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) $ ?
Best Answer
$K$ 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.