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

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.

• Please slow down on your questions, particularly given that that you're asking two back to back "I have no clue" questions. Mar 19, 2021 at 18:44
• Sorry, I have just been doing some questions today and I couldn't figure out these two. Since I couldn't find any similar questions online, so I decided to post them here Mar 19, 2021 at 18:46
• Okay, but please try to add more relevant context: what have you most recently covered? Do you fully understand the terminology, e.g., degree of transcendence? Try to include anything that might be relevant, even if you cannot yet connect it. We just want to see you as invested as we are in helping you. Mar 19, 2021 at 18:49
• If the degree is $3$ can you find a cubic with the roots $x, y, z$? Mar 19, 2021 at 18:51

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 $$\operatorname{tr}.\operatorname{deg}_E F=\operatorname{tr}.\operatorname{deg}_E G.$$

• Nice answer and crystal clear! Thank you :) Mar 19, 2021 at 19:03
• @14159 my pleasure, happy it helped!! :) Mar 19, 2021 at 19:05