The dual numbers are symmetric Frobenius and nonsemisimple over every field

Every class in $A$ has a unique form $a+bx$, so $(1,x)$ is a basis, $x^2=0$, and the multiplication is associative as a quotient of $k[x]$. The matrix of $\beta (u,v)=ϵ(uv)$ in this basis is the displayed matrix; its determinant is $-1$, which is nonzero even in characteristic two. Thus $\beta$ is symmetric and nondegenerate, so $ϵ$ is a Frobenius functional. The right-dual basis to $(1,x)$ for $\beta$ is $(x,1)$, so the inverse-pairing tensor is $x\otimes 1+1\otimes x$. Applying $\mathrm{\Delta }(a)=(a\otimes 1)(x\otimes 1+1\otimes x)$ gives the two formulas.

Direct computation verifies the coalgebra laws. The two counits on $\mathrm{\Delta }(1)$ give $ϵ(1)x+ϵ(x)1=1$ and $1ϵ(x)+xϵ(1)=1$; on $\mathrm{\Delta }(x)$ each gives $xϵ(x)=x$. Coassociativity on $1$ gives on both sides $1\otimes x\otimes x+x\otimes 1\otimes x+x\otimes x\otimes 1,$ and on $x$ gives $x\otimes x\otimes x$ on both sides.

To check both Frobenius identities it is enough by bilinearity to use the four ordered basis pairs. For $(1,1)$ both sides are $\mathrm{\Delta }(1)$; for $(1,x)$ both sides are $\mathrm{\Delta }(x)$; for $(x,1)$ both sides are $x\otimes x$; for $(x,x)$ both sides are $0$, since $x^2=0$. These are exactly $\mathrm{\Delta }(ab)=(a\otimes 1)\mathrm{\Delta }(b)=\mathrm{\Delta }(a)(1\otimes b)$, the two typed identities after inserting the vector-space associator. Thus $(A,\mu ,\eta ,\mathrm{\Delta },ϵ)$ is a symmetric Frobenius object.

Let $J=kx$. It is a nonzero proper left submodule of $A$, since $x\cdot x=0$ and scalars preserve $J$. If $L\subseteq A$ is any simple left submodule, multiplication by $x$ maps $L$ into $L$. Its image $xL$ is a submodule, so simplicity gives $xL=0$ or $xL=L$. The latter would imply $L=xL=x^{2}L=0$, contradicting simplicity; hence $xL=0$. For $a+bx\in L$, this implies $0=x(a+bx)=ax$, so $a=0$ and $L\subseteq J$. Therefore every simple submodule of $A$ is contained in the one-dimensional space $J$; their sum cannot equal the two-dimensional module $A$. By [F3], $A$ is not semisimple as a left module over itself.