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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.