Grothendieck Groups and Graded Cartan Pairings — Examples

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] $\mathrm{\square }$

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.

By [F1], $A=\mathbb{C}$ is a field containing the embedded copy of $\mathbb{R}$, and every element of $A$ is uniquely $a+bi$ with $a,b\in \mathbb{R}$. By [F3] every nonzero element of the field $A$ is a unit with two-sided inverse, so $A$ is a division ring; by [F4], with the embedding of $\mathbb{R}$ in the commutative field $A$, it is a unital associative $\mathbb{R}$-algebra whose scalar action is restriction of complex multiplication. By [F2], $1,i$ is an $\mathbb{R}$-basis, so ${\dim }_{\mathbb{R}}A=2$.

Let $N\le S=A$ be a submodule of the regular module [F6] and suppose $0\mathrm{\ne }x\in N$. For $y\in A$ the element $y{x}^{-1}$ lies in $A$, so closure under scalars [F6] gives $y=(y{x}^{-1})x\in N$. Hence $N=A$, and the only submodules of $S$ are $0$ and $S$. Since $S\mathrm{\ne }0$, [F7] makes $S$ simple.

Let $T$ be a simple left $A$-module [F7] and choose $0\mathrm{\ne }t\in T$. The map $\phi :A\to T$, $\phi (x)=xt$, is $A$-linear: $\phi (x+y)=(x+y)t=xt+yt$ and $\phi (ax)=(ax)t=a(xt)$ for $a,x,y\in A$, by the module axioms [F5], which is exactly the two clauses of [F8]. If $\phi (x)=0$ with $x\mathrm{\ne }0$, then $t=1\cdot t=(x^{-1}x)t=x^{-1}(xt)=0$ by [F5], a contradiction; hence $\ker \phi =0$ and $\phi$ is injective. Its image is a submodule of $T$ containing $t\mathrm{\ne }0$, so simplicity [F7] forces $\mathrm{im}\phi =T$. Therefore $\phi$ is an isomorphism of $A$-modules and $T\cong A=S$: up to isomorphism, $S$ is the only simple left $A$-module.

The regular module $P=A$ is projective in the sense of [F12]: if $q:E\to M$ is a surjective $A$-module homomorphism and $f:A\to M$ is $A$-linear, choose $e\in E$ with $q(e)=f(1)$; then $\tilde{f}(x)=xe$ defines an $A$-linear map $\tilde{f}:A\to E$ by [F5], and $q(\tilde{f}(x))=xq(e)=xf(1)=f(x)$ for all $x\in A$, using the $A$-linearity of $f$ [F8]. The identity map $A\to A$ is surjective with kernel $0$, which is superfluous: if $N\le A$ and $N+0=A$, then $N=A$. Hence the identity $P\to S$ is a projective cover of $S$ in the sense of [F13], with projective finite-dimensional source $P=A=S$ of $\mathbb{R}$-dimension $2$ by [F2].

Evaluation at $1$ is an $\mathbb{R}$-linear bijection $\mathrm{ev}:{\mathrm{Hom}}_A(P,S)\to A$, $\mathrm{ev}(f)=f(1)$; here ${\mathrm{Hom}}_A(P,S)$ is an $\mathbb{R}$-vector subspace of ${\mathcal{L}}_{\mathbb{R}}(A,A)$ by [F15], and $\mathrm{ev}$ is $\mathbb{R}$-linear because addition and scalar multiplication of homomorphisms are pointwise. It is injective: if $f(1)=0$, then $f(x)=xf(1)=0$ for every $x\in A$ by the second clause of [F8]. It is surjective: for $\lambda \in A$ the map $x\mapsto x\lambda$ is $A$-linear by [F5] and has value $\lambda$ at $1$. Hence ${\mathrm{Hom}}_A(P,S)\cong A=\mathbb{C}$ as $\mathbb{R}$-vector spaces, and [F16] with ${\dim }_{\mathbb{R}}\mathbb{C}=2$ from [F2] gives ${\dim }_{\mathbb{R}}{\mathrm{Hom}}_A(P,S)=2$.

The algebra $A$ is finite-dimensional over $k=\mathbb{R}$ by step 1.1, and by step 1.3 the single module $S=A$ represents all simple left $A$-modules; step 1.4 supplies a finite-dimensional projective cover $\mathrm{i}\mathrm{d}:P\to S$ of that representative. Applying [F14] to this data: the split Grothendieck group $K_0(A)$ of finite-dimensional projective left $A$-modules [F18] is the free abelian group with basis $[P]$, the cover $P$ is indecomposable, and every finite-dimensional projective left $A$-module is a finite direct sum of copies of $P$ [F11]. If a nonzero finite-dimensional projective $X$ is isomorphic to ${⨁}_{i=1}^{m}P$ with $m\ge 0$, then $m\ge 1$ because $X\mathrm{\ne }0$; if $m\ge 2$, then $X=P\oplus {⨁}_{i=2}^{m}P$ exhibits $X$ as a direct sum of two nonzero submodules, contradicting indecomposability. Hence $m=1$ and $X\cong P$: up to isomorphism, $P$ is the only indecomposable finite-dimensional projective left $A$-module.

We verify the hypotheses of [F19] for the full subcategory $\mathcal{M}$ of finite-dimensional left $A$-modules [F18]. First, $\mathcal{M}$ is abelian: the category of all left $A$-modules is abelian [F10]; inside $\mathcal{M}$ the zero module and finite biproducts exist, a finite biproduct of finite-dimensional modules having finite-dimensional underlying space by [F26]; and kernels, images and cokernels of $A$-linear maps of finite-dimensional modules are again finite-dimensional — kernels and images are $\mathbb{R}$-linear subspaces of finite-dimensional spaces, hence finite-dimensional and of no larger dimension by [F25], while a cokernel $N\mathrm{/}\mathrm{im}f$ is the image of the quotient map, so [F9] and [F17] give ${\dim }_{\mathbb{R}}N={\dim }_{\mathbb{R}}(\mathrm{im}f)+{\dim }_{\mathbb{R}}(N\mathrm{/}\mathrm{im}f)$ — so the abelian-category clauses of [F10] hold in the full subcategory. Second, $\mathcal{M}$ is essentially small: for each $n\ge 0$ the module structures on the $\mathbb{R}$-vector space ${\mathbb{R}}^n$ are given by the $\mathbb{R}$-bilinear maps $A\times {\mathbb{R}}^n\to {\mathbb{R}}^n$ satisfying the axioms [F5], and these maps form a set; every finite-dimensional module is isomorphic to one of these models after choosing an $\mathbb{R}$-basis. Third, every object of $\mathcal{M}$ has finite length in the sense of [F23], by induction on ${\dim }_{\mathbb{R}}M$: for $M=0$ the empty chain is a composition series [F24]; for $M\mathrm{\ne }0$ choose a proper submodule $N\le M$ of maximal $\mathbb{R}$-dimension [F6] among the finite set of dimensions of proper submodules (the zero submodule is proper because $M\mathrm{\ne }0$). If $N<N^{\mathrm{\prime }}<M$, then $N^{\mathrm{\prime }}\mathrm{/}N\mathrm{\ne }0$ and the quotient map $N^{\mathrm{\prime }}\to N^{\mathrm{\prime }}\mathrm{/}N$ is $\mathbb{R}$-linear with kernel $N$, so [F17] gives ${\dim }_{\mathbb{R}}{N}^{\mathrm{\prime }}={\dim }_{\mathbb{R}}N+{\dim }_{\mathbb{R}}(N^{\mathrm{\prime }}\mathrm{/}N)>{\dim }_{\mathbb{R}}N$, contradicting maximality among proper submodules. If $M\mathrm{/}N$ had a proper nonzero submodule $U$, its inverse image $N^{\mathrm{\prime }}$ under the quotient map $q:M\to M\mathrm{/}N$ would be a submodule by [F6], [F8] and [F9]; surjectivity of $q$ and $\ker q=N$ would give $N<N^{\mathrm{\prime }}<M$, which was just excluded. Since $N<M$, the quotient $M\mathrm{/}N$ is nonzero and therefore simple [F7]. Also ${\dim }_{\mathbb{R}}N<{\dim }_{\mathbb{R}}M$ by [F25], so by induction $N$ has a finite composition series, and appending the top object $M$, whose quotient $M\mathrm{/}N$ is simple, gives a composition series of $M$ in the sense of [F24]. Since by step 1.3 the single module $S$ represents all simple classes, [F18] and [F19] give $G_0(A)\cong \mathbb{Z}[S]$, with the class $[S]$ of the regular module as the only basis element.

By steps 2.1 and 2.2, $[P]$ is the single basis class of $K_0(A)$ and $[S]$ the single basis class of $G_0(A)$, so the well-defined pairing of [F21] takes the value $⟨[P],[S]{⟩}_A=h(P,S)={\dim }_{\mathbb{R}}{\mathrm{Hom}}_A(P,S)=2$ of [F20] and step 1.5. By [F22] the pairing matrix in these bases has the single entry ${\dim }_k{\mathrm{End}}_A(S)={\dim }_{\mathbb{R}}{\mathrm{End}}_{\mathbb{C}}(\mathbb{C})$, and the evaluation argument of step 1.5 with $P=S=A$ identifies ${\mathrm{End}}_A(S)\cong A=\mathbb{C}$ as $\mathbb{R}$-vector spaces, of dimension $2$ by [F2]; so the entry is $2$. The bases $[P]$ and $[S]$ would be dual exactly if this single matrix entry were $1$, which it is not. In particular the hypothesis of [F22] that ${\mathrm{End}}_A(S_i)=k$ for every $i$ is false here: ${\mathrm{End}}_A(S)\cong \mathbb{C}$ has $\mathbb{R}$-dimension $2$, so it is not the scalar field $k=\mathbb{R}$, and the dual-basis conclusion fails for this input. Hence that hypothesis cannot be dropped from the theorem. [F1, F2, F16, F20, F21, F22, step 2.1, step 2.2, step 1.5, algebra] $\mathrm{\square }$