Proof. We define $s$ pointwise. Let $x:X$ . Using a Lemma of KrausFootnote ⁵ and the $p_{ij}$ , we get a factorization

– which defines a unique value $s(x):P(x)$ .

Proof. After a variable substitution $X \mapsto X + a$ , we can assume that $f(0)$ is regular. Now let $g : A[X]$ be given with $fg = 0$ . Then in particular $f(0) g(0) = 0$ , so $g(0) = 0$ . By induction, all coefficients of $g$ vanish.

Proof. Choose generators

\begin{equation*} \frac {g_{i1}}{1},\dots, \frac {g_{ik_i}}{1} \end{equation*}

for each $I_i$ . These generators will still generate $I_i$ , if we multiply any of them with any power of the unit $\frac{f_i}{1}$ . Now

\begin{equation*} A_{f_if_j}\cdot I_i\subseteq A_{f_if_j}\cdot I_j \end{equation*}

means that for any $g_{ik}$ , we have a relation

\begin{equation*} (f_if_j)^l g_{ik}=\sum _{l}h_{l}g_{jl}\end{equation*}

for some power $l$ and coefficients $h_{l}:A$ . This means that $f_i^lg_{ik}$ is contained in $I_j$ . Multiplying $f_i^lg_{ik}$ with further powers of $f_i$ or multiplying $g_{jl}$ with powers of $f_j$ does not change that. So we can repeat this for all $i$ and $k$ to arrive at elements $\tilde{g_{ik}}:A$ , which generate an ideal $I\subseteq A$ with the desired properties.

Proof. A cokernel is a set-quotient by an effective relation, so the projection map is surjective. On the other hand, if $f$ is surjective and we are in the situation:

then we can construct a map $\varphi :B\to C$ as follows. For $x:B$ , we define the type of possible values $\varphi (x)$ in $C$ as

\begin{equation*} \sum _{z:C}\exists _{y:A}(f(y)=x) \wedge g(y)=z \end{equation*}

which is a proposition by algebraic calculation. By surjectivity of $f$ , this type is inhabited and therefore contractible. So we can define $\varphi (x)$ as its center of contraction.

Proof. By Lemma3.1.2, we have a natural equivalence

\begin{equation*} X\to \mathrm {Spec}\; (R^X) \end{equation*}

and by SQC, the natural map

\begin{equation*} A\to R^{\mathrm {Spec}\;A} \end{equation*}

is an equivalence. We therefore have a contravariant equivalence between the category of finitely presented $R$ -algebras and the category of affine schemes. In particular, $\mathrm{Spec}$ is an embedding.

Proof. If $\mathrm{Spec}\;A = \emptyset$ , then $A = R^{\mathrm{Spec}\;A} = R^\emptyset = 0$ by (SQC). If $A = 0$ , then there are no homomorphisms $A \to R$ since $1 \neq 0$ in $R$ by (Loc).

Proof. Part (a) is the special case $n = 1$ of (b). For (b), consider the $R$ -algebra $A \;\colon \!\!\!\equiv\; R/{}(x_1, \dots, x_n)$ . Then the set $\mathrm{Spec}\;A \equiv{\mathrm{Hom}}_{R\text{-}\mathrm{Alg}}(A, R)$ is a proposition (i.e.,, it has at most one element), and, more precisely, it is equivalent to the proposition $x = 0$ . By Proposition2.2.2, the negation of this proposition is equivalent to $A = 0$ and thus to $(x_1, \dots, x_n) = R$ . Using (Loc), this is the case if and only if one of the $x_i$ is invertible.

For (c), we instead consider the algebra $A \;\colon \!\!\!\equiv\; R_x \equiv R[\frac{1}{x}]$ . Here we have $A = 0$ if and only if $x$ is nilpotent, while $\mathrm{Spec}\;A$ is the proposition $\mathrm{inv}(x)$ . Thus, we can finish by Proposition2.2.2, together with part (a) to go from $\lnot \mathrm{inv}(x)$ to $\lnot \lnot (x = 0)$ .

Proof. Let $f : R[X]$ be given. Since our goal is a proposition, we can assume we have a bound $n$ on the degree of $f$ , so

\begin{equation*} f = \sum _{i = 0}^n c_i X^i \rlap {.} \end{equation*}

Since our goal is even double-negation stable, we can assume $c_n = 0 \lor c_n \neq 0$ and by induction $f = 0$ (in which case we are done) or $c_n \neq 0$ . If $n = 0$ we are done, setting $\alpha \;\colon \!\!\!\equiv\; c_0$ . Otherwise, $f$ is not invertible (using $0 \neq 1$ by (Loc)), so $R[X]/(f) \neq 0$ , which by (SQC) means that $\mathrm{Spec}(R[X]/(f)) = \{ x : R \mid f(x) = 0 \}$ is not empty. Using the double-negation stability of our goal again, we can assume $f(a) = 0$ for some $a : R$ and factor $f = (X - a_1) f_{n - 1}$ . By induction, we get $f = \alpha \cdot (X - a_1) \dots (X - a_n)$ . Finally, we decide each of the finitely many propositions $a_i = a_j$ , which we can assume is possible because our goal is still double-negation stable, to get the desired form $f = \alpha \cdot{(X - \widetilde{a}_1)}^{e_1} \dots{(X - \widetilde{a}_n)}^{e_n}$ with distinct $\widetilde{a}_i$ .

Proof. The natural map $X\to \mathrm{Spec}\;(R^X)$ is given by mapping $x:X$ to the evaluation homomorphism at $x$ . There merely is an $A$ such that $X=\mathrm{Spec}\;A$ . Applying $\mathrm{Spec}$ to the canonical map $A\to R^{\mathrm{Spec}\;A}$ , yields an equivalence by SQC. This is a (one sided) inverse to the map above. So we have $X=\mathrm{Spec}\;(R^X)$ .

Proof.

\begin{align*} D(f) &= \sum _{x : X} D(f)(x) = \sum _{x : \mathrm {Spec}\;A} \mathrm {inv}(f(x))\\& = \sum _{x : {\mathrm {Hom}}_{R\text {-}\mathrm {Alg}}(A, R)} \mathrm {inv}(x(f)) = {\mathrm {Hom}}_{R\text {-}\mathrm {Alg}}(A[f^{-1}], R) = \mathrm {Spec}\;A[f^{-1}] \end{align*}

Proof. It is enough to show the statement for $U=D(g)$ , $g:A_f$ . Then

\begin{equation*} g=\frac {h}{f^k}\rlap {.}\end{equation*}

Now $D(hf)$ is an affine-open in $X$ that coincides with $U$ :

Let $x:X$ , then $(hf)(x)$ is invertible, if and only if both $h(x)$ and $f(x)$ are invertible. The latter means $x:D(f)$ , so we can interpret $x$ as a homomorphism from $A_f$ to $R$ . Then $x:D(g)$ means $x(g)$ is invertible, which is equivalent to $x(h)$ being invertible, since $x(f)^k$ is invertible anyway.

Proof. Since $D(f)=\mathrm{Spec}\;A_f$ , by Proposition2.2.2, we know $D(f)=\emptyset$ , if and only if, $A_f=0$ . The latter is equivalent to $f$ being nilpotent.

Proof. Since $D(g_1, \dots, g_n) = \{\, x \in \mathrm{Spec}\;A \mid x \notin V(g_1, \dots, g_n) \,\}$ ,Footnote ⁷ the inclusion $D(f) \subseteq D(g_1, \dots, $ $g_n)$ can also be written as $D(f) \cap V(g_1, \dots, g_n) = \varnothing$ , that is, $\mathrm{Spec}((A/(g_1, \dots, g_n))$ $[f^{-1}]) = \varnothing$ . By (SQC) this means that the finitely presented $R$ -algebra $(A/(g_1, \dots, g_n))[f^{-1}]$ is zero. And this is the case if and only if $f$ is nilpotent in $A/(g_1, \dots, g_n)$ , that is, if $f \in \sqrt{(g_1, \dots, g_n)}$ , as stated.

Proof. By the universal property of the tensor product $A \otimes _R B$ .

Proof. The maps $f:X\to Z$ , $g:Y\to Z$ are induced by $R$ -algebra homomorphisms $f^*:A\to R$ and $g^*:B\to R$ . Let

\begin{equation*} (h,k,p) : \mathrm {Spec}\;A \times _{\mathrm {Spec} C} \mathrm {Spec} B \end{equation*}

with $p:h\circ f^*=k\circ g^*$ . This defines a $R$ -cocone on the diagram

Since $A\otimes _C B$ is a pushout in $R$ -algebras, there is a unique $R$ -algebra homomorphism $A\otimes _C B \to R$ corresponding to $(h,k,p)$ .

Proof. We recall $\mathrm{Spec}\;(A \otimes B) = \mathrm{Spec}\;A \times \mathrm{Spec}\;B$ from Lemma3.2.1 and calculate as follows.

\begin{align*}& A \otimes B = (\mathrm{Spec}\;(A \otimes B) \to R) = (\mathrm{Spec}\;A \times \mathrm{Spec}\;B \to R) = (\mathrm{Spec}\;A \to (\mathrm{Spec}\;B \to R)) = (\mathrm{Spec}\;A \to B) \\& \quad c \mapsto (\chi \mapsto \chi (c)) \mapsto ((\varphi, \psi ) \mapsto (\varphi \otimes \psi )(c)) \mapsto (\varphi \mapsto (\psi \mapsto (\varphi \otimes \psi )(c))) \mapsto (\varphi \mapsto (\varphi \otimes B)(c)) \end{align*}

The last step is induced by the identification $B = (\mathrm{Spec}\;B \to R),\, b \mapsto (\psi \mapsto \psi (b))$ , and we use the fact that $\psi \circ (\varphi \otimes B) = \varphi \otimes \psi$ .

Proof. The set of eventually vanishing sequences $\mathbb{N} \to R$ is in bijection with the set $R[X]$ of polynomials, by taking the entries of a sequence as the coefficients of a polynomial. So the family of sequences $s$ is equivalently a family of polynomials $s : \mathrm{Spec}\;A \to R[X]$ . Now we apply Lemma3.3.1 with $B = R[X]$ to see that such a family corresponds to a polynomial $p : A[X]$ . Note that for a point $x : \mathrm{Spec}\;A$ , the homomorphism

\begin{equation*} x \otimes R[X] : A[X] = A \otimes R[X] \to R \otimes R[X] = R[X] \end{equation*}

simply applies the homomorphism $x$ to every coefficient of a polynomial, so we have $(s(x))_n = x(p_n)$ . This concludes our argument because the coefficients of $p$ , just like any polynomial, form an eventually vanishing sequence.

Proof. Given a function $f : \mathrm{Spec}\;A \to \mathbb{N}$ , we construct the family $s : \mathrm{Spec}\;A \to (\mathbb{N} \to R)$ of eventually vanishing sequences given by

\begin{equation*} s(x)(n) \;\colon \!\!\!\equiv\; \begin {cases} \,1 &\text {if $n \lt f(x)$}\\ \,0 &\text {else} \rlap {.} \end {cases} \end{equation*}

Since $0 \neq 1 : R$ by Loc, we in fact have $s(x)(n) = 0$ if and only if $n \geq f(x)$ . Then the claim follows from Lemma3.3.2.

Proof. By Theorem3.3.3 we merely have a bound $N:\mathbb{N}$ for $f$ and we let $e_i$ be the indicator function of $f^{-1}(i)$ .

Proof. By Z-choice, there merely exists a cover $\mathrm{Spec}\;A = \bigcup _{i = 1}^n D(f_i)$ and functions $p_i : D(f_i) \to \mathbb{N}$ such that $p_i(x) \in P(x)$ for all $x : D(f_i)$ . By Theorem3.3.3, every $p_i : D(f_i) = \mathrm{Spec}\;A[f_i^{-1}] \to \mathbb{N}$ is merely bounded by some $N_i : \mathbb{N}$ , and then $\mathrm{max}(N_1, \dots, N_n) \in P(x)$ for all $x : \mathrm{Spec}\;A$ .

Proof. For a finitely generated ideal $I = (x_1, \dots, x_n)$ , the proposition $I = (0)$ is indeed a closed proposition, since it is equivalent to $x_1 = 0 \land \dots \land x_n = 0$ . It is also evident that we get all closed propositions in this way. What remains to show is that

\begin{equation*} I = (0) \Rightarrow J = (0) \qquad \text {iff}\qquad J \subseteq I \rlap . \end{equation*}

For this we use synthetic quasicoherence. Note that the set $\mathrm{Spec}\;R/I ={\mathrm{Hom}}_{R\text{-}\mathrm{Alg}}(R/I, R)$ is a proposition (has at most one element), namely it is equivalent to the proposition $I = (0)$ . Similarly, ${\mathrm{Hom}}_{R\text{-}\mathrm{Alg}}(R/J, R/I)$ is a proposition and equivalent to $J \subseteq I$ . But then our claim is just the equation

\begin{equation*} {\mathrm {Hom}}(\mathrm {Spec} R/I, \mathrm {Spec} R/J) = {\mathrm {Hom}}_{R\text {-}\mathrm {Alg}}(R/J, R/I) \end{equation*}

which holds by Proposition2.2.1, since $R/I$ and $R/J$ are finitely presented $R$ -algebras if $I$ and $J$ are finitely generated ideals.

Proof. The inclusion $V(f_1, \dots, f_n) \subseteq V(g_1, \dots, g_m)$ means a map $\mathrm{Spec}\;(A/(f_1, \dots, f_n)) \to \mathrm{Spec}\;(A/(g_1, \dots, g_m))$ over $\mathrm{Spec}\;A$ . By Proposition2.2.1, this is equivalent to a homomorphism $A/(g_1, \dots, g_m) \to A/(f_1, \dots, f_n)$ , which in turn means the stated inclusion of ideals.

Proof. By Z-choice and boundedness, there is a cover $D(f_1),\dots, D(f_l)$ , such that on each $D(f_i)$ , $C$ is the vanishing set of functions

\begin{equation*} g_1,\dots, g_n:D(f_i)\to R\rlap {.} \end{equation*}

By Lemma4.1.3, the ideals generated by these functions agree in $A_{f_i f_j}$ , so by Lemma1.3.8, there is a finitely generated ideal $I\subseteq A$ , such that $A_{f_i}\cdot I$ is $(g_1,\dots, g_n)$ and $C=\mathrm{Spec}\;A/I$ .

Proof. Indeed, by Proposition2.2.3, the proposition $\mathrm{inv}(f_1) \lor \dots \lor \mathrm{inv}(f_n)$ is the negation of ${f_1 = 0} \land \dots \land{f_n = 0}$ .

of Proposition 4.2.3. For unions, we can just append lists. For intersections, we note that invertibility of a product is equivalent to invertibility of both factors. Double-negation stability follows from Proposition4.2.2.

Proof. If $U(y)$ is an open proposition for all $y : Y$ , then $U(f(x))$ is an open proposition for all $x : X$ .

Proof. We show a proposition, so we can assume $\iota : X\to{\mathbb{A}}^n$ is a subtype. Then for $x,y:X$ , $x\neq y$ is equivalent to $\iota (x)\neq \iota (y)$ . But for $x,y:{\mathbb{A}}^n$ , $x\neq y$ is the open proposition that $x-y\neq 0$ .

Proof. By Proposition4.2.3, $\neg \neg (x=y)$ implies $\prod _{U:X\to{\mathrm{Open}}}U(x)\to U(y)$ . For the other implication, $\neg (x=y)$ is open by Lemma4.2.5, so we get a contradiction.

Proof. Let $L(x)$ be the type of finite lists of elements of $R$ , such that one of them being invertible is equivalent to $U(x)$ . By assumption, we know

\begin{equation*}\prod _{x:X}\lVert L(x) \rVert \rlap {.}\end{equation*}

So by Z-choice, we have $s_i:\prod _{x:D(f_i)}L(x)$ . We compose with the length function for lists to get functions $l_i:D(f_i)\to \mathbb{N}$ . By Theorem3.3.3, the $l_i$ are bounded. Since we are proving a proposition, we can assume we have actual bounds $b_i:\mathbb{N}$ . So we get functions $\tilde{s_i}:D(f_i)\to R^{b_i}$ , by append zeros to lists which are too short, hat is, $\widetilde{s}_i(x)$ is $s_i(x)$ with $b_i-l_i(x)$ zeros appended.

Then one of the entries of $\widetilde{s}_i(x)$ being invertible is still equivalent to $U(x)$ . So if we define $g_{ij}(x)\;\colon \!\!\!\equiv\; \pi _j(\widetilde{s}_i(x))$ , we have functions on $D(f_i)$ , such that

\begin{equation*} D(g_{i1},\dots, g_{ib_i})=U\cap D(f_i) \rlap {.} \end{equation*}

By Lemma3.1.7, this is enough to solve the problem on all of $X$ .

Proof. Let $X_i=\mathrm{Spec}\;A_i$ be a finite affine cover of $X$ . It is enough to show that the restriction $V_i$ of $V$ to $X_i$ is qc-open. $U_i\;\colon \!\!\!\equiv\; X_i\cap U$ is qc-open in $X_i$ , since $X_i$ is qc-open. By Theorem4.2.7, $U_i$ is affine-open in $X_i$ , so $U_i=D(f_1,\dots, f_n)$ . $V_i\cap D(f_j)$ is affine-open in $D(f_j)$ , so by Lemma3.1.7, $V_i\cap D(f_j)$ is affine-open in $X_i$ . This implies $V_i\cap D(f_j)$ is qc-open in $X_i$ and so is $V_i=\bigcup _{j}V_i\cap D(f_j)$ .

Proof.

1. (a) Apply Lemma4.2.8 to the point $\mathrm{Spec}\;R$ .

2. (b) Apply the above pointwise.

Proof. We can assume $P = (f_1 = \dots = f_n = 0)$ and $Q = (\mathrm{inv}(g_1) \lor \dots \lor \mathrm{inv}(g_m))$ . Then we have:

\begin{align*} (P \to Q) &= \quad \text{Proposition 2.2.3 for $g_1, \dots, g_m$}\\ (P \to \lnot (g_1 = \dots = g_m = 0)) &= \\ \lnot (f_1 = \dots = f_n = g_1 = \dots = g_m = 0) &= \quad \text{Proposition 2.2.3 for $f_1, \dots, f_n, g_1, \dots, g_m$}\\ (\mathrm{inv}(f_1) \lor \dots \lor \mathrm{inv}(f_n) \lor \mathrm{inv}(g_1) \lor \dots \lor \mathrm{inv}(g_m) &= \quad \text{Proposition 2.2.3 for $f_1, \dots, f_n$}\\ \lnot P \lor Q & \end{align*}

Proof. By Proposition4.2.3, open proposition are always double-negation stable, which settles one implication. For the implication

\begin{equation*} \left (\prod _{U:X\to {\mathrm {Open}}}U(x)\to U(y)\right ) \Rightarrow \neg \neg (x=y) \end{equation*}

we can assume that $x$ and $y$ are both inside an open affine $U$ and use that the statement holds for affine schemes by Lemma4.2.6.

Proof. As is shown for example here, such a pushout is always 0-truncated. Let $U_1,\dots, U_n$ be a cover of $X$ and $V_1,\dots, V_m$ be a cover of $Y$ . By Lemma4.2.8, $U_i\cap U$ is open in $Y$ , so we can use (large) pushout-recursion to construct a subtype $\tilde{U_i}$ , which is open in the pushout and restricts to $U_i$ on $X$ and $U_i\cap U$ on $Y$ . Symmetrically, we define $\tilde{V_i}$ and in total get an open finite cover of the pushout. The pieces of this new cover are equivalent to their counterparts in the covers of $X$ and $Y$ , so they are affine as well.

Proof. Using Theorem4.2.7.

Proof. Any open subtype of $X$ is also open in $A$ . So it is enough to show that any affine open $U_i$ of $X$ , has affine intersection with $A$ . But $U_i\cap A$ is closed in $U_i$ and therefore affine by Lemma4.1.4.

Proof. Any affine scheme is merely embedded into ${\mathbb{A}}^n$ for some $n:\mathbb{N}$ . The proposition $x=y$ for elements $x,y:{\mathbb{A}}^n$ is equivalent to $x-y=0$ , which is equivalent to all entries of this vector being zero. The latter is a closed proposition.

Proof. Let $x,y:X$ and $U\subseteq X$ be an affine open containing $x$ . Then, $U(y)\wedge x=y$ is equivalent to $x=y$ , so it is enough to show that $U(y)\wedge x=y$ is a scheme. As a open subscheme of the point, $U(y)$ is a scheme and $(x:U(y))\mapsto x=y$ defines a closed subtype by Lemma5.5.1. But this closed subtype is a scheme by Proposition5.4.3.

Proof. We start with an affine $X=\mathrm{Spec}\;A$ and $Y_x=\mathrm{Spec}\;B_x$ . Locally on $U_i = D(f_i)$ , for a Zariski-cover $f_1,\dots, f_l$ of $X$ , we have $B_x=\mathrm{Spec}\;R[X_1,\dots, X_{n_i}]/(g_{i,x,1},\dots, g_{i,x,m_i})$ with polynomials $g_{i,x,j}$ . In other words, $B_x$ is the closed subtype of ${\mathbb{A}}^{n_i}$ where the functions $g_{i,x,1},\dots, g_{i,x,m_i}$ vanish. By Lemma3.2.2, the product

\begin{equation*} V_i\;\colon \!\!\!\equiv\; U_i\times {\mathbb {A}}^{n_i}\end{equation*}

is affine. The type $(x:U_i)\times \mathrm{Spec}\;B_x\subseteq V_i$ is affine, since it is the zero set of the functions

\begin{equation*} ((x,y):V_i)\mapsto g_{i,x,j}(y) \end{equation*}

Furthermore, $W_i\;\colon \!\!\!\equiv\; (x:U_i)\times \mathrm{Spec}\;B_x$ is open in $(x:X)\times Y_x$ , since $W_i(x)$ is equivalent to $U_i(\pi _1(x))$ , which is an open proposition.

This settles the affine case. We will now assume that $X$ and all $Y_x$ are general schemes. We pass again to a cover of $X$ by affine open $U_1,\dots, U_n$ . We can choose the latter cover, such that for each $i$ and $x:U_i$ , the $Y_{\pi _1(x)}$ are covered by $l_i$ many open affine pieces $V_{i,x,1},\dots, V_{i,x,l_i}$ (by Theorem3.3.3). Then $W_{i,j}\;\colon \!\!\!\equiv\; (x:U_i)\times V_{i,x,j}$ is affine by what we established above. It is also open. To see this, let $(x,y):((x:X)\times Y_x)$ . We want to show that $(x,y)$ being in $W_{i,j}$ is an open proposition. We have to be a bit careful, since the open proposition $V_{i,x,j}$ is only defined, for $x:U_i$ . So the proposition we are after is $(z:U_i(x,y))\times V_{i,z,j}(y)$ . But this proposition is open by Lemma4.2.9.

Proof. By univalence, there is an equivalence

\begin{equation*} \left (\sum _{Y:\mathrm {Type}}(Y\to X)\right )\simeq (X\to \mathrm {Type})\rlap {.} \end{equation*}

From left to right, the equivalence is given by turning a $f:Y\to X$ into $x\mapsto \mathrm{fib}_f(x)$ , from right to left is given by taking the dependent sum. So we just have to note that both constructions preserve schemes. From left to right, this is Theorem5.6.4, from right to left, this is Theorem5.6.1.

Proof. Restriction of Corollary5.6.2.

Proof. The type $X\times _Z Y$ is given as the following iterated dependent sum:

\begin{equation*} \sum _{x:X}\sum _{y:Y}f(x)=g(y)\rlap {.}\end{equation*}

The innermost type, $f(x)=g(y)$ is the equality type in the scheme $Z$ and by Proposition5.5.2 a scheme. By applying Theorem5.6.1 twice, we prove that the iterated dependent sum is a scheme.

Proof. Let $L$ be a line. We merely have $x:V\setminus \{0\}$ such that

\begin{equation*} \prod _{y:V}\left (L (y) \Leftrightarrow \exists c:R.y=c\cdot x\right )\end{equation*}

We may replace the “ $\exists$ ” with a “ $\sum$ ,” since $c$ is uniquely determined for any $x,y$ . This means we can construct the map $\alpha \mapsto \alpha \cdot x:R\to L$ , and it is an equivalence.

Proof. Let $U_i([x_0:\dots :x_n])\;\colon \!\!\!\equiv\; (x_i\neq 0)$ . This is well defined since the proposition is invariant under multiplication by a unit. Furthermore, $U_i$ is open and the $U_i$ form a cover, by the generalized field property (Proposition2.2.3).

So what remains to be shown is that the $U_i$ are affine. We will show that $U_i={\mathbb{A}}^n$ . As an intermediate step, we have:

\begin{equation*} U_i=\{(x_0,\dots, x_n):{\mathbb {A}}^{n+1}\mid x_i=1\} \end{equation*}

by mapping $[x_0:\dots :x_n]$ with $x_i\neq 0$ to $\left (\frac{x_0}{x_i},\dots, \frac{x_n}{x_i}\right )$ and conversely, $(x_0,\dots, x_n)$ with $x_i=1$ to $[x_0:\dots :x_{i-1}:1:x_{i+1}:\dots :x_n]\in U_i$ .

But then, $\{(x_0,\dots, x_n):{\mathbb{A}}^{n+1}\mid x_i=1\}$ is equivalent to ${\mathbb{A}}^n$ by leaving out the $i$ th component, so the $U_i$ are affine.

Proof. Let $\varphi :\mathbb{P}^n\to \{\text{lines in ${\mathbb{A}}^{n+1}$}\}$ be given by mapping $[x_0:\dots :x_n]$ to $\langle (x_0,\dots, x_n)\rangle \subseteq{\mathbb{A}}^{n+1}$ , that is, the line generated by the vector $x\;\colon \!\!\!\equiv\; (x_0,\dots, x_n)$ . The map is well defined, since multiples of $x$ generate the same line.

Then $\varphi$ is surjective, since for any line $L\subseteq{\mathbb{A}}^{n+1}$ , there merely is a nonzero $x\in L$ , that we can take as a preimage. To conclude, we note that $\varphi$ is also an embedding. So let $\varphi ([x])=\varphi ([y])$ . Then, since $\langle x\rangle =\langle y\rangle$ , there is a $\lambda \in R^\times$ , such that $x=\lambda y$ , so $[x]=[y]$ .

Proof. $[x]$ and $[y]$ are equal, if and only if there merely is a $\lambda :R^\times$ , such that $\lambda x = y$ . By calculation, if there is such a $\lambda$ , we always have $x_iy_j=y_ix_j$ .

So let $x_iy_j=y_ix_j$ for all $i\neq j$ . Then, in particular, there are $i,j$ such that $x_i\neq 0$ and $y_j\neq 0$ . If $i=j$ , we define $\lambda \;\colon \!\!\!\equiv\; \frac{x_i}{y_i}$ . If $i\neq j$ , we have $x_iy_j=y_ix_j$ and therefore $y_i\neq 0$ and $x_j\neq 0$ , so we can also set $\lambda \;\colon \!\!\!\equiv\; \frac{x_i}{y_i}$ . By calculation, we have $\lambda y=x$ .

The dual statement follows by Proposition2.2.3.

Proof. The first two statements hold in general for inequality. For the third statement, let $x,y,z:\mathbb{P}^n$ . Note that if $x=z$ and $z=y$ , it follows that $x=y$ . So we have $\neg (x=y)\Rightarrow \neg (x=z\wedge z=y)$ . By Lemma6.1.7, $x=y$ and $x=z\wedge z=y$ are both equivalent to the statement that some vector with components in $R$ is zero, so we can replace negated equality, with existence of a nonzero element, or more explicitly, the following are equivalent:

\begin{align*} &\neg (x=y)\Rightarrow \neg (x=z\wedge z=y) \\ &\neg \left (\prod _{i\neq j}x_iy_j=y_ix_j\right ) \Rightarrow \neg \left (\prod _{i\neq j}x_iz_j=z_ix_j \wedge \prod _{i\neq j}y_iz_j=z_iy_j \right ) \\[3pt] & \left (\bigvee _{i\neq j}x_iy_j\neq y_ix_j\right ) \Rightarrow \left (\bigvee _{i\neq j}x_iz_j\neq z_ix_j \vee \bigvee _{i\neq j}z_iy_j\neq y_iz_j\right ) \\[3pt] & (x\neq y) \Rightarrow (x\neq z) \vee (z\neq y) \end{align*}

Proof. Consider the affine cover of $\mathbb{P}^1 = U_0 \cup U_1$ as in the proof of Theorem6.1.5. Both $U_0$ and $U_1$ are isomorphic to ${\mathbb{A}}^1$ and the intersection $U_0 \cap U_1$ is ${\mathbb{A}}^1 \setminus \{0\}$ , embedded in $U_0$ by $x \mapsto x$ and in $U_1$ by $x \mapsto \frac{1}{x}$ . So we have a pushout square as follows.

If we apply the functor $X \mapsto R^X$ to this diagram, we obtain a pullback square of $R$ algebras, and we can insert the known $R$ algebras for the affine schemes involved.

Here, the different variable names $X$ and $Y$ indicate the resulting homomorphisms. Now it is an algebraic computation, understanding the elements of $R[X, Y]/(1 - XY)$ as Laurent polynomials, to see that the pullback is the algebra $R$ , so we have $R^{\mathbb{P}^1} = R$ as desired.

Proof. What we want to prove is a proposition, so we can assume chosen $a, b \in{\mathbb{A}}^{n+1} \setminus \{0\}$ with $p = [a]$ , $q = [b]$ . Then we set

\begin{equation*} f([x, y]) \;\colon \!\!\!\equiv\; [xa + yb] \rlap {.}\end{equation*}

Let us check that $xa + yb \neq 0$ . By Proposition2.2.3, we have that $x$ or $y$ is invertible and both $a$ and $b$ have at least one invertible entry. If $xa = - yb$ , then it follows that $x$ and $y$ are both invertible and therefore $a$ and $b$ would be linearly equivalent, contradicting the assumption $p \neq q$ . Of course $f$ is also well-defined with respect to linear equivalence in the pair $(x, y)$ .

Proof. This is a special case of Lemma6.1.8, but we can also give a very direct proof: Let $p = [a]$ with $a \in{\mathbb{A}}^{n+1} \setminus \{0\}$ . By Proposition2.2.3, there is an $i \in \{0, \dots, n\}$ with $a_i \neq 0$ . If $i = 0$ then $p \neq [0 : 1 : 0 : \dots ]$ , if $i \geq 1$ then $p \neq [1 : 0 : 0 : \dots ]$ .

Proof. Let $f : \mathbb{P}^n \to R$ be given. For any two distinct points $p \neq q : \mathbb{P}^n$ , we can apply Lemma6.2.2 and (merely) find a map $\widetilde{f} : \mathbb{P}^1 \to R$ with $\widetilde{f}([0 : 1]) = f(p)$ and $\widetilde{f}([1 : 0]) = f(q)$ . Then we see $f(p) = f(q)$ by Lemma6.2.1. In particular, we have $f([1 : 0 : 0 : \dots ]) = f([0 : 1 : 0 : \dots ])$ . And then, by Lemma6.2.3, we get $f(p) = f([1 : 0 : 0 : \dots ])$ for every $p : \mathbb{P}^n$ .

Proof. For $U = D(f)$ , this follows from Lemma2.2.4 because $D(\alpha \cdot{(X - a_1)}^{e_1} \dots{(X - a_n)}^{e_n}) = D((X - a_1) \dots (X - a_n))$ . In general, we have $U = D(f_1) \cup \dots \cup D(f_n)$ by Theorem4.2.7, so we do not get (that $U = \emptyset$ or) a list of elements $a_1, \dots, a_n : R$ such that $U ={\mathbb{A}}^1 \setminus \{ a_1, \dots, a_n \}$ . Then we cannot get rid of any duplicates in the list.

Proof. By Lemma2.2.4, we can assume

\begin{align*} U \cup V &= D((X - a_1) \dots (X - a_k)) \rlap{,}\\ U &= D((X - a_1) \dots (X - a_k) (X - b_1) \dots (X - b_l)) \rlap{,}\\ V &= D((X - a_1) \dots (X - a_k) (X - c_1) \dots (X - c_m)) \rlap{,}\\ U \cap V &= D((X - a_1) \dots (X - a_k) (X - b_1) \dots (X - b_l) (X - c_1) \dots (X - c_m)) \rlap{,} \end{align*}

where all linear factors are distinct. Then every function $f : U \cap V \to R^\times$ can by (SQC), Lemma2.2.4 and comparing linear factors not be written in the form

\begin{equation*} f = \alpha \cdot {(X - a_1)}^{e_1} \dots {(X - a_k)}^{e_k} {(X - b_1)}^{e'_1} \dots {(X - b_l)}^{e'_l} {(X - c_1)}^{e''_1} \dots {(X - c_m)}^{e''_m} \end{equation*}

with $\alpha : R^\times$ , $e_i, e'_i, e''_i : \mathbb{Z}$ . Other linear factors cannot appear, since they do not represent invertible functions on $U \cap V$ . Now we can write $f = gh$ as desired, for example, with

\begin{align*} g &= \alpha \cdot{(X - a_1)}^{e_1} \dots{(X - a_k)}^{e_k}{(X - b_1)}^{e'_1} \dots{(X - b_l)}^{e'_l} \rlap{,}\\ h &={(X - c_1)}^{e''_1} \dots{(X - c_m)}^{e''_m} \rlap{.} \end{align*}

Proof. Let $T$ be an $R^\times$ -torsor on ${\mathbb{A}}^1$ , that is, for every $x :{\mathbb{A}}^1$ , $T_x$ is a set with a free and transitive $R^\times$ action and $\lVert T_x \rVert$ . By (Z-choice), we get a cover of ${\mathbb{A}}^1$ by open subsets ${\mathbb{A}}^1 = \bigcup _{i = 1}^n U_i$ and local sections $s_i : (x : U_i) \to T_x$ of the bundle $T$ . From this, we cannot construct a global section by induction on $n$ : Given any two local sections $s_i, s_j$ defined on $U_i, U_j$ , let $f : U_i \cap U_j \to R^\times$ be the unique function with $f(x)s_i(x) = s_j(x)$ for all $x : U_i \cap U_j$ . Then by Lemma6.3.4, we not find $g : U_i \to R^\times$ , $h : U_j \to R^\times$ such that the sections $x \mapsto g(x)s_i(x)$ and $x \mapsto{h(x)}^{-1}s_j(x)$ , defined on $U_i$ respectively $U_j$ , agree on $U_i \cap U_j$ . This yields a section $\widetilde{s} : (x : U_i \cup U_j) \to T_x$ by Lemma1.2.2 and we can replace $U$ and $V$ by $U \cup V$ in the cover. Finally, when we get to $n = 1$ , we have $U_1 ={\mathbb{A}}^1$ and the global section $s_1 : (x : X) \to T_x$ .

Proof. Given a line bundle $\mathcal{L}$ , we can construct an $R^\times$ torsor

\begin{equation*} x \mapsto \mathcal {L}_x \setminus \{0\} \rlap {.} \end{equation*}

Note that there is a well-defined $R^\times$ action on $M \setminus \{0\}$ for every $R$ module $M$ , and the action on $\mathcal{L}_x \setminus \{0\}$ is free and transitive and we have $\lVert \mathcal{L}_x \setminus \{0\} \rVert$ since we merely have $\mathcal{L}_x = R$ as $R$ modules. By Theorem6.3.5, there not is a global section of this torsor, so we have a section $s : (x :{\mathbb{A}}^1) \to \mathcal{L}_x$ with $s(x) \neq 0$ for all $x :{\mathbb{A}}^1$ . But this means that the line bundle $\mathcal{L}$ is trivial, since we can build an identification $\mathcal{L}_x = R$ by sending $s(x)$ to $1$ .