Symmetric Polynomials and the Fundamental Theorem of Symmetric Functions — Examples

The leading monomial of $f$ is $x^3$. The elementary monomial prescribed by [L1] is $e_1^3$, and $e_1^3=x^3+3x^{2}y+3x{y}^2+y^3$.

Subtracting gives $f-e_1^3=-2x^{2}y-2x{y}^2=-2xy(x+y)=-2e_1{e}_2$, whose leading multidegree is smaller than $(3,0)$.

Hence $f=e_1^3-2e_1{e}_2$ in $\mathbb{Z}[x,y]$. Both sides have integer coefficients, so expanding the right side recovers the original polynomial over every commutative ring; there the remainder $-2e_1{e}_2$ of step 2.1 may itself vanish, as it does in characteristic $2$, in which case the algorithm halts one step earlier at the same answer.

The leading multidegree of $P$ is $(3,0,0)$ with coefficient $1$, so [L1] first subtracts $e_1^3$. Its expansion is ${\sum }_{\mathrm{s}\mathrm{y}\mathrm{m}}{x}^3+3{\sum }_{\mathrm{s}\mathrm{y}\mathrm{m}}{x}^{2}y+6xyz$.

The remainder is $-{\sum }_{\mathrm{s}\mathrm{y}\mathrm{m}}{x}^{2}y-3xyz$, whose leading multidegree is $(2,1,0)$. Since $e_1{e}_2={\sum }_{\mathrm{s}\mathrm{y}\mathrm{m}}{x}^{2}y+3xyz$, the next prescribed subtraction is $-e_1{e}_2$ and the remainder becomes zero.

Therefore $P=e_1^3-e_1{e}_2$. Direct expansion verifies the equality and exhibits both lexicographic decreases.

For $f(x,y)=x^2+x{y}^{100}$, the tuple $(2,0)$ is lexicographically larger than $(1,100)$, so $x^2$ is the leading monomial.

Under $x=t$ and $y=t^{100}$, the polynomial becomes $t^2+t^{10001}$, whose highest power comes from $x{y}^{100}$, not from $x^2$.

Thus the fixed weighted substitution reverses these two terms and does not detect the lexicographic leader. Degree-dependent separated weights can encode any one finite support, but this fixed choice cannot replace lexicographic descent.

At $k=1$, [L1] gives $e_1=p_1$. At $k=2$, it gives $2e_2=e_1{p}_1-p_2$, hence $p_2=e_1^2-2e_2$.

At $k=3$, [L1] gives $3e_3=e_2{p}_1-e_1{p}_2+p_3$; substituting step 1.1 yields $p_3=e_1^3-3e_1{e}_2+3e_3$.

At $k=4$, $e_4=0$, so [L1] gives $0=e_3{p}_1-e_2{p}_2+e_1{p}_3-p_4$. Substitution from steps 1.1 and 2.1 gives the displayed formula for $p_4$.

When $6$ is invertible, so are $1,2,3$, and the first three Newton identities recursively solve for the $e_i$, as asserted by [L2].

Over ${\mathbb{F}}_2$, [L1] gives $p_1=e_1$ and $p_2=e_1{p}_1=e_1^2$, because $2=0$.

Hence every polynomial in $p_1,p_2$ belongs to the proper subring ${\mathbb{F}}_2[e_1]$.

The element $e_2$ is not in ${\mathbb{F}}_2[e_1]$, since an equality $e_2=Q(e_1)$ would be a nonzero polynomial relation between $e_1$ and $e_2$, contrary to [L2].

Thus $p_1,p_2$ do not generate the two-variable symmetric ring over ${\mathbb{F}}_2$, refuting the universal statement and showing why the factorial-unit hypothesis is necessary.

Listing all monomials of total degrees $0,1,2,3$ gives the displayed values of $h_0,h_1,h_2,h_3$.

Here $E(-t)=1-e_{1}t+e_2{t}^2$. Comparing the coefficient of $t^k$ in [L1] gives $h_k-e_1{h}_{k-1}+e_2{h}_{k-2}=0$ for $k\ge 2$.

At $k=1$ the same identity gives $h_1=e_1$, and at $k=2$ it gives $h_2=e_1{h}_1-e_2$, hence $e_2=h_1^2-h_2$. This explicitly realizes the free-generation statement [L2].

Expand $(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta$.

Substitute [L2] into step 1.1 to get $(\alpha -\beta {)}^2=b^2-4c$, and apply [L1] to identify this with $\mathrm{Disc}(f)$.

By [L1], this element vanishes exactly when $\alpha =\beta$. In characteristic two the formula becomes $b^2$, while $f^{\mathrm{\prime }}=b$, so the same repeated-root criterion remains valid; completing a square is a separate issue.

Since $f^{\mathrm{\prime }}(t)=3t^2+p$, the root-product formula inside [L1] gives $\mathrm{Res}(f,f^{\mathrm{\prime }})={\prod }_{r\in \{\alpha ,\beta ,\gamma \}}(3r^2+p)$.

Expanding this product gives $27(\alpha \beta \gamma {)}^2+9p({\alpha }^2{\beta }^2+{\alpha }^2{\gamma }^2+{\beta }^2{\gamma }^2)+3p^2({\alpha }^2+{\beta }^2+{\gamma }^2)+p^3$.

By [L2], ${\alpha }^2+{\beta }^2+{\gamma }^2=-2p$ and ${\alpha }^2{\beta }^2+{\alpha }^2{\gamma }^2+{\beta }^2{\gamma }^2=p^2$, while $(\alpha \beta \gamma {)}^2=q^2$. Substitution in step 2.1 gives $\mathrm{Res}(f,f^{\mathrm{\prime }})=4p^3+27q^2$.

Apply [L1] to obtain $\mathrm{Disc}(f)=-4p^3-27q^2$. Every calculation used integer coefficients, so reduction to any field characteristic is valid.

Since ${\alpha }^2=-a\alpha -b$ and ${\beta }^2=-a\beta -b$, put $u=c-a$ and $v=d-b$ to obtain $g(\alpha )=u\alpha +v$ and $g(\beta )=u\beta +v$.

Multiply and use [L2]: $g(\alpha )g(\beta )=u^2\alpha \beta +uv(\alpha +\beta )+v^2=b{u}^2-auv+v^2$.

Substitution of $u=c-a$ and $v=d-b$ gives the displayed formula, and [L1] identifies it as the resultant.

For $f=(t-1)(t-2)$ and $g=(t-2)(t-4)$ the formula gives $0$, as the shared root predicts. For the same $f$ and $g=t^2+1$ it gives $10$, so over $\mathbb{Q}$ the polynomials have no common root.