The dual numbers have Cartan map multiplication by two

Write a polynomial as $p(x)={\sum }_{n\ge 0}{a}_n{x}^n$, with only finitely many nonzero coefficients, and let $I=x^{2}k[x]$. Multiplication by $x^2$ shifts coefficients two places, so elements of $I$ have zero constant and linear coefficients; conversely, any polynomial with those two coefficients zero is in $I$. Sums and differences remain multiples of $x^2$, and multiplying by any polynomial on either side again gives a multiple of $x^2$; thus $I$ is a two-sided ideal. Modulo $I$ every polynomial has the representative $a_0+a_{1}x$, since $p(x)-(a_0+a_{1}x)=x^2{\sum }_{n\ge 2}{a}_n{x}^{n-2}$, and that representative is unique. By [F5]–[F9], $A=k[x]\mathrm{/}I$ is the quotient $k$-algebra with this multiplication. Thus $1,\epsilon$ form a $k$-basis, ${\dim }_{k}A=2$, and ${\epsilon }^2=0$.

The category $A\text{-}\mathbf{M}\mathbf{o}\mathbf{d}$ is abelian by [F10]. Its full subcategory ${\mathcal{M}}_{\mathrm{f}\mathrm{d}}(A)$ of finite-dimensional modules is closed under kernels and cokernels, since kernels are subspaces and cokernels are quotients of finite-dimensional vector spaces. Finite biproducts are finite-dimensional, and the coimage-to-image isomorphism remains in this full subcategory; hence ${\mathcal{M}}_{\mathrm{f}\mathrm{d}}(A)$ is abelian. It is essentially small: on $k^n$, an unital $A$-action is determined by a matrix $T\in M_n(k)$ with $T^2=0$. For each $n$ these matrices form a set, and every $n$-dimensional module is isomorphic to one of these models after choosing a basis. Their union over $n\ge 0$ is a set, so the isomorphism classes form a set. This object-by-object argument makes no simultaneous choice of bases.

By step 1.1, every element of $A$ is $a+b\epsilon$. If $a\mathrm{\ne }0$, then $a+b\epsilon$ is a unit, with inverse $a^{-1}-a^{-2}b\epsilon$; if $a=0$, the element is nilpotent or zero and is not a unit. Hence the nonunits are exactly the proper ideal $(\epsilon )$, and every maximal left ideal is $(\epsilon )$, since a proper left ideal contains no unit. The quotient $A\mathrm{/}(\epsilon )\cong k$ is a field, so $(\epsilon )$ is maximal. Any simple left module $S$ is cyclic: for $0\mathrm{\ne }s\in S$, the map $A\to S$, $a\mapsto as$, is onto, and its kernel is a maximal left ideal. It follows that $S\cong A\mathrm{/}(\epsilon )\cong k$. Thus $k$ is the unique simple isomorphism class.

Every object $M$ of ${\mathcal{M}}_{\mathrm{f}\mathrm{d}}(A)$ has finite length. The zero module has the empty composition series. For nonzero $M$, choose a proper submodule $N$ of maximal $k$-dimension; it exists because $0$ is proper and possible dimensions lie in a finite set. The quotient $M\mathrm{/}N$ is nonzero, and a proper nonzero submodule of it would lift to a proper submodule of $M$ strictly containing $N$. Thus $M\mathrm{/}N$ is simple. Induction on ${\dim }_{k}M$ gives a finite composition series for $N$; appending $M\mathrm{/}N$ gives one for $M$.

Define the augmentation $\pi :A\twoheadrightarrow k$ by $\pi (a+b\epsilon )=a$; its kernel is $(\epsilon )$. The source $A$ is projective: given a surjection $q:E\twoheadrightarrow M$ and $f:A\to M$, choose $e\in E$ with $q(e)=f(1)$ and define $\tilde{f}(a)=ae$; then $q\tilde{f}(a)=af(1)=f(a)$. If a submodule $N\le A$ satisfies $N+(\epsilon )=A$, write $1=n+c\epsilon$ with $n\in N$. Then $n=1-c\epsilon$ is a unit, with inverse $1+c\epsilon$, so $N=A$. Thus the kernel is superfluous and [F13] makes $\pi$ a projective cover of the unique simple $k$.

Steps 2.1, 3.1 and 2.2 verify that ${\mathcal{M}}_{\mathrm{f}\mathrm{d}}(A)$ is an essentially small abelian category of finite-length objects with exactly one simple isomorphism class, represented by $k$. By [F2], its Grothendieck group is the free abelian group on $[k]$: $G_0(A)\cong \mathbb{Z}[k]$.

The algebra $A$ is finite-dimensional by step 1.1, and its only simple isomorphism class is $k$ by step 2.2. The cover in step 3.2 is $A\twoheadrightarrow k$. Applying [F3] to this one representative shows that $K_0(A)\cong \mathbb{Z}[A]$.

The ideal $(\epsilon )=k\epsilon$ is a submodule of $A$. The map $k\to k\epsilon$, $c\mapsto c\epsilon$, is an $A$-module isomorphism, because $\epsilon$ acts by zero on both modules. The quotient $A\mathrm{/}(\epsilon )$ is also isomorphic to $k$. Therefore $0\to k\epsilon \to A\to A\mathrm{/}(\epsilon )\to 0$ is short exact, and [F15] gives $[A]=[k\epsilon ]+[A\mathrm{/}(\epsilon )]=2[k]$ in $G_0(A)$. The Cartan map of [F1] sends the projective class $[A]$ to this same module class. By steps 4.1–4.2, this is multiplication by $2$ from $\mathbb{Z}[A]$ to $\mathbb{Z}[k]$; its image is $2\mathbb{Z}[k]$, which is proper. Hence the Cartan map is not an isomorphism. [F1, F5, F6, F7, F15, step 1.1, step 2.2, step 4.1, step 4.2, algebra] $\mathrm{\square }$