Coideals, ring ideals, Hopf ideals, and subbialgebras

The line $J=kx$ is a coideal: $1\otimes x$ lies in the image of $H\otimes J\to H\otimes H$, $x\otimes 1$ lies in the image of $J\otimes H\to H\otimes H$, and $\epsilon (x)=0$. Linearity gives both inclusions for every scalar multiple of $x$. It is not a ring ideal because $x\in J$ but $x\cdot x=x^2\mathrm{\notin }kx$, as the distinct monomials $x,x^2$ are linearly independent by [F8].

The set $(x^2)$ is a ring ideal by [F5], but it is not a coideal. Let $q:k[x]\to k[x]\mathrm{/}(x^2)$ be the quotient map. The coideal image condition would force $(q\otimes q)\mathrm{\Delta }(x^2)=0$; instead $\mathrm{\Delta }(x^2)=1\otimes x^2+2x\otimes x+x^2\otimes 1$, so $(q\otimes q)\mathrm{\Delta }(x^2)=2\bar{x}\otimes \bar{x}\mathrm{\ne }0$. By [F4] $\bar{x}\mathrm{\ne }0$, and by [F6] $\bar{x}\otimes \bar{x}$ is a nonzero tensor-basis element; $2\mathrm{\ne }0$ by the characteristic hypothesis.

The ideal $(x)$ is a Hopf ideal. It is a two-sided ideal, $\epsilon (x)=0$, and each term of $\mathrm{\Delta }(x)=1\otimes x+x\otimes 1$ lies in the sum of the images of $H\otimes (x)$ and $(x)\otimes H$. Since $\mathrm{\Delta }$ is an algebra map, for any $f\in k[x]$, $\mathrm{\Delta }(xf)=\mathrm{\Delta }(x)\mathrm{\Delta }(f)$ lies in the sum of those images: in each product of a term with $x$ in one factor and a term of $\mathrm{\Delta }(f)$, the factor containing $x$ remains in $(x)$. Also $S(xf)=S(f)S(x)=-S(f)x\in (x)$ by [F7], and $\epsilon (xf)=0$. Thus all conditions in [F3] hold.

In $k[\mathbb{Z}]$, the elements $t^n$ for $n\in \mathbb{Z}$ form a basis by [F2], so those with $n\ge 0$ form a basis of $K:=\mathrm{span}\{t^n:n\ge 0\}$. Restrict multiplication and unit to $K$, define ${\mathrm{\Delta }}_K$ linearly on this basis by ${\mathrm{\Delta }}_K(t^n)=t^n\otimes t^n$, and let ${\epsilon }_K$ be the restriction of $\epsilon$. For $m,n\ge 0$, $t^m{t}^n=t^{m+n}$, ${\mathrm{\Delta }}_K(t^{m+n})=(t^m\otimes t^m)(t^n\otimes t^n)$, and ${\epsilon }_K(t^{m+n})={\epsilon }_K(t^m){\epsilon }_K(t^n)=1$, with ${\mathrm{\Delta }}_K(1)=1\otimes 1$ and ${\epsilon }_K(1)=1$; bilinearity extends these equations to $K$, so ${\mathrm{\Delta }}_K$ and ${\epsilon }_K$ are unital algebra maps. Coassociativity and both counit identities hold on each basis element because $t^n$ is group-like, and hence hold on all of $K$ by linearity. The inclusion $K\hookrightarrow k[\mathbb{Z}]$ preserves multiplication, unit, coproduct, and counit on the basis and therefore is a bialgebra map. Thus $K$ is a subbialgebra by [F3], proved directly from its structure maps without using the AC-qualified tensor-subspace shorthand. But $S(t)=t^{-1}\mathrm{\notin }K$, since all $t^n$, $n\in \mathbb{Z}$, are linearly independent; hence $K$ is not a Hopf subalgebra.