The finite dual of the primitive polynomial Hopf algebra

Every finite-codimensional ideal $I⊴k[x]$ is nonzero, since [F7] gives the infinite linearly independent family $1,x,x^2,\dots$. Choose a nonzero element of $I$ of least degree and scale it to a monic polynomial $p$. Dividing any $f\in I$ by $p$, the remainder is in $I$ and has smaller degree, so it is zero; hence $I=(p)$. If $I=k[x]$, a functional vanishing on it is zero, which is the empty sum in the claimed direct sum.

The primitive coproduct sends a polynomial $p(x)$ to $p(X+Y)$ in $k[X,Y]$. Writing $X=a+u$, $Y=b+v$, the value of $({\delta }_{a,r}\ast {\delta }_{b,s})(p)$ is the coefficient of $u^r{v}^s$ in $p(a+b+u+v)$. Expanding $p(a+b+z)={\sum }_n{\delta }_{a+b,n}(p)z^n$, that coefficient is $\left(\genfrac{}{}{0px}{}{r+s}{r}\right){\delta }_{a+b,r+s}(p)$; hence ${\delta }_{a,r}\ast {\delta }_{b,s}=\left(\genfrac{}{}{0px}{}{r+s}{r}\right){\delta }_{a+b,r+s}$, with binomial coefficients interpreted in $k$.

For a proper cofinite ideal, $p$ has positive degree. Repeatedly applying [F1] and dividing by $x-a$ whenever $a$ is a root factors $p$, up to a nonzero scalar, as ${\prod }_{a\in E}(x-a{)}^{m_a}$ for a finite set $E\subset k$ of distinct roots and positive integers $m_a$. If $a\mathrm{\ne }b$, then $u=(x-a)\mathrm{/}(b-a)$ and $v=-(x-b)\mathrm{/}(b-a)$ satisfy $u+v=1$, with $u\in (x-a)$ and $v\in (x-b)$. Expanding $1=(u+v{)}^{m+n-1}$ shows $(x-a{)}^m+(x-b{)}^n=k[x]$, since every term contains either at least $m$ factors $u$ or at least $n$ factors $v$. Thus the powers for distinct roots are pairwise comaximal.

By [F3], $k[x]\mathrm{/}(p)$ is isomorphic to ${\prod }_{a\in E}k[x]\mathrm{/}((x-a{)}^{m_a})$. In each factor [F2] gives the basis $1,(x-a),\dots ,(x-a{)}^{m_a-1}$, whose dual basis consists of the coefficient functionals ${\delta }_{a,0},\dots ,{\delta }_{a,m_a-1}$. Pulling these functionals back to $k[x]$ expresses any $f$ vanishing on $(p)$ as a finite sum of ${\delta }_{a,r}$. Conversely, ${\delta }_{a,r}$ vanishes on the cofinite ideal $((x-a{)}^{r+1})$, so it is in $A^{\circ }$. The local dual bases in the CRT product are independent, so the sum over distinct $a$ is direct.

For $f,g\in k[x]$, expand $f={\sum }_i{\delta }_{a,i}(f)(x-a{)}^i$ and $g={\sum }_j{\delta }_{a,j}(g)(x-a{)}^j$. The coefficient of $(x-a{)}^r$ in $fg$ is ${\sum }_{i+j=r}{\delta }_{a,i}(f){\delta }_{a,j}(g)$; by [F5], this gives ${\mathrm{\Delta }}^{\circ }{\delta }_{a,r}={\sum }_{i+j=r}{\delta }_{a,i}\otimes {\delta }_{a,j}$.

Since $1\in k[x]$ has coefficient $1$ at degree $0$ about every $a$ and zero at higher degrees, ${\epsilon }^{\circ }({\delta }_{a,r})={\delta }_{a,r}(1)$ is $1$ for $r=0$ and $0$ otherwise. The convolution unit is ${\epsilon }_A$, evaluation at $0$, namely ${\delta }_{0,0}$. Finally, for $p(x)$, the coefficient of $(x-a{)}^r$ in $p(-x)$ is $(-1{)}^r{\delta }_{-a,r}(p)$, so $S^{\circ }{\delta }_{a,r}=(-1{)}^r{\delta }_{-a,r}$.

Steps 1.1, 2.1, and 3.1 identify the finite dual as the stated direct sum, while steps 1.2, 4.1, and 4.2 compute the convolution product, coproduct, unit, counit, and antipode. Together with the finite-dual structure of [F5], these establish all the displayed formulas.