AG review: elements of ring as function on affine scheme, nillradical functor and Gelfand transformation.

 

AG reviewGelfand Transformation.Diagonal Algebra Gelafnd Transformation and Fourier Transformation

AG review

In AG 4, we view $f\in A$$f\in A$ as function on $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$. Indeed, it define a ring homomorphism as follows:

Let $A$$A$ be a commutative ring, $k(x):=\mathrm{Frac}(A/{\mathfrak{p}}_x)$$k(x):=\mathrm{Frac(}A/\mathfrak p_x)$. Define $k:=\coprod _{x\in \mathrm{Spec}(A)}k(x)$$k:=\coprod_{x\in\mathrm{Spec}(A)}k(x)$.

We could consider ${\chi }_A:f\in A⟼\hat{f}:A\to k$$\chi_A :f\in A\longmapsto \hat{f}:A\to k$ via $\hat{f}({\mathfrak{p}}_x):=f\mathrm{mod}{\mathfrak{p}}_x$$\hat f(\mathfrak p_x):= f\mod \mathfrak p_x$.

$$\begin{array}{}\text{(1)}& \widehat{f+g}=\hat{f}+\hat{g},\widehat{fg}=\hat{f}\hat{g}\end{array}$$

The kernel of $\chi$$\chi$ is ${\mathfrak{n}}_A$$\mathfrak n_A$. Hence $\mathrm{Im}\chi \cong A/{\mathfrak{n}}_A$$\mathrm{Im}\chi\cong A/\mathfrak n_A$, if $A$$A$ is reduced, then $\chi$$\chi$ is injective, $A\cong \mathrm{Im}\chi$$A\cong\mathrm{Im}\chi$​​​.

Hence we could view $A/{\mathfrak{n}}_A$$A/\mathfrak n_A$ as a subring of the function ring on $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$​.

Notice that for a ring homomorphism $\varphi :A\to B$$\phi:A\to B$, if $a\in {\mathfrak{n}}_A$$a\in\mathfrak n_A$, then $f(a)\in {\mathfrak{n}}_B$$f(a)\in\mathfrak n_B$.

Hence we get a functor $\mathfrak{n}:\mathrm{CRing}\to \mathrm{ReCRing}$$\mathfrak n:\mathrm{CRing}\to\mathrm{ReCRing}$.

$$\begin{array}{}\text{(2)}& \begin{array}{c}\begin{array}{ccc}A& \stackrel{\varphi }{\to }& B\\ {\pi }_A\downarrow & & {\pi }_B\downarrow & \\ A/{\mathfrak{n}}_A& \stackrel{\phi }{\to }& B/{\mathfrak{n}}_B\end{array}\end{array}\end{array}$$

Proposition. $\mathfrak{n}$$\mathfrak n$ is the left adjoint of the inclusion functor $\iota :\mathrm{ReCRing}\to \mathrm{CRing}$$\iota:\mathrm{ReCRing}\to\mathrm{CRing}$.

Proof. We only need to prove that $\mathrm{Hom}(A/{\mathfrak{n}}_A,B)\cong \mathrm{Hom}(A,\iota (B))$$\mathrm{Hom}(A/\mathfrak n_A,B)\cong\mathrm{Hom}(A,\iota (B))$, here $B$$B$ is reduced.

It is true because for all $\varphi \in \mathrm{Hom}(A,\iota (B))$$\phi\in\mathrm{Hom}(A,\iota(B))$, ${\mathfrak{n}}_A\subseteq \ker \varphi$$\mathfrak n_A\subseteq\mathrm{ker}\phi$. $◻$$\square$

Corollary: The functor $\mathfrak{n}$$\mathfrak n$ preserve colimit, in particular, coproduct.

Rember that the coproduct is tensor product here, hence we have:

$$\begin{array}{}\text{(3)}& \mathfrak{n}(A{\otimes }_{\mathbb{Z}}B)\cong A/{\mathfrak{n}}_A{\otimes }_{\mathbb{Z}}B/{\mathfrak{n}}_B\end{array}$$

Remark.

Reader may relate it to the torsion module and torsion free module. That is, for an integral domain $R$$R$, consider

$$\begin{array}{}\text{(4)}& M⟼M_{tf}:=M/M_{tor},M_{tor}:=\{m\in M,\mathrm{\exists }r\in R,rm=0\}\end{array}$$

And we also have

$$\begin{array}{}\text{(5)}& \mathrm{Hom}(M_{tf},N)\cong \mathrm{Hom}(M,\iota (N))\end{array}$$

Remark. Readers may remind that $\mathrm{Spec}(A)\cong \mathrm{Spec}(A/{\mathfrak{n}}_A)\cong \mathrm{Spec}(\mathrm{Im}({\chi }_A))$$\mathrm{Spec}(A)\cong\mathrm{Spec}(A/\mathfrak n_A)\cong\mathrm{Spec}(\mathrm{Im}(\chi_A))$​.

The reason is, $f-g\in {\mathfrak{n}}_A$$f-g\in\mathfrak n_A$ iff ${\chi }_A(f)={\chi }_A(g)$$\chi_A(f)=\chi_A(g)$, implies V(f)=V(g).

Gelfand Transformation.

For a $K-$$K-$Alg $A$$A$, Consider ${\mathrm{Hom}}_K(A,K)$$\mathrm{Hom}_{K}(A,K)$ and $a\in A$$a\in A$, we could define ${\chi }_A:A\to K^{{\mathrm{Hom}}_K(A,K)}$$\chi_A:A\to K^{\mathrm{Hom}_K(A,K)}$

$$\begin{array}{}\text{(6)}& {\chi }_A(a)=\hat{a},\hat{a}(f)=f(a)\end{array}$$

Again

$$\begin{array}{}\text{(7)}& \widehat{a+b}=\hat{a}+\hat{b},\widehat{ab}=\hat{a}\hat{b}\end{array}$$

Diagonal Algebra

Definition. Let $A$$A$ be a finite $K-$$K-$algebra of degree $n$$n$. Set $X={\mathrm{Hom}}_{K-\mathrm{Alg}}(A,K)$$X=\mathrm{Hom}_{K-\mathrm{Alg}}(A,K)$. The following conditions are equivalent:

$(i)$$(i)$ A is isomorphic to an algebra of functions on set finite set, i.e. $K^{[n]}$$K^{[n]}$.

$(ii)$$(ii)$ The Gelfand transformation is an algebra isomorphism.

$(iii)$$(iii)$ $\mathrm{\forall }a\in A,a\ne 0,\mathrm{\exists }\zeta \in X,\zeta (a)\ne 0$$\forall a\in A,a\ne 0,\exists\zeta\in X,\zeta(a)\ne 0$.

$(iv)$$(iv)$ $\mathrm{Card}(X)=n$$\mathrm{Card}(X)=n$.

$(v)$$(v)$ $\mathrm{Card}(X)\ge n$$\mathrm{Card}(X)\ge n$​.

A algebra $A$$A$ satisfies one of the conditions will be called diagonal algebra.

Proof.

$(i)⟹(ii)$$(i)\implies (ii)$. Notice that $A\cong K^{[n]}⟹X\cong {\mathrm{Hom}}_{K-\mathrm{Alg}}(K^{[n]},K)$$A\cong K^{[n]}\implies X\cong\mathrm{Hom}_{K-\mathrm{Alg}}(K^{[n]},K)$. Then $X\cong [n]$$X\cong [n]$ since the $K-$$K-$algebra homomorphism has to be projection map. Let $g$$g$ be a $K-$$K-$algebra homomorphism to $K$$K$, then $g((1,1,1,...,1))=1$$g((1,1,1,...,1))=1$.

Hence $A\cong K^X$$A\cong K^X$. We only need to check that the Gelfand transformation is injective,

Let $a\ne b\in K^X$$a\ne b\in K^{X}$, then there exists a ${\pi }_j$$\pi_j$ such that $\hat{a}({\pi }_j)={\pi }_j(a)\ne \hat{b}({\pi }_j)={\pi }_j(b))$$\hat a(\pi_j)=\pi_j(a)\ne\hat b(\pi_j)=\pi_j(b))$​. Hence the Gelfand transformation is injective.

$(ii)⟹(iii)$$(ii)\implies (iii)$. Since the Gelfand transformation is an isomorphism, then $\hat{a}\ne 0$$\hat a\ne 0$​​.

$(iii)⟹(v)$$(iii)\implies (v)$. $(iii)$$(iii)$ shows that the kernel of the Gelfand transformation is $0$$0$, hence $A\to K^X$$A\to K^X$ is injective.

$(ii)⟹(iv)$$(ii)\implies (iv)$. Obviously.

$(iv)⟹(v)$$(iv)\implies (v)$​​. Obviously.

$(v)⟹(iv),(i)$$(v)\implies (iv),(i)$. Notice that $X\cong \mathrm{MaxSpec}(A)$$X\cong\mathrm{MaxSpec}(A)$. It has to be finite since by Chinese Remainder Theorem, $A\to A/\bigcap {\mathfrak{m}}_i\to \prod A/{\mathfrak{m}}_i$$A\to A/\bigcap{\mathfrak m_i}\to \prod A/\mathfrak m_i$ is surjective, hence $\dim \prod A/{\mathfrak{m}}_i=\mathrm{Card}(X)\le n$$\dim \prod A/\mathfrak m_i=\mathrm{Card}(X)\le n$.

By Chinese remainder theorem, $A$$A$ quotient the intercetion of all the maximal ideal:

$$\begin{array}{}\text{(8)}& A\to K^X\end{array}$$

is surjective. Hence $n\ge \mathrm{Card}(X)$$n\ge\mathrm{Card}(X)$. Hence it is an isomorphism. $◻$$\square$

Gelafnd Transformation and Fourier Transformation

Let $G=(\mathbb{Z}/n\mathbb{Z},+)$$G=(\mathbb Z/n\mathbb Z,+)$ and consider the group vector space $({\mathbb{C}}^G,\ast )$$(\mathbb C^G,*)$. Here we define $e_x\ast e_y=e_{x+y}$$e_x*e_y=e_{x+y}$ on the basis.

Hence

$$\begin{array}{}\text{(9)}& f\ast g=\sum _{n\in G}f(n)e_n\ast \sum _{m\in G}g(m)e_m=\sum _{x\in G}\sum _{m+n=x}f(n)g(m)e_x=\sum _{x\in G}\sum _{y\in G}f(y)g(x-y)e_x\end{array}$$

i.e.

$$\begin{array}{}\text{(10)}& {\displaystyle f\ast g(x)=\sum _{y\in G}f(y)g(y-x)}\end{array}$$

Now consider $X={\mathrm{Hom}}_{\mathbb{C}-\mathrm{Alg}}({\mathbb{C}}^G,\mathbb{C})$$X=\mathrm{Hom}_{\mathbb C-\mathrm{Alg}}(\mathbb C^G,\mathbb C)$​, We want to shows that $({\mathbb{C}}^G,\ast )$$(\mathbb C^G,*)$ is diagonal, hence the Gelfand Transformation will be an isomorphism between $({\mathbb{C}}^G,\ast )$$(\mathbb C^G,*)$ and $({\mathbb{C}}^X,\cdot )$$(\mathbb C^X,\cdot)$.

For all $\phi \in {\mathbb{C}}^G$$\varphi\in\mathbb C^G$, there exists a unique linear map $\zeta :{\mathbb{C}}^G\to \mathbb{C}$$\zeta: \mathbb C^G\to\mathbb C$ such that $\zeta (e_x)=\phi (x)$$\zeta(e_x)=\varphi(x)$.

$\zeta$$\zeta$ is an algebra homomorphism if and only if $\zeta (e_0)=1,\zeta (e_x\ast e_y)=\zeta (e_x)\zeta (e_y)=\phi (x+y)=\phi (x)\phi (y)$$\zeta(e_0)=1,\zeta(e_x*e_y)=\zeta(e_x)\zeta(e_y)=\varphi(x+y)=\varphi(x)\varphi(y)$.

Conversely, for all the algebra homomorphism $\zeta :{\mathbb{C}}^G\to \mathbb{C}$$\zeta:\mathbb C^G\to \mathbb C$, there exists a unique function $\phi \in {\mathbb{C}}^G$$\varphi\in\mathbb C^G$ such that $\zeta (e_x)=\phi (x)$$\zeta(e_x)=\varphi(x)$. The function $\phi$$\varphi$ is a group homomorphism since:

$$\begin{array}{}\text{(11)}& \phi (x+y)=\zeta (e_{x+y})=\zeta (e_x\ast e_y)=\zeta (e_x)\zeta (e_y)=\phi (x)\phi (y)\end{array}$$

Hence we have

$$\begin{array}{}\text{(12)}& X={\mathrm{Hom}}_{\mathbb{C}-\mathrm{Alg}}({\mathbb{C}}^G,\mathbb{C})\cong {\mathrm{Hom}}_{\mathrm{Grp}}(G,{\mathbb{C}}^{\ast })\cong [n]\end{array}$$

Let ${\theta }_k=e^{{\displaystyle \frac{2\pi ik}{n}}}$$\theta_k=e^{\dfrac{2\pi i k}{n}}$ be a $n-$$n-$th root of 1, then $x⟼{\theta }^x$$x\longmapsto\theta^x$ gives us a $n$$n$ distinct group homomorphism ${\phi }_k(x)=e^{{\displaystyle \frac{2\pi ikx}{n}}}$$\varphi_k(x)=e^{\dfrac{2\pi i kx}{n}}$ from $G$$G$ to $\mathbb{C}$$\mathbb C$. Hence ${\mathbb{C}}^G$$\mathbb C^G$ is diagnoal, the Gelfand Transformation $\chi :({\mathbb{C}}^G,\ast )\to ({\mathrm{Hom}}_{\mathrm{Set}}(X,\mathbb{C}),\cdot )$$\chi:(\mathbb C^G,*)\to (\mathrm{Hom}_{\mathrm{Set}}(X,\mathbb C),\cdot)$ is an isomorphism.

$$\begin{array}{}\text{(13)}& \hat{f}({\zeta }_k)={\zeta }_k(f)={\zeta }_k(\sum _{x\in G}f(x)e_x)=\sum _{x\in G}{\zeta }_k(f(x)){\zeta }_k(e_x)=\sum _{x\in G}f(x){\phi }_k(x)=\sum _{x\in G}f(x)e^{{\displaystyle \frac{2\pi ikx}{n}}}\end{array}$$