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I was reading $ 16.3 $ SPLITTING FIELD in Algebra by Artin, where I met the following:

For every element $ \beta $ of $ K $, there is a polynomial $ p(u_1,\cdots, u_n) $ with coefficients in $ F $, such that $ p(\alpha_1, \cdots, \alpha_n)=\beta $, where field $ K=F(\alpha_1, \cdots, \alpha_n) $ is the splitting field for $ f $ over field $ F $, and $ f(x)=(x-\alpha_1)\cdots (x-\alpha_n) $ with $ \alpha_i\in K $.

Why can we express $ \beta $ by $ p(\alpha_1, \cdots, \alpha_n) $ ?

  • What do you know about $K$? I suppose that it is an algebraic extension gotten by adjoining the $\alpha_i$. Then every element of $F(\alpha_i)$ is of the form $p(\alpha_i)$. – 4-ier Aug 22 '18 at 05:54
  • Is it clear that $\beta$ can be expressed by a rational function $r(\alpha_1,\ldots, \alpha_n)$ and the question is why we can take polynomial expression? – Lior B-S Aug 22 '18 at 06:11
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    To put it succinctly...because all rings $R$ with $F\subseteq R\subseteq K$ are fields. – C Monsour Aug 22 '18 at 06:43

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$K$ is the minimal field containing all the $\alpha_i$. The subset of all elements that can be expressed as $p(\alpha_1,\dots,\alpha_n)$ is a subfield of $K$. (It's easy to check that it is closed under addition and multiplication and additive inverse, since so are polynomials with coefficients in $F$. It's closed under taking multiplicative inverses because in any algebraic extension any element's inverse is an $F$-linear combination of its positive powers...see the next paragraph if this is not a familiar fact.) But since $K$ is minimal with this property, a subfield with this property must be all of $K$.

To explain the argument about not needing to be able to explicitly take inverses, note that if, for $\gamma\in K$, $x^m+b_{n-1}x^{m-1}+\cdot\cdot\cdot+b_0$ is the minimal polynomial for $\gamma$ over $F$, then $\gamma^{-1}=-(1/b_0)\gamma^{m-1}-(b_{m-1}/b_0)\gamma^{m-2}-\cdot\cdot\cdot-(b_1/b_0)$. So closure of a subring of $K$ under addition and multiplication also gives closure under multiplicative inverse. (The inverses of $b_0$ are not a problem since $b_0\in F$ and we already have inverses of everything in $F$.) Or to put it another way, any subring of $K$ containing $F$ is also a subfield. Thus, since the set of all $p(\alpha_1,\dots,\alpha_n)$ with $p\in F[x]$ is obviously a subring it is also a subfield.

C Monsour
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  • Thank you! So actually the second paragraph is no longer necessary, we only need to show $ F\subset F[ \alpha_1, \cdots, \alpha_n ]\subset K $ and because $ F[ \alpha_1, \cdots, \alpha_n ] $ is a ring and $ F $ and $ K $ are fields and we are done. –  Aug 22 '18 at 06:59
  • If you had said "....because $F$ is a field and $K$ is an algebraic extension", you'd be right. (The theorem that all intermediate rings are fields is only for algebraic extensions, not transcendental ones.) The second paragraph is a proof of this result for those unfamiliar with it, so whether you need it depends on how much you already know. – C Monsour Aug 22 '18 at 07:03
  • Yes, but I think $ K $ should be a finite extension of $ F $ to guarantee it is a finite dimensional vector space over $ F $. See: https://math.stackexchange.com/q/2879987/578483 –  Aug 22 '18 at 07:31
  • It's a mistake for you to assume that just because a property holds for finite extensions that it does not also hold in some other cases. In this instance, the fact that every intermediate ring of an infinite-dimensional algebraic field extension is a field is an easy corollary of that fact for finite extensions. Not that it matters for solving your posted problem, where you only need the fact for finite extensions. – C Monsour Aug 22 '18 at 07:37
  • I am sorry, I don't think the second paragraph is making any point. To show that $ F[\alpha_1, \cdots, \alpha_n] $ is a field, we should see that $ F[\alpha_1] $ is a field and $ F[\alpha_1, \cdots, \alpha_n]=F[\alpha_1][\alpha_2, \cdots \alpha_n] $, by induction we are done. –  Aug 22 '18 at 07:42
  • As I said before, if you already know $F[\alpha_1]$ is a field you don't need the second paragraph. – C Monsour Aug 22 '18 at 07:47
  • Oh, I see. Could you give a hint for proving the generalized case when we only require algebraic field extension? It seems we can not use rank-nullity theorem any more in that case? –  Aug 22 '18 at 07:47
  • It's trivial. Let $K/F$ be an algebraic extension and let $R$ be an intermediate ring. Let $\alpha\in R$. Then $\alpha^{-1}\in F[\alpha]\subseteq R$ since $F[\alpha]$ is finite-dimensional for all $\alpha\in K$ by the definition of "algebraic". Thus $R$ contains the inverses of all its elements, which was all we needed to show. – C Monsour Aug 22 '18 at 07:52