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The actual-degree condition in the resultant theorem matters when coefficients depend on a parameter. TakeThe nominal degree-one Sylvester matrix is $\left(\begin{array}{cc}t& 1\\ t& 2\end{array}\right)$, whose determinant is $t$. Its specialization at $t=0$ therefore vanishes. But the specialized polynomials are the nonzero constants $1$ and $2$, which have no common root at all.

For $t\ne 0$, the roots of a polynomial are $-1/t$ and $-2/t$; neither has a finite limit as $t\to 0$. Both leading coefficients vanish at the exceptional parameter. The fixed-size determinant has retained the original degree format even though the specialized polynomial degrees have dropped. Recompute using the actual specialized polynomials and check the original equations before accepting a zero of a parameter-dependent resultant as a solution.

Consider the original pair of polynomial equationsTreat them as polynomials in $y$ with coefficients in $\mathbb{R}[x]$. Their Sylvester matrix and resultant areEvery common solution must therefore have $x=-1$ or $x=1$. Back-substitution into $F=0$ gives $y=x^2=1$, and both pairs also satisfy $G=0$. The complete real solution set isHere elimination and direct substitution agree: subtracting $F$ from $G$ gives $2x^2-2$, exactly the same condition.

In general, first solve the one-variable resultant equation, then recover and check the other coordinate in both original polynomial equations. A real coordinate candidate can correspond only to nonreal values of the eliminated variable. Also, vanishing leading coefficients require the separate check in degree drop in a specialized resultant. The general elimination-and-back-substitution method is described in the lecture's §3.1.^[ref]

For genuine degrees three and two, the resultant is the Sylvester matrix determinantEquivalently, if $r_1,r_2,r_3$ are the roots of $p$, counted with multiplicity, then $R(p,q)=a^2{\prod }_{i}q(r_i)$. Thus $R(p,q)\ne 0$ means that the polynomials have no common root, or equivalently are coprime over $\mathbb{C}$. One can also see this from the matrix: its singularity gives a nonzero relation $Ap+Bq=0$ with $\mathrm{deg}A<2$, $\mathrm{deg}B<3$. If $p,q$ were coprime, $p\mid B$ would force $B=0$, then $A=0$, a contradiction. Conversely, a common factor supplies such a relation after dividing $p,q$ by that factor. When a leading coefficient is zero, use the resultant for the actual polynomial degrees rather than assuming the displayed fixed-degree criterion unchanged.

An algebraic addition theorem for a meromorphic function $f$ means a nonzero polynomial $P(U,V,W)$ withidentically where the values are finite. A local relation extends meromorphically to the other regular arguments.

For a polynomial $f$ of positive degree $d$, let $z_i$ be the $d$ roots of $f(z)-U$ and $w_j$ the $d$ roots of $f(w)-V$. FormIts coefficients are separately symmetric polynomials in the two sets of roots. The Fundamental theorem of symmetric polynomials and the fixed nonzero leading coefficient of $f$ show they are polynomials in $U,V$. The expression is monic of degree $d^2$ in $W$, so it is nonzero. If $U=f(z)$, $V=f(w)$, one factor vanishes at $W=f(z+w)$. This proves the addition relation, including at exceptional multiple-root values by polynomial identity. A constant polynomial instead has the relation $W-f(0)=0$.

For the elliptic case, let $E=\mathbb{C}/\Lambda$. A nonconstant elliptic function $f$ defines a finite morphism $E\to {\mathbb{P}}^1$, so the function field of $E$ is a finite algebraic extension of $\mathbb{C}(f)$. The elliptic function-field decomposition expresses it as $\mathbb{C}(\wp ,{\wp }^{\prime })$, withThe Weierstrass addition formula makes translation algebraic:Differentiating and using the cubic differential equation also makes ${\wp }^{\prime }(z+w)$ rational in the four separate values. Therefore $f(z+w)$ lies in a finite algebraic extension of $\mathbb{C}(f(z),f(w))$. Its algebraic equation, after clearing denominators, is the required polynomial relation. This sketches why every elliptic function, not only $\wp$, has an algebraic addition theorem. Constants are already covered.

For the final assertion, consider $f(z)=e^{e^z}$, an entire function. If $c$ is a period, then $e^{e^z(e^c-1)}=1$ for every $z$. The continuous exponent must be a constant integer multiple of $2\pi i$; its derivative forces $e^c=1$. Hence its periods are exactly $2\pi i\mathbb{Z}$, making it a simply periodic function.

Suppose it had an algebraic addition theorem $P$. There are only finitely many values $V_0$ for which $P(U,V_0,W)$ is identically zero: a nonzero coefficient polynomial in $V$ already bounds this exceptional set. Choose a positive irrational $\alpha$ with $e^{\alpha }$ outside this set, and set $c=\log \alpha$. Specializing $w=c$ gives a nonzero polynomial $Q(U,W)=P(U,e^{\alpha },W)$ and, with $t=e^z$,All exponents with nonzero coefficients are distinct, by irrationality of $\alpha$. Divide by the exponential with largest real exponent and let $t\to +\infty$; its nonzero coefficient would have to tend to zero. This contradiction proves that simple periodicity does not imply an algebraic addition theorem. It is the simply periodic entire function without an algebraic addition theorem example.

The resultant turns a shared-root of a polynomial question into a calculation using only coefficients. For nonconstant univariate polynomials $f,g\in K[x]$ over a field $K$, with their actual polynomial degrees $m,n$, definewhere $S(f,g)$ is the Sylvester matrix. ThenHere $\overline{K}$ is an algebraic closure. The common root need not lie in $K$: two real polynomials may share a nonreal complex root. The subscript identifies which variable is eliminated.^[ref]

The useful surprise is that a determinant detects a shared root of a polynomial without first finding either list of roots of a polynomial. If the coefficients depend on a parameter, the resultant itself is a polynomial in that parameter. Its zeros pick out possible exceptional parameter values. Elimination theory uses this to reduce a pair of polynomial equations in two variables to a single necessary equation in one variable; eliminating a variable with a resultant gives an exact example below.

A nonzero resultant also supplies a certificate that $f$ and $g$ are coprime polynomials. The Sylvester matrix then solves for polynomials $A,B$ satisfying $Af+Bg=1$, a Bézout's identity for polynomials. Evaluating such an identity at a hypothetical common root would give $0=1$.

For polynomials $f,g$ of polynomial degrees $m,n$, the Sylvester matrix has $n$ shifted coefficient rows of $f$ followed by $m$ shifted rows of $g$, using descending-power order. It is a square matrix of size $m+n$. Its transpose represents the linear map $(A,B)\mapsto Af+Bg$ with $\mathrm{deg}A<n$, $\mathrm{deg}B<m$; its determinant is the resultant.

For an original parameter example, choose $f(x)=x^2-1$ and $g_t(x)=x-t$. ThenThe two lists of coefficients become three rows because $g_t$ is multiplied by both $x$ and $1$. Thus the common-root parameter values are $t=-1$ and $t=1$. At $t=1/2$, the resultant is $-3/4$, so the two polynomials have no common root.

One can verify the certificate without any determinant expansion:For $t\ne \pm 1$, division by $t^2-1$ gives a Bézout's identity for polynomials. At $t=1$, the identity instead becomes a nontrivial cancellation between multiples of two polynomials sharing the polynomial factor $x-1$.

The general proof needs only a short linear algebra outline. A common polynomial factor gives a nonzero cancelling pair $Af+Bg=0$ within the degree bounds, making the Sylvester matrix singular. Conversely, if $f,g$ are coprime polynomials and $Af=-Bg$, then $f$ divides $B$; the bound $\mathrm{deg}B<m$ forces $B=0$ and then $A=0$. Thus the linear map has zero kernel of a linear map exactly in the coprime case. Over an algebraic closure, a nonconstant common polynomial factor is equivalent to a common root.

In 1840, James Joseph Sylvester described forming the shifted coefficient rows and taking a signed sum of products selected by permutations. In modern notation, this is the Sylvester matrix and its determinant. The construction turns the nonlinear common-root of a polynomial question into linear algebra. His paper's word derivatives concerns elimination, rather than the formal derivative used below.^[ref]