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So I am currently studying Algebraic Number theory and a theorem in the Book states the following:

Let $L/K$ be a field extension. Then $\alpha \in L$ is algebraic over $K$ if and only if there is a finite extension of $K$ inside $L$ containing $\alpha$.

I will give a sketch of the proof below and I am stuck on one particular bit which I need help with. So, we prove the "if" part by showing all finite extensions are algebraic.

For the "only if" part we consider the subspace ( we call it $M$) spanned by {$1,\alpha,\alpha^2,\alpha^3,...$} and show that this subspace is in fact finite dimensional over K and is also a subring of L. (This subspace is nothing but $K(\alpha)$, but that notation hasn't been introduced yet which is why we are doing all this) Now this is the part where it is confusing me, to show that this subspace is a field, all we have to show is the existence of multiplicative inverse of an element. So we pick an $x \in M$, non-zero and define a multiplication-by-x map given by $m_{x}: M \rightarrow M$ where $m_{x}(y) = x.y$ $\forall y \in M$. We then show that this map is injective and then by the rank nullity theorem deduce that it is surjective which implies that $x$ has an inverse. So the map being injective means that dimension of kernel of this linear operator is $0$ but how would that mean that the map is surjective? Apologies for the lengthy post, I suppose I could have just asked a shorter question but I felt it necessary to provide some context.

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  • $\begingroup$ Remember that you are working on the $K$-vectorial space $M$ which is finite dimensional so for $f$ a $K$-linear application, $f$ injective iff $f$ surjective iff $f$ bijective. $\endgroup$ – Clément Guérin Apr 14 '15 at 14:23
  • $\begingroup$ If $V$ is a finite dimensional vector space, any linear map $V\to V$ which is injective is also surjective: rank-nullity theorem. $\endgroup$ – egreg Apr 14 '15 at 14:23
  • $\begingroup$ ok that certainly helps, I have another question however, if suppose we have a linear map from $U \rightarrow V$ then would the same result apply or is it only for linear operators? $\endgroup$ – user1314 Apr 14 '15 at 14:26
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You are mentioning the rank-nullity theorem. It says that, if $f\colon V\to W$ is a linear map and $V$ is finite dimensional, then $$ \dim V=\dim\ker f+\dim\operatorname{im}f $$ In the particular case when $W=V$ and $f$ injective, we get $$ \dim V=\dim\ker f+\dim\operatorname{im}f=\dim\operatorname{im}f $$ so $f$ is surjective.

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