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I am trying to see why

$$\mathbb{Z}[s_1, s_2, s_3]$$ is a unique factorisation domain where: $$\begin{split} s_1&=x+y+z\\ s_2&=xy+yz+zx\\ s_3&=xyz\\ \end{split}$$

I encountered this while solving a problem for an exam. I want to show that $\mathbb{Z}[s_1, s_2, s_3]$ is integrally closed in its field of fractions, but I am not sure how to prove it is a UFD.

Christian E. Ramirez
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permutation_matrix
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  • Can you show that it is isomorphic to a polynomial ring in $3$ variables over the integers? – Arkady Dec 29 '22 at 11:48
  • I understand what the map should be, but I cannot show that it is injective. equivalently I need to show that $s_1, s_2$ and $s_3$ are algebraically independent. – permutation_matrix Dec 29 '22 at 11:56
  • It may be worthwhile to take a look at elementary symmetric polynomials. Your question is a special case of the more general theory. – nilsw Dec 29 '22 at 15:31

2 Answers2

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We show that the symmetric polynomials are algebraically independent in general.

Notation. $\sigma_{k, n}$ will denote the $k$-th symmetric polynomial in the $n$-variables $x_1, \ldots, x_n$. (Here, $1 \leqslant k \leqslant n$.)
For example, $\sigma_{1,n} = x_1 + \cdots + x_n$, and $\sigma_{n, n} = x_1 \cdots x_n$.

By induction on $n$, we show that the map $$\varphi_n : \Bbb Z[x_1, \ldots, x_n] \to \Bbb Z[\sigma_{1, n}, \ldots, \sigma_{k, n}]$$ defined by

$$x_i \mapsto \sigma_{i, n}$$ is injective. ($n = 3$ is the case you are interested in.)

For $n = 1$, the map is the identity map, and we are done.

Assume now that the result holds for $n - 1$.
For the sake of contradiction, assume that $\varphi_n$ is not injective. Pick $f \in \ker(\varphi_n)$ such that $f$ has minimal degree in $x_n$. We can write $$f(x_1, \ldots, x_n) = f_0(x_1, \ldots, x_{n - 1}) + \cdots + f_d(x_1, \ldots, x_{n - 1}) x_n^d.$$ Note that $f_0(x_1, \ldots, x_{n - 1}) \in \Bbb Z[x_1, \ldots, x_{n - 1}]$ is not the zero polynomial. (Why? We are using the minimality of $d$ and the fact that $x_n$ is a nonzerodivisor.)

Now, since $f \in \ker(\varphi_n)$, we have $$f_0(\sigma_{1, n}, \ldots, \sigma_{n - 1, n}) + \cdots + f_d(\sigma_{1, n}, \ldots, \sigma_{n - 1, n}) \sigma_{n,n}^d = 0.$$ Note that the LHS is still a polynomial in $x_1, \ldots, x_n$. Thus, we may put $x_n = 0$ above. Note that $\sigma_{n, n}|_{x_n = 0} = 0$ and $\sigma_{k, n}|_{x_n = 0} = \sigma_{k, n - 1}$ for $1 \leqslant k < n$.
Thus, the above equation becomes $$f_0(\sigma_{1, n - 1}, \ldots, \sigma_{n - 1, n - 1}) = 0,$$ which is a contradiction since $\varphi_{n - 1}$ was supposed to be injective.

Aryaman Maithani
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The field extension $\mathbf Q(x_1,\ldots,x_n)/\mathbf Q(s_1,\ldots,s_n)$ is algebraic since the $x_i$'s are all roots of $(T-x_1)\cdots(T-x_n) = T^n - s_1T^{n-1} + \cdots + (-1)^ns_n$, which has coefficients in $\mathbf Q(s_1,\ldots,s_n)$.

For a tower of fields $L/K/M$, the transcendence degree is additive: $$ {\rm tr \ deg.}(L/M) = {\rm tr \ deg.}(L/K) + {\rm tr \ deg.}(K/M) $$ as cardinal numbers. If $L/K$ is algebraic, then $ {\rm tr \ deg.}(L/K) = 0$, and thus ${\rm tr \ deg.}(L/M) = {\rm tr \ deg.}(K/M)$. Applying this to the tower of fields $\mathbf Q(x_1,\ldots,x_n)/\mathbf Q(s_1,\ldots,s_n)/\mathbf Q$, we get $$ {\rm tr \ deg.}(\mathbf Q(x_1,\ldots,x_n)/\mathbf Q) = {\rm tr \ deg.}(\mathbf Q(s_1,\ldots,s_n)/\mathbf Q), $$ and the left side is $n$ since $x_1, \ldots, x_n$ are algebraically independent over $\mathbf Q$. Thus $s_1, \ldots, s_n$ are algebraically independent over $\mathbf Q$.

There was nothing special about using $\mathbf Q$ here. The exact same arguments shows that if $x_1, \ldots, x_n$ are algebraically independent over a field $F$ then their elementary symmetric polynomials $s_1, \ldots, s_n$ are also algebraically independent over $F$.

KCd
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