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I am trying to prove that, for a field extension $\mathbf{K}/k$ and $a$ an algebraic element over $k$, $$k(a)=k[a],$$ where $k(a)$ is the subfield of $\mathbf{K}$ generated by $a$ and $k[a]$ is the subring of $\mathbf{K}$ generated by $a$.

I know (please correct me if I am wrong) that $k[a]$ consists of all elements of $\mathbf{K}$ that can be expressed as polynomials in $a$ with coefficients in $k$, while $k(a)$ consists of all elements of $\mathbf{K}$ that can be expressed as the quotient of two such polynomials. What we want to show, I guess, amounts to proving that if $f(x)$ and $g(x)$ are two polynomials with coefficients in $k$ then $\frac{f(a)}{g(a)}$ is also a polynomial (in $a$) with coefficients in $k$. I am not sure how to prove this. What I did is, I considered the minimal polynomial $m$ of $a$ over $k$. Then we have that $f(a)$ and $g(a)$ can both be reduced "modulo $m$" to give polynomials of degree at most $\deg{m}-1$. But then, how can I see what their quotient looks like?

When we want to prove this result considering the extension $\mathbb{R}/\mathbb{Q}$ and the algebraic number $\sqrt{2}$, a quotient of polynomials in $\sqrt{2}$ with coefficients in $\mathbb{Q}$ can be viewed as a polynomial in $\sqrt{2}$ with coefficients in $\mathbb{Q}$ by multiplying both the numerator and denominator with the "conjugate" of the denominator. Is this the idea that has to be used here? If yes, what plays the role of the conjugate? Thanks for any help.

PaulRenet1
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    All you have to do is provve $k[a]$ is a field, as the only field it could possibly be is $k(a)$. Do you see how to prove $a$ is invertible? – Gerry Myerson Jul 05 '14 at 10:49
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    @GerryMyerson Is this why? : Since $a$ is algebraic, there exists a polynomial $P(x)=k_0+k_1x+\cdots+k_n x^n$ such that $P(a)=0$. But then, $a(k_1+\cdots+k_n a^{n-1})=-k_0$ and thus $a[(k_1+\cdots+k_n a^{n-1})(-k_0^{-1})]=1$, proving that $a$ is invertible. – PaulRenet1 Jul 05 '14 at 10:56
  • Yes, that's how you prove $a$ is invertible. – Gerry Myerson Jul 05 '14 at 10:58
  • If $k[a]$ is a field, then it has to be $k(a)$ because it would then be a subfield of $\mathbf{K}$ containing $a$ and $k$ and which is the smallest such (because every other will have to contain $k[a]$ for obvious reasons). Ok, so this is settled. Now, $k[a]$ is indeed a field because since $a$ is invertible so is any power of it and hence also any power of it multiplied by an element of the field $k$. What I can not see is why the sum of invertible elements is invertible. Is this a fact that I am not aware of, or just something particular about this situation that I have to prove? Thanks! – PaulRenet1 Jul 05 '14 at 11:14
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    $k[a]$ is a finite dimensional vector space over $k$, so every element of $k[a]$ is algebraic over $k$, so every element of $k[a]$ is invertible. – Gerry Myerson Jul 05 '14 at 12:03
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    Possible duplicate of Algebraic field extensions: Why $k(\alpha)=k[\alpha]$. – Watson Sep 07 '16 at 09:52

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