Hopf ideals, kernels and quotients of commutative Hopf algebras

(Coefficient criterion.) Let $X,H$ be $k$-vector spaces, let $h_1,\dots ,h_n\in H$ be linearly independent and let $x_1,\dots ,x_n\in X$ satisfy ${\sum }_i{x}_i\otimes h_i=0$ in $X{\otimes }_{k}H$. Then $x_1=\cdots =x_n=0$. Indeed, by [F3] the element ${\sum }_i{e}_{(x_i,h_i)}$ of the free module on $X\times H$ is a finite $k$-linear combination of finitely many bilinearity generators; let $X_0\subseteq X$ and $H_0\subseteq H$ be the spans of the initial $x_i,h_i$ together with all vectors occurring in that finite witness, so that $X_0,H_0$ are finite-dimensional and the same combination exhibits ${\sum }_i{x}_i\otimes h_i=0$ already in $X_0{\otimes }_k{H}_0$. By [F4] the independent list $h_1,\dots ,h_n$ extends to a finite basis $v_1,\dots ,v_N$ of $H_0$ with $v_i=h_i$ for $i\le n$, and $X_0$ has a finite basis $u_1,\dots ,u_M$; the universal property [F3] gives an isomorphism $X_0{\otimes }_k{H}_0\to k^{M\times N}$ with $u_a\otimes v_b\mapsto E_{ab}$ (both composites with the canonical maps are the identity on spanning sets). The image of ${\sum }_i{x}_i\otimes h_i$ is the matrix whose $i$-th column is the coordinate vector of $x_i$ for $i\le n$ and whose other columns vanish, so this matrix is zero by the assumed relation, and each $x_i$ is zero.

(Part (b).) Let $\mathfrak{a}\subseteq A$ be a Hopf ideal and let $q\text{\hspace{0.17em}⁣}:A\to A\mathrm{/}\mathfrak{a}$ be the quotient map. Since $\mathrm{\Delta }(\mathfrak{a})\subseteq \ker (A{\otimes }_{k}A\to A\mathrm{/}\mathfrak{a}{\otimes }_{k}A\mathrm{/}\mathfrak{a})$, there are unique $k$-algebra homomorphisms $\bar{\mathrm{\Delta }}\text{\hspace{0.17em}⁣}:A\mathrm{/}\mathfrak{a}\to A\mathrm{/}\mathfrak{a}{\otimes }_{k}A\mathrm{/}\mathfrak{a}$, $\bar{\epsilon }\text{\hspace{0.17em}⁣}:A\mathrm{/}\mathfrak{a}\to k$ and $\bar{S}\text{\hspace{0.17em}⁣}:A\mathrm{/}\mathfrak{a}\to A\mathrm{/}\mathfrak{a}$ with $\bar{\mathrm{\Delta }}q=(q\otimes q)\mathrm{\Delta }$, $\bar{\epsilon }q=\epsilon$ and $\bar{S}q=qS$, by the universal property of the quotient ring [F2] and because $q\otimes q$, $\epsilon$ and $qS$ kill $\mathfrak{a}$. The three Hopf identities for $(\bar{\mathrm{\Delta }},\bar{\epsilon },\bar{S})$ hold because they hold for $(\mathrm{\Delta },\epsilon ,S)$ and $q,q\otimes q,q\otimes q\otimes q$ are surjective; this makes $A\mathrm{/}\mathfrak{a}$ a Hopf algebra with $q$ a morphism, and any Hopf structure with that property must satisfy the three displayed identities, so it is unique. If $g\text{\hspace{0.17em}⁣}:A\to C$ is a Hopf morphism with $\mathfrak{a}\subseteq \ker g$, then $g$ induces $\bar{g}\text{\hspace{0.17em}⁣}:A\mathrm{/}\mathfrak{a}\to C$ with $\bar{g}q=g$ by [F2]; since $q$ is surjective and $g$ preserves $\mathrm{\Delta },\epsilon ,S$, so does $\bar{g}$, and $\bar{g}$ is the unique such map.

(Kernel of $f\otimes f$.) One has $\ker (f\otimes f)=A\otimes K+K\otimes A$. The inclusion $\supseteq$ is [F5]. For $\subseteq$, write an element of $A{\otimes }_{k}A$ as $x={\sum }_{i=1}^n{a}_i\otimes b_i$ and suppose $(f\otimes f)(x)=0$. Choose a maximal linearly independent subfamily $(f(a_j){)}_{j\in J}$ of the finite list $(f(a_1),\dots ,f(a_n))$; by [F4] every remaining $f(a_i)$ is a finite linear combination ${\sum }_{j\in J}{c}_{ij}f(a_j)$ with $c_{ij}\in k$. Then ${\sum }_{j\in J}f(a_j)\otimes (f(b_j)+{\sum }_{i\mathrm{\notin }J}{c}_{ij}f(b_i))=0$ in $B{\otimes }_{k}B$, so step 1.1 gives $w_j:=b_j+{\sum }_{i\mathrm{\notin }J}{c}_{ij}{b}_i\in K$ for every $j\in J$. Moreover $a_i^{\mathrm{\prime }}:=a_i-{\sum }_{j\in J}{c}_{ij}{a}_j\in K$ for $i\mathrm{\notin }J$, and the identity $x={\sum }_{j\in J}{a}_j\otimes w_j+{\sum }_{i\mathrm{\notin }J}{a}_i^{\mathrm{\prime }}\otimes b_i$ shows $x\in A\otimes K+K\otimes A$.

(Part (a).) If $x\in K$, then $(f\otimes f){\mathrm{\Delta }}_A(x)={\mathrm{\Delta }}_{B}f(x)=0$, so ${\mathrm{\Delta }}_A(x)\in \ker (f\otimes f)=A\otimes K+K\otimes A$ by step 2.1; further ${\epsilon }_A(x)={\epsilon }_B(f(x))=0$ and $f(S_A(x))=S_B(f(x))=0$, so $S_A(x)\in K$. Hence $K$ is a Hopf ideal of $A$. The same finite-relation argument identifies $U{\otimes }_{k}H$ with its image in $X{\otimes }_{k}H$ for every inclusion $U\subseteq X$: a zero relation has a finite witness; in the resulting finite-dimensional spaces extend a basis of the span of the first factors in $U$ to a basis of the ambient first-factor space, and use the coordinate tensor matrices of step 1.1. Applying this in both factors makes $f(A)\otimes f(A)\to B\otimes B$ injective. For $y=f(x)\in f(A)$ one has ${\mathrm{\Delta }}_B(y)=(f\otimes f){\mathrm{\Delta }}_A(x)\in f(A)\otimes f(A)$, ${\epsilon }_B(y)={\epsilon }_A(x)$ and $S_B(y)=f(S_A(x))\in f(A)$, so $f(A)$ is a Hopf subalgebra of $B$ with the induced structure maps.

(Part (c).) By step 3.1 the image $f(A)$ is a Hopf subalgebra of $B$ and $K$ is a Hopf ideal, so by step 1.2 the quotient $A\mathrm{/}K$ is a Hopf algebra; the map $\bar{\iota }\text{\hspace{0.17em}⁣}:A\mathrm{/}K\to f(A)$, $a+K\mapsto f(a)$, given by the first isomorphism theorem [F2], is a $k$-algebra isomorphism preserving the three structure maps, since $f$ does and $q$ is surjective. Composing this isomorphism with the inclusion $f(A)\hookrightarrow B$ factors $f$ as a surjection followed by an injection of Hopf algebras; any such factorization is unique because the quotient map is an epimorphism and the inclusion is a monomorphism, which also forces the middle isomorphism to be unique.