The graded dual numbers have Cartan polynomial 1+v²

Let $x$ denote the polynomial variable and $I=(x^2)=x^{2}k[x]$. It is an additive subgroup, and multiplication by any polynomial sends $x^{2}q$ to another multiple of $x^2$ on either side; thus it is a two-sided ideal by [F3, F4]. The coefficient of $x^2$ in $x^2$ is $1\mathrm{\ne }0$, so [F2] gives every $f\in k[x]$ a unique division remainder $a+bx$ modulo $I$. Hence each element of $A=k[x]\mathrm{/}I$ has a unique form $a+b\epsilon$, and $\{1,\epsilon \}$ is a $k$-basis with ${\epsilon }^2=0$. By [F5] and [F6] the coset multiplication is well defined, and [F7] makes $A$ a unital ring. The composite $k\to k[x]\to A$ is unital and multiplicative: the first map is the constant-polynomial homomorphism [F3], and the quotient map preserves sums, products, and identity by the coset operations in [F5] and [F7]. The quotient is commutative because $k[x]$ is commutative, so this map has central image. Thus [F8] makes $A$ a two-dimensional unital $k$-algebra. Define $A_0=k1$, $A_2=k\epsilon$, and $A_d=0$ for $d\mathrm{\notin }\{0,2\}$. The multiplication rules $1\cdot 1=1$, $1\cdot \epsilon =\epsilon \cdot 1=\epsilon$, and ${\epsilon }^2=0$ verify [F9]. The quotient $S=A\mathrm{/}(\epsilon )$ is one-dimensional over $k$ and concentrated in degree zero.

Let $T$ be any nonzero finite-dimensional graded-simple left $A$-module, with graded-simple as in [F16]. The submodule $\epsilon T$ is graded since $\epsilon$ is homogeneous. If $\epsilon T\mathrm{\ne }0$, simplicity gives $\epsilon T=T$, whence $T=\epsilon T={\epsilon }^{2}T=0$, a contradiction; thus $\epsilon T=0$. The action factors through $A\mathrm{/}(\epsilon )=k$, so each homogeneous component $T_d$ is a graded submodule. Simplicity forces exactly one component to be nonzero. That component has dimension one over $k$, since if its dimension exceeded one, the span of any nonzero vector would be a proper nonzero graded submodule. Hence $T\cong S\{r\}$ for its unique nonzero degree $r$. This also proves $S$ is graded-simple. Distinct $r$ give distinct supports, so there is exactly one graded-simple shift orbit, represented by $S$.

The regular graded module $A=A\{0\}$ is a finite direct sum of shifts of itself, so [F11] makes it projective in ${\mathrm{GrMod}}_0(A)$. It is finite-dimensional by step 1.1, hence is a finite graded projective.

The quotient map $\pi :A\twoheadrightarrow S$ is degree-zero and has kernel $k\epsilon$. If a graded submodule $N\le A$ satisfies $N+k\epsilon =A$, then taking degree-zero components gives $N_0=A_0=k1$, since $(k\epsilon {)}_0=0$. Thus $1\in N$, so $N=A$. By [F12], $\pi$ is a finite graded projective cover of $S$. To verify indecomposability directly, suppose $A=U\oplus V$ for nonzero graded submodules. Since $\pi$ is nonzero, one restriction, say $\pi {\mathrm{\mid }}_U$, is nonzero; its image is a nonzero graded submodule of the graded-simple $S$, hence is all of $S$. Therefore every element of $A$ differs from an element of $U$ by an element of $\ker \pi$, so $U+\ker \pi =A$. Superfluity forces $U=A$, contradicting $V\mathrm{\ne }0$. Thus $A$ is graded-indecomposable.

Apply [F17] to the unique simple shift orbit from step 2.1 and its cover $A\twoheadrightarrow S$ from step 3.1. It gives $[S]$ as an $R$-basis of $G_0^{\mathrm{g}\mathrm{r}}(A)$ and $[A]$ as an $R$-basis of $K_0^{\mathrm{g}\mathrm{r}}(A)$.

Identify $S$ with $k$ via the quotient map. Define $j:S\{2\}\to A$ by $j(\lambda )=\lambda \epsilon$. It is degree-zero because the degree-zero element of $S$ lies in degree two after shifting, and $\epsilon$ has degree two. For $a=\alpha +\beta \epsilon \in A$ and $\lambda \in k$, the quotient action on $S$ gives $j(a\lambda )=\alpha \lambda \epsilon$, while $aj(\lambda )=(\alpha +\beta \epsilon )\lambda \epsilon =\alpha \lambda \epsilon$ because ${\epsilon }^2=0$; hence $j$ is $A$-linear. It is injective since $\epsilon \mathrm{\ne }0$ by the unique normal form, and its image is $k\epsilon =\ker \pi$. Thus $0\to S\{2\}\stackrel{j}{\to }A\stackrel{\pi }{\to }S\to 0$ is exact. By [F14], $[A]=[S\{2\}]+[S]$ in $G_0^{\mathrm{g}\mathrm{r}}(A)$. Now [F10] and [F15] give $[S\{2\}]=v^2[S]$, while [F13] says the graded Cartan map sends $[A]$ to this same class in $G_0^{\mathrm{g}\mathrm{r}}(A)$. Therefore $c_A^{\mathrm{g}\mathrm{r}}([A])=(1+v^2)[S]$, as claimed.