Let $k$ be a field of characteristic $0$. For $j \ge 0$, let $p_j = t_1^j + \dots + t_n^j \in k[t_1, \dots, t_n]$. Prove that $p_1, \dots, p_n$ generate the $k$-algebra of symmetric polynomials in $k[t_1, \dots, t_n]$, and are algebraically independent over $k$.
As to what I've tried so far, I've messed around with the Newton identities, but they haven't really got me anywhere. Am I going up the right alley? Or is there some other approach I should be taking?
On
Recall that $s_1, s_2, \dots, s_n$ generate the $k$-algebra of symmetric functions, where $$\prod_{i=1}^n (x-t_i) = x^n - s_1x^{n-1} + \dots + (-1)^{n-1}s_{n-1}x + (-1)^ns_n.$$By Exercise IV.11(c) in Lang's Algebra, $${{R'}\over{R}} = \sum_{i=1}^n {1\over{x-t_i}}.$$One can expand$${1\over{x-t_i}} = {1\over{x}}{1\over{1 - t_i/x}}$$as a geometric series to get$$x^{-1} + t_ix^{-2} + (t^i)^2x^{-3} + \dots = \sum_{k=1}^\infty x^{-i}(t_j)^{i-1}.$$$($If one is worried about convergence, one can either assume that $|x/t_i| < 1$ for all $i$, or one can just claim this is $``$just algebra,$"$ and work in an $x^{-1}$-adic completion.$)$
Combining the coefficients, we get$${{R'}\over{R}} = \sum_{i=1}^\infty x^{-i}p_j,$$where for consistency $p_0 = n$. Multiplying through by $R$ we get$$nx^{m-1} - (n-1)s_1x^{n-2} + \dots + (-1)^{n-1}s_{n-1}$$$$ = \left(x^n - s_1x^{n-1} + \dots + (-1)^{n-1}s_{n-1}x + (-1)^ns_n\right)\left(\sum_{i=1}^\infty x^{-i} p_j\right).$$Taking the coefficients of $x^i$ for $-1 \le i \le n-2$ and canceling out powers of $-1$, we get$$(i+1)s_{n-i-1} = ns_{n-i-1} - p_1s_{n-i-2}+ \dots + (-1)^{n-i-1}p_{n-i-1}.$$Letting $k=n-i-1$ $($so $i=n-k-1)$, the above becomes $$(n-k)s_k = ns_k - p_1s_{k-1} + \dots + (-1)^kp_k,$$where $1 \le k \le n$. Notice we have $$s_k = 1/(n-k)\left(p_1s_{k-1} - \dots -(-1)^n p_k\right),$$so by induction the $p_1, \dots, p_j$ generate $s_j$ and thus $p_1, \dots, p_n$ are an algebra basis for the symmetric functions since they generate an algebra basis. $($This uses $(n-k)$ invertible in the field, so one needs either characteristic $0$, or characteristic $> n$.$)$ Assigning weight $i$ to $p$, notice that the number of monomials in the $p_i$'s of a fixed weight equals the number of monomials of the same weight in the $s_i$'s $($since there is a weight preserving bijection between corresponding monomials$)$. But any algebraic relation in the $p_i$'s can be assumed to involve monomials of the same weight. Now both the monomials in the $p_i$'s and of the $s_i$'s of a given weight generate the symmetric functions of the same weight, and have the same cardinality. But we already know the monomials in the $s_i$'s form a basis, so the ones in the $p_i$'s must also, and thus no relation can occur among monomials in the $p_i$'s, that is the $p_i$ are algebraically independent.
There is a theorem which says that the ring of symmetric polynomials is just $k[\sigma_1, \dots \sigma_n]$, where $\sigma_i$ are the elementary symmetric polynomials. So you need to show that $\sigma_i \in k[p_1, \dots,p_n]$ for every $1 \leq i \leq n$. $\sigma_1$ is definitely there. $p_1^2 = p_2 + (1+1)\sigma_2$, so we can write $\sigma_1 = (p_1^2 - p_2) / (1+1)$, since the characteristic of $k$ is $0$. Likewise using some induction you can show that every $\sigma_i$ is in the $k$ module generated by all the symmetric polynomials constructed on the previous step.
Edit : It's easier then I thought. The inductive hypothesis is that $k[\sigma_1, \dots, \sigma_i] = k[p_1,\dots,p_i]$. Then as the theorem says that $p_{i+1} \in k[\sigma_1, \dots,\sigma_n]$ or actually in $k[\sigma_1, \dots , \sigma_{i+1}]$, because the rest are of higher order. So $p_{i+1} = a\sigma_{i+1} + g(\sigma_1, \dots, \sigma_i)$, where $g \in k[t_1, \dots, t_i]$ and $a \neq 0$. By the inductive hypothesis $g(\sigma_1, \dots, \sigma_i) \in k[p_1, \dots, p_i]$, so $\sigma_{i+1} \in k[p_1, \dots, p_{i+1}]$.