Minimal polynomial of $x$ over $\mathbb{Q}(x+y+z, x^2+y^2+z^2, x^3+y^3+z^3)$

Question

Let $K$ be the field of fractions of $\mathbb{Q}[x,y,z]$, $$\alpha=x+y+z,\quad \beta=x^2+y^2+z^2,\quad \gamma = x^3+y^3+z^3\in K$$ and $M=\mathbb{Q}(\alpha, \beta, \gamma)$.

1. Is $x\in K$ algebraic over $M$? What is the minimal polynomial?

2. What is the degree of transcendence of $M$ over $\mathbb{Q}$?

For part 1 I really have no idea. I tried some combinations but it didn't get me anywhere. Is there a systematic way of approaching this kind of problems?

For part 2 I'm guessing the degree is 3, but I'm not sure how to prove it.

Answer 1

Consider the polynomial $g(t)=(t-x)(t-y)(t-z)$ in $\mathbb{Q}(x,y,z)[t]$. Expanding it out yields \begin{align} g(t)&=t^3-(x+y+z)t^2 +\\ &=(yz+xz+xy)t-xyz. \end{align} Since $x$ is certainly a root of $g$, to show that $x$ is algebraic over $\mathbb{Q}(\alpha,\beta,\gamma)$ it suffices to show that each of the coefficients of $g$ lies in $\mathbb{Q}(\alpha,\beta,\gamma)$. To see this, note:

• $(x+y+z)=\alpha\in\mathbb{Q}(\alpha,\beta,\gamma)$.
• $\alpha^2=\beta+2(yz+xz+xy)$, so $(yz+xz+xy)=(\alpha^2-\beta)\big/2\in\mathbb{Q}(\alpha,\beta,\gamma)$.
• $\alpha^3=3\alpha\beta-2\gamma+6xyz$, so $xyz=(\alpha^3-3\alpha\beta+2\gamma)\big/6\in\mathbb{Q}(\alpha,\beta,\gamma)$.

This shows part $1$ of the problem, and in fact the analogous result holds for $\mathbb{Q}(x_1,\dots,x_n)$ for any $n\in\mathbb{N}$. For more on this I recommend reading about symmetric polynomials.

Now, to compute $\operatorname{tr}.\operatorname{deg}_\mathbb{Q}\mathbb{Q}(\alpha,\beta,\gamma)$, note that $g(t)$ witnesses that all of the generators of $\mathbb{Q}(x,y,z)$, and hence the field $\mathbb{Q}(x,y,z)$ itself, are algebraic over $\mathbb{Q}(\alpha,\beta,\gamma)$. What can you conclude about the respective transcendence degrees of $\mathbb{Q}(\alpha,\beta,\gamma)$ and $\mathbb{Q}(x,y,z)$ over $\mathbb{Q}$? Answer below, but try to figure it out yourself first!

Suppose $S=\{m_1,\dots,m_k\}\subset\mathbb{Q}(\alpha,\beta,\gamma)$ is a transcendence basis for $\mathbb{Q}(\alpha,\beta,\gamma)$ over $\mathbb{Q}$. By definition this means (i) there is no polynomial in $\mathbb{Q}[t_1,\dots,t_k]$ satisfied by $S$, and (ii) $\mathbb{Q}(\alpha,\beta,\gamma)$ is algebraic over $\mathbb{Q}(S)$. Now we claim that $S$ is in fact a transcendence basis for $\mathbb{Q}(x,y,z)$. Indeed, condition (i) holds immediately, and condition (ii) follows from the fact that an algebraic extension of an algebraic extension is algebraic; we have shown above that $\mathbb{Q}(x,y,z)$ is algebraic over $\mathbb{Q}(\alpha,\beta,\gamma)$, so it follows that $\mathbb{Q}(x,y,z)$ is algebraic over $\mathbb{Q}(S)$, as desired. In particular, we have $$\operatorname{tr}.\operatorname{deg}_\mathbb{Q}\mathbb{Q}(\alpha,\beta,\gamma)=\operatorname{tr}.\operatorname{deg}_\mathbb{Q}\mathbb{Q}(x,y,z)=3,$$ just as you suspected. Note that this argument works much more generally, and we have, for any triple of fields $E\subseteq F\subseteq G$, if $G$ is algebraic over $F$, then $$\operatorname{tr}.\operatorname{deg}_E F=\operatorname{tr}.\operatorname{deg}_E G.$$