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  "body": "\n## Proof\n\nWe use the lexicographic order on exponents. Existence is by induction on the leading exponent; uniqueness compares leading terms.\n\n1. Let $R$ be a [[commutative ring|0GYGHKX834]] and $A = R[x_1, \\dots, x_n]$ the [[polynomial ring|0CES91M79A]]. For $\\alpha \\in \\N^n$, write $x^\\alpha = x_1^{\\alpha_1} \\cdots x_n^{\\alpha_n}$ and $\\abs{\\alpha} = \\alpha_1 + \\dots + \\alpha_n$.\n2. Order $\\N^n$ lexicographically: $\\alpha \u003e \\beta$ if $\\alpha \\neq \\beta$ and the first nonzero entry of $\\alpha - \\beta$ is positive. This is a total order. If $\\alpha \u003e \\beta$, then $\\alpha + \\gamma \u003e \\beta + \\gamma$ for every $\\gamma \\in \\N^n$.\n3. For $f = \\sum_\\alpha c_\\alpha x^\\alpha \\neq 0$, its leading exponent $\\operatorname{LE}(f)$ is the largest $\\alpha$ with $c_\\alpha \\neq 0$. Its leading coefficient is $c_{\\operatorname{LE}(f)}$.\n4. Let $f, g \\neq 0$ with $\\alpha = \\operatorname{LE}(f)$, $\\beta = \\operatorname{LE}(g)$, and let $g$ have leading coefficient $1$. Each monomial of $fg$ comes from some $x^{\\alpha'} x^{\\beta'}$ with $\\alpha' \\le \\alpha$ and $\\beta' \\le \\beta$. By step 2, $\\alpha' + \\beta' \\le \\alpha + \\beta$.\n5. Equality $\\alpha' + \\beta' = \\alpha + \\beta$ holds only for $\\alpha' = \\alpha$, $\\beta' = \\beta$: if $\\alpha' \u003c \\alpha$, then $\\alpha' + \\beta' \u003c \\alpha + \\beta' \\le \\alpha + \\beta$. So $fg$ has leading exponent $\\alpha + \\beta$ and the leading coefficient of $f$.\n6. The [[elementary symmetric polynomial|0ZMY8441GW]] $e_k$ has leading exponent $(1, \\dots, 1, 0, \\dots, 0)$, with $k$ ones, and leading coefficient $1$.\n7. For $a \\in \\N^n$, let $e^a = e_1^{a_1} \\cdots e_n^{a_n}$. By steps 5 and 6, $e^a$ has leading coefficient $1$ and leading exponent $\\mu(a) = (a_1 + \\dots + a_n, \\; a_2 + \\dots + a_n, \\; \\dots, \\; a_n)$.\n8. The map $\\mu$ is injective, since $a_n = \\mu(a)_n$ and $a_k = \\mu(a)_k - \\mu(a)_{k+1}$. Its image is the set of $\\lambda$ with $\\lambda_1 \\ge \\lambda_2 \\ge \\dots \\ge \\lambda_n$: for such $\\lambda$, take $a_k = \\lambda_k - \\lambda_{k+1}$ with $\\lambda_{n+1} = 0$.\n\nExistence.\n\n9. Let $f$ be [[symmetric|06CTFY2DEZ]], and write $f = \\sum_d f_d$, where $f_d$ collects the terms with $\\abs{\\alpha} = d$. A permutation of the variables maps monomials of degree $d$ to monomials of degree $d$. So each $f_d$ is symmetric, and it suffices to treat a symmetric $f$ whose monomials all have degree $d$.\n10. For such $f$, we use induction on $\\operatorname{LE}(f)$ in the finite totally ordered set $\\set{\\alpha \\in \\N^n : \\abs{\\alpha} = d}$. If $f = 0$, take $g = 0$.\n11. Let $f \\neq 0$, with $\\lambda = \\operatorname{LE}(f)$ and leading coefficient $c$. Suppose that $\\lambda_i \u003c \\lambda_{i+1}$ for some $i$. Let $\\lambda'$ be $\\lambda$ with entries $i$ and $i+1$ swapped, so $\\lambda' \u003e \\lambda$.\n12. Swapping $x_i$ and $x_{i+1}$ in $f$ moves the coefficient of $x^\\lambda$ to $x^{\\lambda'}$. Since $f$ is symmetric, $f$ has coefficient $c \\neq 0$ at $x^{\\lambda'}$. This contradicts the maximality of $\\lambda$, so $\\lambda_1 \\ge \\dots \\ge \\lambda_n$.\n13. By step 8, $\\lambda = \\mu(a)$ for some $a \\in \\N^n$. Let $h = f - c \\, e^a$. Each $e_k$ has all monomials of degree $k$, so $e^a$ has all monomials of degree $\\sum_k k a_k = \\abs{\\lambda} = d$.\n14. So $h$ is symmetric with all monomials of degree $d$. All monomials of $f$ and of $e^a$ have exponent at most $\\lambda$, and the coefficient of $x^\\lambda$ in $h$ is $c - c = 0$.\n15. So $h = 0$ or $\\operatorname{LE}(h) \u003c \\lambda$. By induction, $h = g_1(e_1, \\dots, e_n)$ for some $g_1 \\in R[y_1, \\dots, y_n]$. Then $f = g(e_1, \\dots, e_n)$ with $g = g_1 + c \\, y_1^{a_1} \\cdots y_n^{a_n}$.\n\nUniqueness.\n\n16. If $g(e_1, \\dots, e_n) = g'(e_1, \\dots, e_n)$, then $g - g'$ maps to $0$. So it suffices to show: if $g = \\sum_a c_a y^a \\neq 0$, then $g(e_1, \\dots, e_n) = \\sum_a c_a e^a \\neq 0$.\n17. The exponents $\\mu(a)$ with $c_a \\neq 0$ are distinct by step 8. Let $a^*$ be the one with the largest $\\mu(a)$.\n18. The term $c_{a^*} e^{a^*}$ has coefficient $c_{a^*}$ at $x^{\\mu(a^*)}$, by step 7. Every other term $c_a e^a$ has all exponents at most $\\mu(a) \u003c \\mu(a^*)$, so it contributes $0$ there.\n19. So $g(e_1, \\dots, e_n)$ has coefficient $c_{a^*} \\neq 0$ at $x^{\\mu(a^*)}$. Hence it is nonzero, and $g$ is unique.\n\n## Idea\n\nThe leading monomial of a symmetric polynomial has decreasing exponents, and exactly one product of the $e_k$ has that same leading monomial with coefficient $1$. Subtracting the right multiple of it lowers the leading monomial, so the process ends. Different products of the $e_k$ have different leading monomials, so no nontrivial combination of them can vanish.\n",
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