Project:Sandbox

Algebra 3

Assignment 2

u5805609

Question 1.

(a) Let $$R$$ be a ring, $$T$$ be a subset of $$R[x_{1},\ldots,x_{n}]$$, and $$C$$ and $$D$$ are arbitrary $$R$$-algebras.

The functor $$Z_{T}:\text{\textbf{R-Alg}\rightarrow\textbf{Sets}}$$ can be defined on objects by $$Z_{T}(C)=\{y\in C^{n}|\bar{f}(y)=0,\forall f\in T\}$$, where $$\bar{f}(y)$$ is just the inclusion of $$f\in T$$ evaluated at $$y\in C^{n}$$.

On morphisms, if $$g\in\hom(C,D)$$, then $$Z_{T}(f):Z_{T}(C)\rightarrow Z_{T}(D),\,(y_{1},\ldots,y_{n})\mapsto(g(y_{1}),\ldots g(y_{n}))$$. Since $$f$$ is a polynomial in $$T$$, then it is the ’same’ in $$C$$ and $$D$$, that is $$f(\psi(y_{1}),\ldots,\psi(y_{n}))=\psi(f(y_{1},\ldots,y_{n}))=\psi(0)=0$$, so $$Z_{T}(f)$$ does map $$Z_{T}(C)\rightarrow Z_{T}(D)$$.

For $$\text{id}_{C}:C\rightarrow C$$, $$Z_{T}(\text{id}_{C})(y_{1},\ldots,y_{n})=(\text{id}_{C}(y_{1}),\ldots,\text{id}_{C}(y_{n}))=(y_{1},\ldots y_{n})$$, so $$Z_{T}(\text{id}_{C})$$ is the identity on $$Z_{T}(C)$$.

Suppose $$f:C\rightarrow D$$ and $$g:D\rightarrow E$$ are morphisms. $$Z_{T}(g\circ f)(y_{1},\ldots,y_{n})=(g(f(y_{1})),\ldots,g(f(y_{n})))$$ and $$(Z_{T}(g)\circ Z_{T}(f))(y_{1},\ldots y_{n})=Z_{T}(g)(Z_{T}(f)(y_{1},\ldots,y_{n}))=Z_{T}(g)(f(y_{1}),\ldots f(y_{n}))=(g(f(y_{1})),\ldots g(f(y_{n})))=Z_{T}(g\circ f)(y_{1},\ldots,y_{n})$$setsfegaefgaefgaeg

Therefore $$Z_{T}$$ is a functor.

(b) Let $$A$$ be a given $$R$$-algebra, and $$C$$ and $$D$$ be arbitrary $$R$$-algebras.

Then $$\text{Spec}_{R}(A):\text{\textbf{R-Alg}\ensuremath{\rightarrow\textbf{Sets}}}$$ on objects is $$\spec_{R}(A)(C)=\hom(A,C)$$.

On morphisms, if $$g\in\hom(C,D)$$, then $$\spec_{R}(A)(g):\hom(A,C)\rightarrow\hom(A,D)$$, $$f\mapsto g\circ f$$. Since compositions of homomorphisms are homomorphisms, then $$\spec(A)(g)$$ is actually in $$\hom(A,D)$$, so $$\spec(A)$$ does map $$\hom(A,C)$$ to $$\hom(A,D)$$.

For $$\text{id}_{C}:C\rightarrow C$$, $$\spec_{R}(A)(\id_{C}):f\mapsto\id_{C}\circ f=f$$, so $$\spec_{R}(A)(\id_{C})=\id_{Spec_{R}(A)(C)}$$. If $$f:C\rightarrow D$$ and $$g:D\rightarrow E$$ are morphisms, then $$\spec(A)(g\circ f):\psi\mapsto(g\circ f)\circ\psi=g\circ f\circ\psi$$, and $$(\spec(A)(g)\circ\spec(A)(f))(\psi)=\spec(A)(g)(\spec(A)(f)(\psi))=\spec(A)(g)(f\circ\psi)=g\circ(f\circ\psi)=g\circ f\circ\psi=\spec(A)(g\circ f)$$. Therefore $$\spec(A)$$ is a functor.

(c) Suppose $$A=R[x_{1},\ldots,x_{n}]/(T)$$. WTS $$\spec_{R}(A)$$ is isomorphic to $$Z_{T}$$.

$$\begin{array}{ccc} h & \mapsto & g\of h\\ \spec(A)(C) & \xrightarrow{\spec(A)(g)} & \spec(A)(D)\\ \downarrow\eta_{C} & & \downarrow\eta_{D}\\ Z_{T}(C) & \xrightarrow{Z_{T}(g)} & Z_{T}(D)\\ (y_{1},\ldots,y_{n}) & \mapsto & (g(y_{1}),\ldots,g(y_{n})) \end{array}$$

Define an isomorphism of functors $$\eta_{C}:\spec(A)(C)=\hom(A,C)\rightarrow Z_{T}(C)=\{y\in C^{n}|\bar{f}(y)=0,\forall f\in T\}$$ by $$\psi(:A\rightarrow C)\mapsto(\psi(x_{1}),\ldots,\psi(x_{n}))\in C^{n}$$, where $$x_{i}\in A$$ are the generators.

To show $$\eta$$ is an isomorphism of functors it is equivalent to show $$\eta_{C}$$ is isomorphic for every $$C$$.

Suppose that $$\psi$$ and $$\theta$$ have the same image, that is $$(\psi(x_{1}),\ldots,\psi(x_{n}))=(\theta(x_{1}),\ldots,\theta(x_{n}))$$. But $$R$$-algebra homomorphism are uniquely determined by where the generators are sent and $$x_{1},\ldots,x_{n}$$ are all the generators, so $$\psi=\theta$$, so $$\eta_{C}$$ is injective.

Suppose that $$(y_{1},\ldots,y_{n})\in Z_{T}(C)$$. Define $$\psi(x_{i})=y_{i}$$, $$\psi:A\rightarrow C$$. Then since $$\bar{f}((\psi(a_{1}),\ldots\psi(a_{n}))=0$$ for all $$a_{i}\in A$$ and $$f\in T$$, $$\psi$$ is a homomorphism. Then $$\eta_{C}$$ is an isomorophism.

To show $$\eta$$ is a morphism of functors, consider $$(Z_{T}(g)\of\eta_{C})(h)=Z_{T}(g)(\eta_{C}(h))=Z_{T}(g)((h(x_{1}),\ldots,h(x_{n}))=(g(h(x_{1})),\ldots g(h(x_{n}))$$, and $$(\eta_{D}\of\spec(A)(g))(h)=\eta_{D}(g\of h)=(g(h(x_{1})),\ldots,g(h(x_{n})))$$, so the diagram commutes, so $$\eta$$ is an isomorphism of functors. Then $$\spec_{R}(A)$$ and $$Z_{T}$$ are isomorphic.

Question 2.

(a)

Qt&lt;a&gt; := FunctionField( Rationals ); P&lt;x,y,z&gt; := PolynomialRing( Qt, 3); I := ideal&lt; P | x*y, z&gt;; IsRadical(I); J := ideal&lt; P | z-y&gt;; IsRadical(J); Gives output:

true true (b) Adding the following code, where $$I\cup J=$$”I meet J”,

IsRadical(I meet J); Generators(I meet J); gives output

true [    x*y - x*z,     y*z - z^2 ] (c) Adding the following code

IsRadical(I meet J); Generators(I meet J); gives output

false [    y*z - z^2,     x*y^2 - x*y*z ] (d) If $$J=(z-x-y)$$ then the code

Qt&lt;a&gt; := FunctionField( Rationals ); P&lt;x,y,z&gt; := PolynomialRing( Qt, 3); I := ideal&lt; P | x*y, z&gt;; IsRadical(I); J := ideal&lt; P | z-x-y&gt;; IsRadical(J);

IsRadical(I meet J); Generators(I meet J);

IsRadical(I*J); Generators(I*J); gives output, (but without comments in )

true (I radical) true (J radical) true (I union J radical) [    x^2*y + x*y^2 + y^2*z - y*z^2,     x*z + y*z - z^2 ] (Generators of I union J) true (IJ radical) [    x*z + y*z - z^2,     x^2*y + x*y^2 - x*y*z ] (Generators of IJ) The zero locus of $$I$$ (in $$\boldsymbol{Q}^{3}$$) is $$Z(I)=\{(x,y,z)|z=0,xy=0\}=\{x\text{ and }y\text{ axes}\}$$, and $$Z((z-y))=Z(J_{1})$$ is the plane $$z=y$$, and $$Z((z-y-x))=Z(J_{2})$$ is the plane $$z=x+y$$. $$Z(IJ)=Z(I\cap J)=Z(I)\cup Z(J)$$ did not rely on the ambient field being $$\boldsymbol{C}$$. Then in $$Z(IJ_{1})$$, the $$y$$ axis line is ’included twice’, that is it is contained in both $$Z(I)$$ and $$Z(J_{1}$$). In $$Z(IJ_{2})$$, the only subset ’included twice’ is the origin point.

Question 3.

(a) Let $$B=\{\bar{1}\}\cup\{\bar{x}^{m}|m\geq1\}\cup\{\bar{y}^{n}|n\geq1\}$$. WTS that $$B$$ is a $$\boldsymbol{C}$$-linear basis for $$A=\boldsymbol{C}[x,y]/(xy)$$. $$A$$ is a $$\boldsymbol{C}$$-algebra with the obvious inclusion map. $$B$$ is a basis if it spans $$A$$ and is independent.

Let $$\psi:\boldsymbol{C}[x,y]\twoheadrightarrow\boldsymbol{C}[x,y]/(xy)$$ be the quotient map, which is always surjective. So for an element $$\bar{f}\in A$$, there exists $$f\in\boldsymbol{C}[x,y]$$ such that $$\psi(f)=\bar{f}$$. $$f$$ can be expressed in $$\boldsymbol{C}[x,y]$$ as $$f=\sum_{i,j\geq0}^{n}c_{ij}x^{i}y^{j}$$, $$c_{ij}\in\boldsymbol{C}$$. Then $$\psi(f)=\psi(\sum_{i,j\geq0}^{n}c_{ij}x^{i}y^{j})=\sum_{i,j\geq0}^{n}c_{ij}\psi(x)^{i}\psi(y)^{i}=\sum_{i,j\geq0}^{n}c_{ij}\bar{x}^{i}\bar{y}^{j}$$. But in $$A$$ $$\bar{x}\bar{y}=0$$, so there are no terms $$\bar{x}^{i}\bar{y}^{j}$$ unless $$i$$ or $$j$$ is 0, i.e.$$\psi(f)=c_{00}+\sum_{i=1}^{n}d_{i}\bar{x}^{i}+\sum_{j=1}^{n}e_{i}\bar{y}^{j}$$. Then $$B$$ spans $$A$$.

Suppose $$c+\sum_{i=1}^{n}d_{i}\bar{x}^{i}+\sum_{j=1}^{n}e_{i}\bar{y}^{j}=0$$, $$c,d_{i},e_{i}\in\boldsymbol{C}$$. If $$c,d_{i},e_{i}=0$$ necessarily, then $$B$$ is independent. $$c+\sum_{i=1}^{n}d_{i}\bar{x}^{i}+\sum_{j=1}^{n}e_{i}\bar{y}^{j}=\psi(c+\sum_{i=1}^{n}d_{i}x^{i}+\sum_{j=1}^{n}e_{i}y{}^{j})=0$$, so $$c+\sum_{i=1}^{n}d_{i}x^{i}+\sum_{j=1}^{n}e_{i}y{}^{j}=pxy$$, $$p\in\boldsymbol{C}[x,y]$$. But there are no $$xy$$ terms, so $$p=0$$, so $$c+\sum_{i=1}^{n}d_{i}x^{i}+\sum_{j=1}^{n}e_{i}y{}^{j}=0$$, which implies $$c=d_{i}=e_{i}=0$$, since $$x^{i}$$ and $$y^{j}$$ are independent in $$\boldsymbol{C}[x,y]$$.

To multiply basis elements, for $$i,j\geq1$$, $$\bar{x}^{i}\bar{x}^{j}=\psi(x^{i})\psi(x^{j})=\bar{x}^{i+j}$$, $$\bar{y}^{i}\bar{y}^{j}=\bar{y}^{i+j}$$, and $$\bar{x}^{i}\bar{y}^{j}=\bar{x}\bar{y}\bar{x}^{i-1}\bar{y}^{j-1}=\overline{xy}\bar{x}^{i-1}\bar{y}^{j-1}=0$$.

(b) Let $$R$$ be a ring and let $$f\in R[x]$$ be a monic polynomial of degree $$d$$. Let $$B=\{\bar{1},\ldots,\bar{x}^{d-1}\}$$. Let $$\psi:R[x]\twoheadrightarrow R[x]/(f)$$ be the surjective quotient map.

Suppose $$\bar{g}\in R[x]/(f)$$. Then there exists $$g\in R[x]$$ such that $$\psi(g)=\bar{g}$$, $$g=\sum_{i}r_{i}x^{i}$$, $$r_{i}\in R$$. Since $$f$$ is monic, by the polynomial division algorithm $$g=qf+p$$, with $$\deg p<\deg f=d$$, so $$p=\sum_{i}^{d-1}s_{i}x^{i}$$. Then $$\bar{g}=\psi(qf+p)=\psi(q)\psi(f)+\psi(\sum_{i}^{d-1}s_{i}x^{i})=\psi(q)0+\sum_{i}^{d-1}s_{i}\bar{x}^{i}=\sum_{i}^{d-1}s_{i}\bar{x}^{i}$$, so $$B$$ spans $$R[x]/(f)$$.

To show $$B$$ is independent, suppose $$0=\sum_{i}^{d-1}a_{i}\bar{x}^{i}=\sum_{i}^{d-1}a_{i}\psi(x)^{i}=\psi(\sum_{i}^{d-1}a_{i}x^{i})$$. Then $$\sum_{i}^{d-1}a_{i}x^{i}=bf=b\sum_{i}^{d}c_{i}x^{i}$$. Since $$\{x^{n}|n\geq0\}$$ is a $$R$$-linear basis of $$R[x]$$, then for the coefficent of $$x^{d}$$, $$bc_{d}=b=0$$, since $$f$$ is monic. Then $$a_{i}=0$$, so $$B$$ is independent.

(c) Let $$A=\boldsymbol{C}[x,y]/(x^{2}+y^{2}-1)$$. But $$A=\boldsymbol{C}[x][y]/(y^{2}+(x^{2}-1))$$, so by b), since $$y^{2}+(x^{2}-1)$$ is a monic polynomial of degree 2 in $$y$$, then a $$\boldsymbol{C}[x]$$-linear basis for $$A$$ is $$\{1,\bar{y}\}$$, so $$a=c_{1}+c_{1}\bar{y}$$, with $$c\in\boldsymbol{C}[x]$$ for all $$a\in A$$. A basis for $$\boldsymbol{C}[x]$$ is $$\{x^{n}|n\geq0\}$$, so $$c=\sum_{i}b_{i}x^{i}$$, so for $$a\in A$$, $$a=\sum_{i}b_{i}\bar{x}^{i}+\sum_{i}d_{i}\bar{x}^{i}\bar{y}$$ with $$b,d\in\boldsymbol{C}$$. Then a $$\boldsymbol{C}$$-linear basis of $$A$$ is $$\{1,\bar{x}^{n},\bar{x}^{m}\bar{y}\}$$ for all $$n,m\geq0$$.

To multiply basis elements, $$\bar{x}^{n}\bar{x}^{m}=\bar{x}^{n+m}$$, $$\bar{x}^{n}(\bar{x}^{m}\bar{y})=\bar{x}^{n+m}\bar{y}$$, and $$(\bar{x}^{n}\bar{y})(\bar{x}^{m}\bar{y})=\bar{x}^{n+m}\bar{y}^{2}=\bar{x}^{n+m}(1-\bar{x}^{2})=\bar{x}^{n+m}-\bar{x}^{n+m+2}$$.

Question 4.

(a) Suppose $$\eta:F\rightarrow G$$, $$\epsilon:G\rightarrow H$$ are morphisms of functors, $$F,G,H:C\rightarrow D$$, where $$C$$ and $$D$$ are categories. Then

$$\begin{array}{ccc} F(X) & \xrightarrow{F(f)} & F(Y)\\ \downarrow\eta_{X} & & \downarrow\eta_{Y}\\ G(X) & \xrightarrow{G(f)} & G(Y) \end{array}$$

and the same diagram for $$\epsilon$$ commutes. So $$\eta_{Y}\circ F(f)=G(f)\circ\eta_{X}$$, and $$\epsilon_{Y}\circ G(f)=H(f)\circ\epsilon_{X}$$. Then $$\epsilon\circ\eta$$ is a morphism of functors if

$$\begin{array}{ccc} F(X) & \xrightarrow{F(f)} & F(Y)\\ \downarrow(\epsilon\circ\eta)_{X} & & \downarrow(\epsilon\circ\eta)_{Y}\\ H(X) & \xrightarrow{H(f)} & H(Y) \end{array}$$

commutes. The composition $$\epsilon\circ\eta$$ is defined by sending $$X$$ to the composite map $$\epsilon_{X}\circ\eta_{X}$$. Then $$H(f)\circ(\epsilon\circ\eta)_{X}=H(f)\circ\epsilon_{X}\circ\eta_{X}=\epsilon_{Y}\circ G(f)\circ\eta_{X}=\epsilon_{Y}\circ\eta_{Y}\circ F(f)=(\epsilon\circ\eta)_{Y}\circ F(f)$$, that is it the diagram commutes.

(b) Suppose $$F:C\rightarrow D$$ is a functor and $$\id_{F}:F\rightarrow F$$ is the identity morphism, so $$\id_{F,X}=\id_{F(X)}$$ is the identity map $$F(X)\rightarrow F(X)$$. $$\id$$ is a functor if

$$\begin{array}{ccc} F(X) & \xrightarrow{F(f)} & F(Y)\\ \downarrow\id_{X} & & \downarrow\id_{Y}\\ F(X) & \xrightarrow{F(f)} & F(Y) \end{array}$$

commutes. By the definition of a category, for every morphism $$f:X\rightarrow Y$$, $$f\circ\id_{Y}=\id_{X}\circ f$$. Then $$F(f)\circ\id_{Y}=F(f)\circ\id_{F(Y)}=\id_{F(X)}\circ F(f)=\id_{X}\circ F(f)$$, so the diagram commutes.

(c) Let $$X$$ and $$Y$$ be functors $$\mathcal{C}\rightarrow\mathcal{D}$$. WTS a morphism of functors $$f:X\rightarrow Y$$ is an isomorphism if and only if for all objects $$C\in\text{Ob}(\mathcal{C})$$, the morphism $$f_{C}:X(C)\rightarrow Y(C)$$ is an isomorhpism.

First, suppose $$f$$ is an isomorphism of functors. Then there exists $$g:Y\rightarrow X$$ such that $$g\circ f=\id_{X}$$ and $$f\circ g=\id_{Y}$$.

By definition, $$f_{C}:X(C)\rightarrow Y(C)$$ is isomorphic if there exists $$h:Y(C)\rightarrow X(C)$$ such that $$h\circ f_{C}=\id_{X(C)}$$ and $$f_{C}\circ h=\id_{Y(C)}$$.

By the definition of composition and the identity morphism from b), $$(g\circ f)_{C}=\id_{X,C}=\id_{X(C)}$$ and $$(g\circ f)_{C}=g_{C}\circ f_{C}$$, and $$(f\circ g)_{C}=\id_{Y,C}=\id_{Y(C)}$$ and $$(f\circ g)_{C}=f_{C}\circ g_{C}$$, so $$g_{C}$$ is the inverse of $$f_{C}$$, so $$f_{C}$$ is isomorphic.

Now, suppose that for all objects $$C\in\text{Ob}(\mathcal{C})$$ the morphism $$f_{C}:X(C)\rightarrow Y(C)$$ is an isomorphism. That is, there exists $$h_{C}:Y(X)\rightarrow X(C)$$ such that $$h_{C}\circ f_{C}=\id_{X(C)}$$ and $$f_{C}\circ h_{C}=\id_{Y(C)}$$, which must also be unique.

Define the morphism of functors $$h:Y\rightarrow X$$ with $$C\in\mathcal{C}$$ associated to $$h_{C}:Y(C)\rightarrow X(C)$$ where $$h_{C}$$ is the inverse of $$f_{C}$$.

Since $$\id_{Y,C}=\id_{Y(C)}$$ and $$\id_{X,C}=\id_{X(C)}$$, and $$(f\of h)_{C}=f_{C}\of h_{C}=\id_{Y,C}$$ and $$(h\of f)_{C}=h_{C}\of f_{C}=\id_{X,C}$$, then $$f\of h=\id_{Y}$$ and $$h\of f=\id_{X}$$, so $$f$$ is an isomorphism, by definition. But is $$h$$ a morphism of functors?

$$\begin{array}{ccc} X(C) & \xrightarrow{X(x)} & X(D)\\ \downarrow f_{C} & & \downarrow f_{D}\\ Y(C) & \xrightarrow{Y(y)} & Y(D) \end{array}$$

That is, $$f_{D}\circ X(x)=Y(y)\circ f_{C}$$. $$\begin{array}{ccc} Y(C) & \xrightarrow{Y(y)} & Y(D)\\ \downarrow h_{C} & & \downarrow h_{D}\\ X(C) & \xrightarrow{X(x)} & X(D) \end{array}$$

Then $$X(x)\of h_{C}=\id_{X(D)}\of X(x)\of h_{C}=h_{D}\of f_{D}\of X(x)\of h_{C}$$. Since $$f_{D}\circ X(x)=Y(y)\circ f_{C}$$, then $$X(x)\of h_{C}=h_{D}\of Y(y)\circ f_{C}\of h_{C}=h_{D}\of Y(y)\circ\id_{Y(C)}=h_{D}\of Y(y)$$, so $$h$$ is a morphism of functors.

Therefore $$f$$ is an isomorphism of functors.

Question 5.

Let $$X:\text{\textbf{Alg}}_{\boldsymbol{R}}\rightarrow\text{\textbf{Sets}}$$ be a functor and $$n$$ be fixed.

(a) $$X(C)=\{M\in\text{Mat}_{n\times n}(C)|M^{2}=0\}$$. If $$f:C\rightarrow D$$ then $$X(f):m_{ij}\mapsto f(m_{ij})$$, i.e. $$f$$ is applied to each entry of $$M$$. For each entry to be 0, $$(M^{2})_{ij}=\sum_{k=1}^{n}m_{ik}m_{kj}=0$$. Then $$(f(M)^{2})_{ij}=\sum_{k=1}^{n}f(m)_{ik}f(m)_{kj}=\sum_{k=1}^{n}f(m_{ik})f(m_{kj})=f(\sum_{k=1}^{n}m_{ik}m_{kj})=0$$, so $$X(f)$$ does map $$X(C)$$ to $$X(D)$$.

$$X(\id_{C}):m_{ij}\mapsto\id_{C}(m_{ij})=m_{ij}$$, so $$X(\id_{C})=\id_{X(C)}$$. If $$g:D\rightarrow E$$, then $$X(f\of g)(m_{ij})=f(g(m_{ij}))$$ and $$(X(f)\circ X(g))(m_{ij})=X(f)(X(g)(m_{ij}))=X(f)(g(m_{ij}))=f(g(m_{ij}))=X(f\of g)(m_{ij})$$,so $$X$$ is actually a functor.

A matrix $$M$$ has $$n^{2}$$ entries and the equations which must hold are that $$M^{2}=0$$, that is $$\sum_{k=1}^{n}m_{ik}m_{kj}=0$$ for all $$1\leq i,j\leq n$$. Then take $$T=(\sum_{k=1}^{n}m_{ik}m_{kj})$$ for all $$i,j$$. Then the algebra representing $$X$$ is $$A=R[m_{1,1},\ldots,m_{n,n}]/(\sum_{k=1}^{n}m_{ik}m_{kj})$$, and the universal object is $$u=\begin{bmatrix}\bar{m}_{1,1} & \ldots & \bar{m}_{1,n}\\ \vdots & & \vdots\\ \bar{m}_{n,1} & \ldots & \bar{m}_{n,n} \end{bmatrix}$$

Then any specialization of $$u$$ is uniquely determined by the map of generators, so is unique.

Then by 1c), $$\spec(A)=Z_{T}=X$$, so $$X$$ is an affine scheme.

(b) $$X(C)=\{M\in\text{Mat}_{n\times n}(C)|MN=NM\}$$, $$N\in\text{Mat}_{n\times n}(R)$$. $$X$$ acts the same way on morphisms as in (a). $$(MN)_{ij}=\sum_{k=1}^{n}M_{ik}N_{kj}$$ so $$(MN)_{ij}-(NM)_{ij}=\sum_{k=1}^{n}M_{ik}N_{kj}-\sum_{k=1}^{n}N_{ik}M_{kj}=\sum_{k=1}^{n}(M_{ik}N_{kj}-N_{ik}M_{kj})=0$$. Since these are polynomial equations, $$(f(M)N)_{ij}-(Nf(M))_{ij}=\sum_{k=1}^{n}f(M)_{ik}N_{kj}-\sum_{k=1}^{n}N_{ik}f(M)_{kj}=f\left(\sum_{k=1}^{n}(M_{ik}N_{kj}-N_{ik}M_{kj})\right)=f(0)=0$$, so $$X(f):X(C)\rightarrow X(D)$$. Then by the same reasoning as in a), $$X$$ is a functor.

Take $$T=(\sum_{k=1}^{n}(M_{ik}N_{kj}-N_{ik}M_{kj}))$$ for all $$1\leq i,j\leq n$$. Then the algebra representing $$X$$ is $$A=R[m_{1,1},\ldots,m_{n,n}]/(\sum_{k=1}^{n}(m_{ik}N_{kj}-N_{ik}m_{kj})$$, and the universal object is

$$u=\begin{bmatrix}\bar{m}_{1,1} & \ldots & \bar{m}_{1,n}\\ \vdots & & \vdots\\ \bar{m}_{n,1} & \ldots & \bar{m}_{n,n} \end{bmatrix}$$

Therefore $$X$$ is an affine scheme, as in a).

(c) $$X(C)=\{f(t)\in C[t]|f(7)=0,\,\deg f=n\}$$. If $$f:C\rightarrow D$$, then $$X(f):\sum_{i}^{n}c_{i}t^{i}\mapsto\sum_{i}^{n}f(c_{i})t^{i}\in D[t]$$. If $$f(7)=0$$, i.e. $$\sum_{i}^{n}c_{i}7^{i}=0$$, then $$X(f)(7)=\sum_{i}^{n}f(c_{i})7^{i}=f(\sum_{i}^{n}c_{i}7^{i})=0$$, so $$X(f):X(C)\rightarrow X(D)$$.

$$X(\id_{C}):\sum_{i}^{n}c_{i}t^{i}\mapsto\sum_{i}^{n}\id(c_{i})t^{i}=\sum_{i}^{n}c_{i}t^{i}$$, so $$X(\id_{C})=\id_{X(C)}.$$F OF G

Take $$T=(\sum_{i}^{n}y_{i}7^{i})$$. Then The algebra representing $$X$$ is $$A[t]$$, $$A=R[y_{1},\ldots,y_{n}]/(\sum_{i}^{n}y_{i}7^{i})$$, and the universal object is

$$u(t)=\sum_{i}^{n}\bar{y}_{i}t^{i}\in A[t]$$

Then $$X$$ is an affine scheme.

(d) $$X(C)=\{x\in C|x\neq0\}$$. If $$f:C\rightarrow D$$ then the obvious way of trying to define $$X$$ is $$X(f)=f|_{X(C)}$$, i.e. by just restricting the domain to exclude 0. But for some $$R$$-algebra homomorphism, $$f(x)=0$$ for non-trivial $$x$$. For example consider the inclusion map $$f:R[x]\rightarrow R[x]/(x^{2})$$. Then $$x^{2}\neq0\in R[x]$$, but $$f(x^{2})=0$$, so there are elements of $$X(R[x])$$ that do not map to $$X(R[x]/(x^{2}))$$, so $$X$$ is not a functor.

(e) $$X(C)=C^{*}$$ where $$C^{*}$$ is the set of invertible elements of $$C$$. If $$f:C\rightarrow D$$, then $$X(f)=f|_{C^{*}}$$, i.e. by restricting the domain to $$C^{*}$$. Then if $$c\in C^{*}$$, $$X(f)(c)X(f)(c^{-1})=X(f)(cc^{-1})=X(f)(1)=1=X(f)(c^{-1})X(f)(c)$$, so $$f$$ does map $$C^{*}$$ to $$D^{*}$$.

$$X(\id_{C})=\id_{C}|_{C^{*}}=\id_{C^{*}}$$. If $$g:D\rightarrow E$$, $$X(g\of f)=g\of f|_{C^{*}}$$. $$(X(g)\of X(f))(a)=X(g)(f(a))$$, which is invertible, so $$g(f(a))=g\of f|_{C^{*}}$$, so $$X$$ is a functor.

Take $$T=(xy-1)$$. Then a representing algebra for $$X$$ is $$A=R[x,y]/(xy-1)$$, with $$u=\bar{x}$$. Then $$\psi(\bar{x})\psi(\bar{y})=\psi(1)=1$$. Then since a homomorphism is determined why where its generators are sent to, $$\bar{x}\mapsto a$$ and $$\bar{y}\mapsto a^{-1}$$ uniquely determine a homomorphism $$A\rightarrow C$$.

Therefore $$X$$ is an affine scheme.

(f) $$X(C)=\{g(t)\in Ct|g(t)^{2}=1+t\}$$. If $$\psi:C\rightarrow D$$ is a homomorphism, then $$X(\psi):\sum a_{n}t^{n}\mapsto\sum\psi(a_{n})t^{n}$$.

If $$g(t)=\sum_{n\geq0}a_{n}t^{n}$$, then $$(g(t))^{2}=a_{0}^{2}+2a_{0}a_{1}t+\sum_{n\geq2}b_{n}t^{n}=1+t$$, so $$a_{0}^{2}=1$$, $$2a_{0}a_{1}=1$$, and $$b_{n}=0$$. Note that each $$b_{n}$$ is a finite polynomial, since $$b_{n}$$ is a polynomial only in terms of $$a_{i}$$ for $$i\leq n$$, since $$a_{n+1}t^{n+1}$$ multiplied by anything will not contribute to the coefficient of $$t^{n}$$.

Then $$2\psi(a_{0})\psi(a_{1})-1=\psi(2a_{0}a_{1}-1)=0$$, and $$\psi(a_{1})^{2}-1=\psi(a_{1}^{2}-1)=0$$, and $$\psi(b_{i})=\psi(0)=0$$, so $$X(\psi)$$ does map $$X(C)$$ to $$X(D)$$. Also $$X(\id):\sum a_{n}t^{n}\mapsto\sum\id(a_{n})t^{n}=\sum a_{n}t^{n}$$, so $$C(\id_{C})=\id_{X(C)}$$. $$X(\psi\circ\phi)(\sum a_{n}t^{n})=\sum\psi(\phi(a_{n}))t^{n}$$, and $$(X(\psi)\circ X(\phi))(\sum a_{n}t^{n})=\sum\psi(\phi(a_{n}))t^{n}$$, so $$X(\psi)\circ X(\phi)=X(\psi\circ\phi)$$.

Then take $$T=(a_{0}^{2}-1,2a_{0}a_{1}-1,b_{i})$$, where $$b_{i}$$ is a polynomial in finitely many $$a_{i}$$, for all $$i\in\boldsymbol{N}$$. Then the algebra representing $$X$$ is $$A=R[a_{i}|i\in\boldsymbol{N}]/(b_{i})$$, and the universal object is $$u=\sum_{n\geq0}\bar{a}_{i}t^{i}$$

Then $$X$$ is an affine scheme.