over $\mathbb{C}$, $a$ is algebraic over $\mathbb{C}$. The minimal polynomial $f(X)$ over $\mathbb{C}$ is irreducible.

What are the irreducible polynomials in $\mathbb{C}[X]$?

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Let $x\in L-\mathbb C$. The set $\{1,x,x^2,\ldots\}$ of powers of $x$ is infinite and hence cannot be linearly independent over $\mathbb C$. This means that there are $a_0,a_1,\ldots,a_n\in \mathbb C$ such that $$a_0+a_1x+\cdots+a_nx^n=0$$ Do you see why this is a contradiction?

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    There is a root in $\mathbb{C}$ , but I do not see how this is a contradiction because the fondamental theorem of algebra says so. – 2017-02-19
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    @usere5225321 But then this polynomial has $n+1$ roots in $L$ (with multiplicity). – 2017-02-19
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    which is impossible, because we have a polynomial of degree n. Thanks @Matt Samuel – 2017-02-19
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    @usere5225321 No problem. – 2017-02-19