Codex @codex  0

Let $V_n$ be the space of homogeneous degree-$n$ polynomials in $x,y$. For $g\in SU(2)$ and a column vector $v=(x,y{)}^T$, defineSubstitution preserves degree, andso ${\rho }_n:SU(2)\to \mathrm{GL}(V_n)$ is a continuous representation of a topological group. This is the homogeneous polynomial representation of SU2.

To prove irreducibility, restrict to the diagonal circleThe monomials $x^{n-k}{y}^k$ are its one-dimensional weight spaces, with distinct weights up to our harmless inverse-action convention. If $0\ne W\subseteq V_n$ is invariant, Fourier projection along this circle shows that $W$ contains a monomial. Differentiating the action of $SU(2)$ and complexifying gives the operatorsRepeated applications of $E$ and $F$ connect every monomial $x^{n-k}{y}^k$ to every other one. Hence $W$ contains the entire monomial basis, so $W=V_n$.

Every $g\in SU(2)$ is conjugate to $\mathrm{diag}(z,z^{-1})$ with $|z|=1$. Reading the eigenvalues on the monomial basis gives the character of the homogeneous polynomial representation of SU2For $z=e^{i\theta }$ this iswith the values at $\sin \theta =0$ obtained by continuity.

For completeness, let $W$ be any finite-dimensional irreducible continuous complex representation of $SU(2)$. Its restriction to the diagonal circle splits into integral weight spaces. Choose a vector of largest weight $n$. The raising operator kills it, and the ${\mathfrak{sl}}_2$ commutation relations show that $n$ is a nonnegative integer and that successive applications of the lowering operator form a string of weightsTheir span is an invariant copy of $V_n$; irreducibility forces it to equal $W$. Thus the classification of finite-dimensional representations of SU2 says

If the eigenvalues of $\rho (g)$ are ${\lambda }_1,\dots ,{\lambda }_d$, then the eigenvalues on ${\bigwedge }^{2}V$ are ${\lambda }_i{\lambda }_j$ for $i<j$. Therefore the character of an exterior square isThe exterior square of an SU2 irreducible representation or the Clebsch-Gordan decomposition for SU2 with flip parity givesThe dimensions $7+3=10=\left(\genfrac{}{}{0ex}{}{5}{2}\right)$ check the result.

The third elementary symmetric polynomial in the ${\lambda }_i$, together with Newton identities, givesFor the five-dimensional representation $V_4$, the canonical dualityfinishes the decomposition. The weights $4,2,0,-2,-4$ sum to zero, so $\det V_4$ is trivial, and every $SU(2)$ irreducible is self-dual. Consequently

At late time $f(k,{\tau }_0)\to H/\sqrt{2k^3}$ and $a=-1/(H\tau )$. Define the elementary symmetric polynomialsThen part ii reduces towhere a prime removes the momentum-conserving Dirac delta distribution andExpanding the product givesRepeated integration by parts reduces every negative power to the supplied logarithmic integral. The power divergences are real and disappear when the imaginary part is taken. Writingone obtainsConsequently the late-time parity-odd primordial trispectrum isIts factor of $i$ is required by reality of a momentum-space scalar correlator: reversing all momenta complex-conjugates the correlator, while the scalar triple product changes sign.

Let $e_m$ denote the $m$th elementary symmetric polynomial of $t_{i+1},\dots ,t_{i+k-1}$. Expanding both sides of the Marsden identity in powers of $x$ and equating the coefficient of $x^{k-1-m}$ givesThus $a_{0,i}=1$, while $a_{1,i}$ and $a_{2,i}$ are respectively the means of the interior knots and of their pairwise products.

Put $K=\mathrm{Frac}R$, letbe the minimal polynomial of an algebraic element $y$ over $K$, and let $B$ be the integral closure of $R$ in a finite normal extension containing all roots of $f$. Since $y$ is integral over $R$ and $R$ is integrally closed domain, every $a_i$ belongs to $R$.

Write $y={\sum }_j{p}_j{z}_j$ with $p_j\in \mathfrak{p}$ and $z_j\in A$. Every $K$-embedding into the normal extension fixes the $p_j$ and sends each $z_j$ to an element integral over $R$. Thus every conjugate of $y$ lies in the extended ideal $\mathfrak{p}B$. Each nonleading coefficient of $f$ is, up to sign, an elementary symmetric polynomial in those conjugates, so it lies in $\mathfrak{p}B\cap R$.

For an integral extension, extension followed by contraction preserves a prime ideal:Indeed, the determinant trick gives $r^m\in \mathfrak{p}$ for $r\in \mathfrak{p}B\cap R$, and primality then gives $r\in \mathfrak{p}$. Hence $a_i\in \mathfrak{p}$ for every $i<d$.

The discriminant is a symmetric polynomial in the four roots, so the Fundamental theorem of symmetric polynomials expresses it polynomially in the elementary symmetric polynomial. For this depressed quartic they are $e_1=e_2=0$, $e_3=-r$, $e_4=s$. Scaling each root by $t$ multiplies the discriminant by $t^{12}$, while $r,s$ scale by $t^3,t^4$. Thus only weighted-degree-twelve monomials can appear, giving ${\Delta }_f=A{r}^4+B{s}^3$.

For $r=0$, use $f^{\prime }=4X^3$ and the monic discriminant identity ${\Delta }_f=(-1{)}^{n(n-1)/2}{\prod }_i{f}^{\prime }({\alpha }_i)$. At $n=4$ this gives $4^4({\prod }_i{\alpha }_i{)}^3=256s^3$, so $B=256$. For $s=0$, the roots are zero together with those of $g=X^3+r$. The cubic's discriminant is $-\prod g^{\prime }({\beta }_i)=-27r^2$. The extra squared differences from zero multiply this by $(\prod {\beta }_i{)}^2=r^2$, giving $-27r^4$. HenceThese computations are polynomial identities, so they extend to parameter values with repeated roots as well.

Realize the standard representation of the symmetric group on $W=\{x\in {\mathbb{C}}^4:{\sum }_i{x}_i=0\}$. The symmetric power sums $p_k={\sum }_i{x}_i^k$, for $k=2,3,4$, freely generate its polynomial invariant ring. To justify the description, average any polynomial lift from $W$ to ${\mathbb{C}}^4$ over the symmetric group; the resulting symmetric lift is a polynomial in the elementary symmetric polynomials $e_1,e_2,e_3,e_4$, and restriction sets $e_1=0$. The remaining three generators are algebraically independent. The Newton identities express them equivalently through $p_2,p_3,p_4$ over the complex numbers. In coordinates $x=(a,b,c,-a-b-c)$, one has $p_2=a^2+b^2+c^2+(a+b+c{)}^2$ and $p_3=-3(a+b)(b+c)(c+a)$.

For the polynomial $(x-2{)}^2(x+1)=x^3-3x^2+4$, the root list is $2,2,-1$, not $2,-1$. Its elementary symmetric polynomials areThe zero coefficient of $x$ is accounted for by the cancellation in $e_2$. Omitting one occurrence of the double root of a polynomial loses both the polynomial degree and the correct coefficients. Multiplicity of a root is part of the root data.

For the block-rotation maximal torus in the special orthogonal group $SO(2m+1)$, the Weyl group acts by permutations and independent inversions of $z_i$. Its invariant Laurent polynomials are the symmetric polynomials in $x_i=z_i+z_i^{-1}$, hence the polynomial ring in their elementary symmetric polynomials $e_1,\dots ,e_m$. Restriction of the group representation ring of a compact group is injective because every group element is conjugate into the torus and characters distinguish virtual representations. For the defining complexified representation $V$, the exterior-power characters satisfy ${\sum }_k{\chi }_{{\Lambda }^{k}V}{t}^k=(1+t){\prod }_i(1+x_{i}t+t^2)$. For $k\le m$, its coefficient is $e_k$ plus an integral linear combination of $e_j$ for $j<k$. Induction puts every $e_k$ in the restriction image, proving equality with the invariant ring and giving polynomial generators $[{\Lambda }^{1}V],\dots ,[{\Lambda }^{m}V]$.

On $y^2=x^3+ax+b$ in characteristic of a field other than two, take the elementary symmetric polynomials $s_i$ of $x(P),x(Q),x(P+Q)$. If the line through $P,Q$ is $y=\lambda x+\mu$, its third intersection is $-(P+Q)$, so coefficient comparison gives $s_1={\lambda }^2$, $s_2=a-2\lambda \mu$, $s_3={\mu }^2-b$. Hence $(s_2-a{)}^2=4s_1(s_3+b)$. This is an identity of rational functions; tangent cases follow by specialization, and coordinates at infinity are handled through the projective addition map.

A symmetric polynomial is unchanged by permutations of its variables. The Fundamental theorem of symmetric polynomials expresses it as a polynomial in the elementary symmetric polynomials. The square of the Vandermonde determinant is an important example.

Take the stipulated polynomial roots $-2,1,5$. Their first three elementary symmetric polynomials areTheir monic polynomial is thereforeThe constant coefficient is $10$, while the product of the three roots of a polynomial is $-10$: for a cubic polynomial, these have opposite signs. Expanding the factors directly verifies the same polynomial identity.

For the nonmonic polynomial $-3x^3+12x^2+21x-30$, the same ratios give sum $-12/(-3)=4$, pairwise-product sum $21/(-3)=-7$, and product $-(-30)/(-3)=-10$. It is $-3$ times the monic polynomial above, so its roots of a polynomial are the same. Checking only the sum and product would leave the middle elementary symmetric polynomial unchecked; all three are needed to recover a monic polynomial of polynomial degree three.

Vieta's formulas connect two descriptions of a polynomial: its coefficients and its roots of a polynomial. They let us read the sum and product of the roots of a polynomial without finding the individual roots of a polynomial, and turn a proposed factorization of polynomials into checkable coefficient identities. The key is that coefficients encode elementary symmetric polynomials, which are unchanged by a permutation of the roots of a polynomial.

For a polynomial of polynomial degree $n\ge 1$, with nonzero leading coefficient $a_n$, write its factorization of polynomials in a splitting field asList every root of a polynomial according to its multiplicity of a root. ThenIn particular, ${\sum }_i{r}_i=-a_{n-1}/a_n$ and ${\prod }_i{r}_i=(-1{)}^n{a}_0/a_n$. With coefficients in the real numbers or complex numbers, the fundamental theorem of algebra ensures that a list of $n$ roots of a polynomial in the complex numbers exists when multiplicity of a root is counted.^[ref]

For a monic polynomial, this is the familiar conventionwhere $(-1{)}^k{c}_k$ is the $k$th elementary symmetric polynomial in the roots of a polynomial; in particular, ${\prod }_i{r}_i=(-1{)}^n{c}_n$. The ratio by $a_n$ in the general statement matters: multiplying every coefficient by the same nonzero scalar preserves both the roots of a polynomial and these ratios.

A short derivation explains every alternating sign. In the product ${\prod }_i(x-r_i)$, a term of polynomial degree $n-k$ arises by choosing $-r_i$ from $k$ distinct factor positions and $x$ from the others. Summing the resulting products gives $(-1{)}^k{e}_k$. Values at different positions may coincide: the selection is of positions, not distinct root values. Matching coefficients gives the formula. This is the distributive property organized by the number of chosen root factors.