Proof. The proof follows mutatis mutandis as the one for schemes over rings.

Proof. To show that $(X,f,\{U_{ij}\},\{\psi_{ij}\}\}$ is a geometric vector bundle, it suffices to prove that $\psi_{ij}:f^{-1}(U_{ij}) \to \mathbb{A}^n_{U_{ij}}$ is an isomorphism. In fact, we may further assume that $U_{i}=\operatorname{Spec} R_i$ and $U_{ij}=\operatorname{Spec} (R_i)_{f_j}$ for some $f_j \in R_i$. Then, it reduces to show that for a free R_i-module M_i,

(8)\begin{equation} (\text{Sym}(M_i))_{f_j}\simeq \text{Sym}((M_i)_{f_j}), \end{equation}

which is true for modules over semirings.

Finally, it is clear that $(X,f,\{U_{ij}\},\{\psi_{ij}\}\}$ is isomorphic to $(X,f,\{U_i\},\{\psi_i\})$.

Proof. The proof follows mutatis mutandis as the one for schemes over rings.

Proof. The proof follows mutatis mutandis as the one for schemes over rings.

Proof. The proof follows mutatis mutandis as the one for schemes over rings.

Proof. The proof follows mutatis mutandis as the one for schemes over rings.

Proof. The proof follows mutatis mutandis as the one for schemes over rings.

Proof. The proof follows mutatis mutandis as the one for schemes over rings.

Proof. Let $y\in Y(A)$. Elements of E_y are morphisms $v: \operatorname{Spec} A\rightarrow E$ such that $\pi\circ v = y$. These are equivalent to pairs of morphisms $v: \operatorname{Spec} A\rightarrow E$ and $u: \operatorname{Spec} A\rightarrow \operatorname{Spec} A$ such that u is the identity map and $\pi\circ v = y \circ \mathrm{id}$. Without the condition that u is the identity, such pairs correspond to morphisms $\eta: \operatorname{Spec} A \rightarrow E\times_Y \operatorname{Spec} A$, so with this condition included, such pairs correspond to morphisms $\eta: \operatorname{Spec} A \rightarrow E\times_Y \operatorname{Spec} A$ whose composition with the projection $p: E\times_Y \operatorname{Spec} A \rightarrow \operatorname{Spec} A$ is the identity. These are the same as global sections of the sheaf associated with the pullback bundle $p: E\times_Y \operatorname{Spec} A \rightarrow \operatorname{Spec} A$. This establishes the first claim.

Let $v\in E_y$. We identify v with the map $s: \operatorname{Spec} A\rightarrow E \times_Y \operatorname{Spec} A$ obtained by applying the universal property of the pullback to v and $\mathrm{id}$. By applying $- \times_{\operatorname{Spec} A} \operatorname{Spec} B$ we obtain a map $t: \operatorname{Spec} B\rightarrow E \times_Y \operatorname{Spec} B$. So, we have the following diagram:

Since all squares are pullback squares, one can easily see that t is the global section corresponding to the element $\phi_*(v)=v\circ \phi^*$.

Proof. Let $U\subseteq X$ be an open subset. A section of E over U may be pulled back to a section of $E\times_X Y$ over $f^{-1}(U)$. Since $f_*\mathcal{G}$ sends U to the sections of $E\times_X Y$ over $f^{-1}(U)$, we have obtained a homomorphism $\mathcal{F}\rightarrow f_*\mathcal{G}$. By adjunction we obtain a homomorphism $f^*\mathcal{F}\rightarrow \mathcal{G}$.

To check this is an isomorphism, we may work locally (or even on stalks), and so we may assume E is the trivial bundle. Choose a basis $v_1,\ldots, v_n$ of sections of E, and equip $E\times_X Y$ with the pulled back basis $w_1, \ldots, w_n$. We may now identify $\mathcal{F}$ with $\mathcal{O}_X^n$ and $\mathcal{G}$ with $\mathcal{O}_Y^n$. The map $\mathcal{F}\rightarrow f_*\mathcal{G}$ we constructed above is the map $\mathcal{O}_X^n \rightarrow f_*\mathcal{O}_Y^n$ sending one basis to the other (i.e., the direct sum of many copies of the canonical map $\mathcal{O}_X\rightarrow f_*\mathcal{O}_Y$. The map $f^*\mathcal{F}\rightarrow \mathcal{G}$ is then the map $f^*\mathcal{O}_X^n\rightarrow \mathcal{O}_Y^n$ obtained as the direct sum of copies of the canonical isomorphism $f^*\mathcal{O}_X\rightarrow \mathcal{O}_Y$.

Proof. First, we consider what the functor of points looks like on objects via the first statement of proposition 3.12.

Let R be a semiring and A be an R-algebra. For any geometric vector bundle E on a scheme Y over R, the set E(A) is the disjoint union of E_y over all $ y \in Y(A)$. Also, for each $y \in Y(A)$, proposition 3.12 says that E_y is the set of global sections of the pullback of E along y.

Moreover, from lemma 3.15, we know that the sections of this pullback bundle are just global sections of the pullback sheaf. It follows that E_y is the set of global sections of $y^*\mathcal{F}$ where $\mathcal{F}$ is the sheaf of sections of E. By the definition of the functor of points of a locally free sheaf, we can write this as $E_y = \mathcal{F}_y$. Then taking the union over all $y \in Y(A)$ gives

\begin{equation*} E(A) = \mathcal{F}(A). \end{equation*}

The case of morphisms is followed by a similar argument.

Proof. The proof follows mutatis mutandis as the one for schemes over rings.

Proof. Let $\mathcal{H}$ be a sheaf on X. Then by the hom-tensor adjunction and the pullback-pushforward adjunction, we obtain

(10)\begin{align} &\mathrm{Hom}_{\mathcal{O}_X}(f^*\mathcal{F}\otimes_{\mathcal{O}_X} f^*\mathcal{G}, \mathcal{H}) \cong \mathrm{Hom}_{\mathcal{O}_X}(f^*\mathcal{F}, \underline{\mathrm{Hom}}_{\mathcal{O}_X}(f^*\mathcal{G}, \mathcal{H})) \cong \end{align}

(11)\begin{align} &\mathrm{Hom}_{\mathcal{O}_Y}(\mathcal{F}, f_*\underline{\mathrm{Hom}}_{\mathcal{O}_X}(f^*\mathcal{G}, \mathcal{H})) \end{align}

and

(12)\begin{equation} \mathrm{Hom}_{\mathcal{O}_X}(f^*(\mathcal{F}\otimes_{\mathcal{O}_Y} \mathcal{G}), \mathcal{H}) \cong \mathrm{Hom}_{\mathcal{O}_Y}(\mathcal{F}\otimes_{\mathcal{O}_Y} \mathcal{G}, f_*\mathcal{H}) \cong \mathrm{Hom}_{\mathcal{O}_Y}(\mathcal{F}, \underline{\mathrm{Hom}}_{\mathcal{O}_Y}(\mathcal{G}, f_*\mathcal{H})). \end{equation}

The result follows by using (9) (with $\mathcal{G}, \mathcal{H}$ in place of $\mathcal{F}, \mathcal{G}$) to compare the right sides of the two equations above.

Proof. This is just a restatement of lemma 3.18. Let A be an R-algebra and let $x\in X(A)$. Then by definition, the fibre of $(\mathcal{F}\otimes_{\mathcal{O}_X}\mathcal{G})(A)\rightarrow X(A)$ over x is the A-module corresponding to $x^*(\mathcal{F}\otimes_{\mathcal{O}_X}\mathcal{G})$, which is isomorphic to $x^*(\mathcal{F})\otimes_A x^*(\mathcal{G})$. This is the tensor product of the fibres of $\mathcal{F}$ and $\mathcal{G}$.

Proof. Since the fibres are A-modules, we have an addition operation

\begin{equation*} +: E(A)\times_{X(A)} E(A) \to E(A), \end{equation*}

which is natural in A because pulling back sections yields an additive map $E(A)\rightarrow E(B)$ for any $f: A\rightarrow B$. The universal property of pullback yields a natural isomorphism

\begin{equation*} (E\times_X E)(A) \cong E(A)\times_{X(A)} E(A). \end{equation*}

Composing gives a natural addition operation $(E\times_X E)(A) \rightarrow E(A)$. By naturality, this must be the functor of points of a morphism of schemes.

The proof for scalar multiplication is similar; the only substantial difference is that we need to check that $A \times E(A)\rightarrow E(A)$ is natural. If $x\in X(A)$ and $f:A\rightarrow B$, this amounts to showing that the composition of $A\times E_x \rightarrow E_x$ with the map $E_x \rightarrow E_{f(x)} \cong E_x \otimes_A B$ (given by $v\mapsto v\otimes 1$) is equal to the composition of the map $A\times E_x \rightarrow B \times E_{f(x)}$ (given by $a,v \mapsto f(a), v\otimes 1$) with $B \times E_{f(x)}\rightarrow E_{f(x)}$, which is clear.

Proof. The proof follows mutatis mutandis as the one for schemes over rings.

Proof. If $v\otimes w$ is in the fibre over x, then it is clear that $gv\otimes gw$ is in the fibre over gx, which establishes compatibility with the projections, showing (2) of definition 4.1.

The linearity of the actions on $\mathcal{E}_x$ and $\mathcal{F}_x$ implies $gv\otimes gw$ depends bilinearly on v and w. By the universal property of the tensor product, there is a unique linear map $\mathcal{E}_x \otimes \mathcal{F}_x \rightarrow \mathcal{E}_{gx} \otimes \mathcal{F}_{gx} $ sending $v\otimes w$ to $gv\otimes gw$. This establishes that the map of interest is well-defined and linear on fibres. This shows (3) of definition 4.1.

One can easily observe that (1) of definition 4.1 follows immediately from the naturality of the actions on $\mathcal{E}$ and $\mathcal{F}$ plus naturality of the multiplication map $G(A)\times G(A)\rightarrow G(A)$. Thus the tensor product is an equivariant vector bundle.

Now let $\mathcal{O}_X$ be the trivial equivariant line bundle (with the basis element denoted 1). We have an isomorphism of vector bundles

\begin{equation*} \mathcal{O}_X\otimes \mathcal{F}\rightarrow\mathcal{F}, \end{equation*}

which on the level of functors of points is given by $1_x \otimes v \mapsto v$. To check if it is an isomorphism of G-equivariant vector bundles (so that $\mathcal{O}_X$ is the tensor product identity), we need to check that $g(1_x \otimes v)$ maps to gv. But this follows immediately from $g(1_x \otimes v) = g1_x \otimes gv = 1_{gx} \otimes gv$.

It remains to show that equivariant line bundles are invertible under tensor product. Let L be an equivariant line bundle. There is a canonical morphism $\underline{\operatorname{Hom}}_{\mathcal{O}_X}(L, \mathcal{O}_X) \otimes L \rightarrow \mathcal{O}_X$. This is clearly an isomorphism for $L = \mathcal{O}_X$, and so for any line bundle it induces isomorphisms on stalks, and hence is an isomorphism.

By lemma 4.2, the functor of points of $\underline{\operatorname{Hom}}_{\mathcal{O}_X}(L, \mathcal{O}_X)$ has fibre over a point $x\in X(A)$ given by $\operatorname{Hom}_A(L_x, A)$. We may then define the action by saying $g\in G(A)$ acts on $\ell\in \operatorname{Hom}_A(L_x, A)$ by

\begin{equation*} (g\ell)(v) = g(\ell(g^{-1}v)) \textrm{for } v\in L_{gx}. \end{equation*}

Let $L^\vee=\underline{\operatorname{Hom}}_{\mathcal{O}_X}(L, \mathcal{O}_X)$. We claim that $L^\vee$ is an equivariant line bundle. In fact, from the definition, one can easily check that $g\ell$ is in the fibre over gx and $(L^\vee)_{x} \to (L^\vee)_{gx}$ is A-linear. The only nontrivial part is the naturality. In other words, for $f:A \to B$, we have to show that the following diagram commutes

First, for $\ell \in \operatorname{Hom}_A(L_x,A)$, we have

\begin{equation*} L_{f^*x}=L_x\otimes_A B \textrm{and }\psi(\ell)=f^*\ell=\ell\otimes \text{id} :L_x\otimes_AB \to B. \end{equation*}

Hence, we have

\begin{equation*} \psi\circ\rho_A (g,\ell)=f^*(g\ell)=(g\ell)\otimes \text{id}. \end{equation*}

So, for $w=\sum v_i\otimes b_i \in L_{f^*x}$ from the linearity, we have

\begin{equation*} \psi\circ\rho_A (g,\ell)(w)=\sum_i g(\ell(g^{-1}v_i)) \otimes b_i. \end{equation*}

On the other hand, we have

\begin{equation*} f^*g(w)=f^*g(\sum_iv_i\otimes b_i)=\sum_i gv_i \otimes b_i. \end{equation*}

Hence, we have

\begin{equation*} \rho_B\circ (\varphi\times\psi)(g,\ell)(w) = (f^*gf^*\ell)(w) = f^*g(f^*\ell(f^*g^{-1}w))=f^*g(f^*\ell(\sum_ig^{-1}v_i\otimes b_i)) \end{equation*}

\begin{equation*} =f^*g (\sum_i \ell(g^{-1}v_i)\otimes b_i) = \sum_i g(\ell(g^{-1}v_i))\otimes b_i. \end{equation*}

This shows that $\psi\circ\rho_A =\rho_B\circ (\varphi\times\psi)$.

The canonical isomorphism

\begin{equation*} L^\vee \otimes L\rightarrow \mathcal{O}_X \end{equation*}

sends $(\ell, v)$ to $\ell(v)$. Since $(g\ell)(gv) = g(\ell(v))$, this isomorphism is equivariant, so the line bundle $L^\vee$ is the inverse under tensor product.

Proof. Suppose $f, g: X \rightarrow Y$ are such that $f_A = g_A$ for all irreducible idempotent semirings A. Take an open affine cover $\{U_i=\operatorname{Spec} A_i\}_{i \in I}$ of X, where A_i are idempotent semirings. Observe that A_i is irreducible, because X is irreducible. The inclusion $\iota_i: U_i \rightarrow X$ is an element $\iota_i \in X(A_i)$. Then

\begin{equation*} f|_{U_i} = f_{A_i}(\iota_i) = g_{A_i}(\iota_i) = g|_{U_i} \end{equation*}

Since this holds for all i, we have f = g.

Now suppose we are given functions $f_A: X(A) \rightarrow Y(A)$ for all irreducible idempotent semirings A, and that this is natural in A. Let U_i, A_i, and ι_i be as before. For each $i, j$, let $\{V_{ijk}\}$ be a cover of $U_i \cap U_j$ by affine open sets and write $V_{ijk} = \mathrm{Spec}(B_{ijk})$. Let $\kappa_{ijk} \in X(B_{ijk})$ be the inclusion $V_{ijk}\rightarrow X$. Let $\phi_{ijk}: V_{ijk} \rightarrow U_i$ be the inclusion. Note that $\iota_{i} \circ \phi_{ijk} = \kappa_{ijk}$.

We obtain maps

\begin{equation*} a_i := f_{A_i}(\iota_i): U_i \rightarrow Y, \quad b_{ijk}:= f_{B_{ijk}}(\kappa_{ijk}): V_{ijk} \rightarrow Y. \end{equation*}

Naturality implies that

\begin{equation*} a_i |_{V_{ijk}} = a_i \circ \phi_{ijk} = f_{B_{ijk}}(\iota_i \circ \phi_{ijk}) = b_{ijk}. \end{equation*}

Similarly, the restriction of a_j to V_ijk is b_ijk. So a_i and a_j agree on V_ijk for all k, which implies they agree on $U_i \cap U_j$. Since we have a collection of morphisms $a_i: U_i \rightarrow Y$ that agree on overlaps, we obtain a morphism $g:X \to Y$ such that $g|_{U_i} = a_i$. In particular, we have

\begin{equation*} g_{A_i}=f_{A_i}. \end{equation*}

It remains to show that $g_A = f_A$ for all irreducible A. We fix a morphism $x: \operatorname{Spec} A \rightarrow X$ and show that

\begin{equation*} f_A(x) = g_A(x). \end{equation*}

Consider first the case where x factors through some U_i, i.e., $x = \iota_i \circ y$ for some $y \in U_i(A)$. By naturality, we have

\begin{equation*} f_A(x) = f_A(\iota_i \circ y) = f_{A_i}(\iota_i) \circ y, \end{equation*}

and similarly for g. Since $g_{A_i} = f_{A_i}$ by construction, $f_A(x) = g_A(x)$ for such x.

We now consider the general case. Cover each $x^{-1}(U_i)$ by open affines, to get a cover $\{W_{ij}=\operatorname{Spec} C_{ij}\}$ of $\operatorname{Spec} A$ such that $x(W_{ij})\subseteq U_i$. Let $\mu_{ij}: W_{ij} \rightarrow \operatorname{Spec} A$ be the inclusion. Let $x_{ij}: W_{ij} \rightarrow X$ be given by

\begin{equation*} x_{ij} = x \circ \mu_{ij},\quad \text{i.e., } x_{ij} = x|_{W_{ij}}. \end{equation*}

We have that x_ij factors through U_i, and hence

\begin{equation*} f_{C_{ij}}(x_{ij}) = g_{C_{ij}}(x_{ij}). \end{equation*}

By rewriting in terms of x and µ_ij and by using naturality, we have

\begin{equation*} f_A(x) \circ \mu_{ij} = f_{C_{ij}}(x \circ \mu_{ij}) = g_{C_{ij}}(x \circ \mu_{ij}) = g_A(x) \circ \mu_{ij}. \end{equation*}

It follows that the maps $f_A(x), g_A(x): \operatorname{Spec} A \rightarrow Y$ agree on each W_ij, and thus are equal.

Proof. I is an ideal by the same proof as the classical case.

Since $a+b \in I$, there exists $n \in \mathbb{N}$ such that $(a+b)^n=0$. Since A is idempotent, we have

\begin{equation*} (a+b)^n = \sum a^kb^{n-k} =0, \end{equation*}

which implies that $a^kb^{n-k}=0$ for all k, as any idempotent semiring is zero-sum-free. In particular $a^n=0$, showing that $a \in I$. Thus I is subtractive.

The claim about primality follows immediately from the definition of an irreducible semiring.

Proof. In the monoid case, we can adjoin a zero element to get to the case of an $\mathbb{F}_1$-algebra with no non-trivial zero-divisors.

Let $x, y\in A \otimes_{\mathbb{F}_1} M$ be such that xy is nilpotent. We wish to show either x or y is nilpotent. Write

\begin{equation*} x = \sum_{m\in M} x_m [m], \end{equation*}

where $x_m \in A$, and we have a similar expansion for y. Then

\begin{equation*} \sum_{m, n \in M} x_m y_n [mn] \end{equation*}

is nilpotent. Thus, from lemma 5.4, $x_m y_n [mn]$ is nilpotent for all $m, n\in M$. Then there is some k such that $(x_m y_n)^k [(mn)^k] = 0$. Since non-zero elements of M form a basis for $A \otimes_{\mathbb{F}_1} M$, either $[(mn)^k] = 0$ or $(x_m y_n)^k = 0$.

If for all $n\in M$ either y_n is nilpotent or $[n]$ is nilpotent, then $y = \sum y_n [n]$ lies in the nilradical, and we are done. Otherwise choose $n\in M$ such that neither y_n nor $[n]$ is nilpotent and choose $m\in M$ arbitrarily. In the case that $x_m y_n$ is nilpotent, since y_n is not, irreducibility of A tells us x_m is nilpotent. In the other case where $[mn]$ is nilpotent, irreducibility of M tells us $[m]$ is nilpotent. Thus for all $m\in M$, $x_m [m]$ is nilpotent and thus $x = \sum x_m [m]$ is nilpotent. This shows irreducibility of $A\otimes_{\mathbb{F}_1}M$.

For the second assertion, if neither A nor M have non-trivial zero-divisors (in particular all nilpotents are trivial), suppose xy = 0 (in particular xy is nilpotent). We may follow the argument above, and in the first case of the previous paragraph, we obtain that for all $n\in M$, either y_n is nilpotent (so $y_n = 0$) or $[n]$ is nilpotent (so $[n] = 0$). Either way $y = \sum y_n [n] = 0$. In the remaining case, as in the previous paragraph, we get that for each m, x_m or $[m]$ is nilpotent, and as in the previous case, we get $x = \sum x_m [m] = 0$.

Proof. Let P be the nilradical, which is a subtractive prime ideal by lemma 5.4. Then $A / P$ is a zero-sum-free semiring with no nontrivial zero-divisors. This implies the map $\phi: A/ P \rightarrow \mathbb{B}$ given by $\phi(0) = 0$ and $\phi(x) = 1$ for x ≠ 0 is a homomorphism. We consider instead the epimorphism $\psi: A \rightarrow \mathbb{B}$ obtained by composing with the quotient map; explicitly $\psi(x)$ is 0 or 1 according to whether x is nilpotent.

Write $x \equiv y$ if x and y are both nilpotent or both non-nilpotent. Since this is equivalent to $\psi(x) = \psi(y)$, ≡ is a congruence. By assumption it contains all generators of the congruence ∼. Thus, if $x \sim y$, then $x \equiv y$, which is the first part of the result.

For the second part of the result, suppose $\bar{x} \bar{y} \in A / \sim$ is nilpotent. Pick lifts $x, y\in A$, and observe that for some k, $(xy)^k \sim 0$. Thus $(xy)^k$ is nilpotent, which implies xy is nilpotent. By irreducibility, either x or y is nilpotent, which implies the same for $\bar{x}$ or $\bar{y}$.

If A has no nontrivial zero-divisors, suppose $\bar{x} \bar{y} = 0$. It is nilpotent, so we may follow the previous paragraph to obtain that x or y is nilpotent. But A has no nontrivial nilpotents, so x = 0 or y = 0 and hence $\bar{x} = 0$ or $\bar{y} = 0$.

Proof. Observe that $\mathbb{T}[M]$ has no nontrivial zero-divisors by proposition 5.6. Let ∼ be the congruence of $\mathbb{T}[M]$ generated by bend relations. Let v_K be the valuation on K.

Let $f \in I \subseteq K[M]$. Write $f = \sum a_i [m_i]$ with the a_i all non-zero. If f has no terms (i.e., f = 0), then the set of bend relations for f is empty and the bend congruence is just the diagonal congruence. If there is only one term then $[m_1] = f / a_1 \in I \cap M$ contradicting the hypothesis. Thus we may assume there are at least two non-zero terms.

For each i, consider the corresponding bend relation

\begin{equation*} \sum_j v_K(a_j) [m_j] \sim \sum_{j \neq i} v_K(a_j) [m_j]. \end{equation*}

Since f contains more than one term, both sums are nonempty. Moreover each term is non-zero. So the bend relation identifies a non-zero element with another non-zero element. Since $\mathbb{T}[M]$ is reduced, the bend relation identifies two non-nilpotents. Thus the result follows from lemma 5.7.

Proof. If we forget the addition structure on B, we can view it as an irreducible $\mathbb{F}_1$-algebra. We can then construct the A-algebra $A \otimes_{\mathbb{F}_1} B$. This is irreducible by proposition 5.6. By abuse of notation, we denote it $A[B]$, since this makes the asymmetry in how we treat A and B more clear.

Define a congruence ∼ on $A[B]$ generated by $[b_1] + [b_2] \sim [b_1 + b_2]$ for $b_1, b_2 \in B$ and $r \sim [r]$ for $r \in R$ (i.e., ∼ identifies the copy of R for B with the one for A). Because the structure maps have trivial kernel, $r^k = 0$ if and only if $[r]^k = 0$, so r is nilpotent if and only if $[r]$ is nilpotent. If $[b_1 + b_2]$ is nilpotent, there is some k such that $[(b_1 + b_2)^k] = 0$ and hence $(b_1 + b_2)^k = 0$, i.e., $b_1 + b_2$ is nilpotent. By lemma 5.4, this implies b ₁ and b ₂ are nilpotents in B so $[b_1], [b_2] \in A[B]$ are nilpotent, and hence $[b_1] + [b_2]$ is nilpotent. Conversely if $[b_1] + [b_2]$ is nilpotent, lemma 5.4 implies each term is nilpotent, so $b_1, b_2$ are nilpotent, and hence $[b_1 + b_2]$ is nilpotent. So no generator of ∼ identifies a nilpotent element with a non-nilpotent. Thus $A[B] / \sim$ is irreducible by lemma 5.7.

$A[B]$ has the universal property that a homomorphism $A[B] \rightarrow S$ corresponds to a pair of a semiring homomorphism $A \rightarrow S$ and an $\mathbb{F}_1$-algebra homomorphism $B \rightarrow S$ (viewed S as a multiplicative monoid with 0). A homomorphism $(A[B] / \sim) \rightarrow S$ corresponds to a homomorphism $\phi: A[B] \rightarrow S$ such that $\phi(r) = \phi([r])$ for all $r\in R$ and $\phi([b_1]) + \phi([b_2]) = \phi([b_1 + b_2])$ for all $b_1, b_2 \in B$. This in turn corresponds to a pair of a homomorphism $\phi_A: A \rightarrow S$ and an $\mathbb{F}_1$ algebra homomorphism $\phi_B: B\rightarrow S$ such that ϕ_A and ϕ_B agree on R and such that ϕ_B preserves addition (i.e., ϕ_B is in fact a semiring homomorphism).

The universal property of tensor products of semirings says that a homomorphism $A \otimes_R B \rightarrow S$ corresponds to a pair of homomorphisms $A \rightarrow S$ and $B\rightarrow S$ which agree on R. Thus, $A \otimes_R B \cong A[B] / \sim$. Since $A[B] / \sim$ is irreducible, so is $A \otimes_R B$.

Proof. If X is irreducible, clearly the graph is connected. Conversely, suppose that the graph is connected. If U_i and U_j are nondisjoint, then $U_i \cap U_j$ is an open subset of the irreducible space U_i and therefore contains the generic point of U_i. Similarly, it contains the generic point of U_j. But $U_i \cap U_j$ is irreducible (as it is an open subset of an irreducible space), so it contains a unique generic point. Thus, if i and j are connected by an edge, U_i and U_j share the same generic point. Since the graph is connected, all of the U_i share the same generic point η.

Let U be an open set. There is some i such that $U \cap U_i \neq \emptyset$. Since $U \cap U_i$ is an open set in the irreducible space U_i, it contains the generic point η. Since all open sets contain η, no two open sets can be disjoint.

Proof. First, we note that the construction of the fibred product in the semiring setting is the same as in the classical case of rings. One can check the same arguments go through mutatis mutandis.

Now, from the construction of the fibred product $X\times_KY$, if U is an open subset of X, then the preimage of U under $\pi_1: X \times_K Y \rightarrow X$ is isomorphic to $U \times_K Y$. Moreover, the product of affine schemes corresponds to the tensor product of semirings.

Let $U_i = \operatorname{Spec} A_i$ be an open affine cover of X and $V_j = \operatorname{Spec} B_j$ be an open affine cover of Y. $X \times_K Y$ has an open cover by open subschemes of the form $\pi_1^{-1}(U_i) = U_i \times_K Y$, and these in turn may be covered by open subschemes of the form $U_i \times_K V_j = \operatorname{Spec} A_i \otimes_K B_j$. By proposition 5.10, each $U_i \times_K V_j$ is irreducible.

Next, we consider the graph associated with this open cover (with an edge linking open subsets if they are not disjoint). This graph has vertices of the form (i, j), where i and j belong to the index sets of the covers $\{U_i \}$ and $\{V_j \}$, respectively.

If U_i and U_k are not disjoint, then $\pi_1^{-1}(U_i \cap U_k) \cong (U_i \cap U_j) \times_K Y$, so the left side is nonempty. In particular, $U_i \times_K Y$ and $U_j \times_K Y$ are not disjoint open subschemes. We may apply this with V_j in place of Y to obtain that $U_i \times_K V_j$ and $U_k \times_K V_j$ are not disjoint open subschemes of $X \times_K V_j \subseteq X \times_K Y$. So for each j we get an edge linking (i, j) to (k, j). Similarly if V_j and V_l are not disjoint then for each i, $U_i \times_K V_j$ intersects $U_i \times_K V_l$. So we get edges linking (i, j) to (i, l).

Thus the product graph of the graphs associated with the covers $\{U_i \}$ and $\{V_j \}$ is a spanning subgraph of the graph associated with the product cover. Since the product of connected graphs is connected, the result follows from lemma 5.12.

Proof. X has an open cover $\{U_{\sigma} \}$ (by nonempty sets) indexed by the cones in Δ in which $U_{\sigma} \subseteq U_\tau$ whenever σ is a face of τ. Now, the result directly follows from lemma 5.12 applied to $G_{\Delta}$.

Proof. We begin with the affine case in which $X = \operatorname{Spec} A$, where we are looking at a homomorphism $\phi: R\oplus S\rightarrow A$. Let $e = \phi(1, 0)$ and $f = \phi(0, 1)$. Since $(1, 0) + (0, 1) = (1, 1)$ and $(1, 0)(0, 1) = (0, 0)$, it follows that (e, f) forms an idempotent pair. By connectedness either e = 0 or f = 0. Without loss of generality consider the latter case. Then $\phi(r, s) = re + sf = re$ so ϕ factors through the projection $R\oplus S\rightarrow R$. Note that unless X is the empty scheme, ϕ cannot factor through both projections because then we would have $\phi(1, 1) = \phi(0, 0)$ contradicting that homomorphisms are unital.

Now consider the non-affine case. Construct a cover by affine open subsets $\{U_i\}$. Let $\phi:X \to \operatorname{Spec}(R\oplus S)$. Then, the restriction $\phi|_{U_i}$ factors through either $\operatorname{Spec} R$ or $\operatorname{Spec} S$. Suppose that for two indices i and j, $\phi|_{U_i}$ factors through $\operatorname{Spec} R$, $\phi|_{U_j}$ factors through $\operatorname{Spec} S$, and $U_i\cap U_j \neq \emptyset$. Then, we obtain a contradiction by letting W be an affine open subset of $U_i\cap U_j$ and observing that $\phi|_W$ factors through both $\operatorname{Spec} R$ and $\operatorname{Spec} S$.

Let U be the union of all U_i such that $\phi|_{U_i}$ factors through $\operatorname{Spec} R$ and V be the union of all U_i such that $\phi|_{U_i}$ factors through $\operatorname{Spec} S$. We just showed U and V are disjoint, and clearly they cover X. So one of them is empty; without loss of generality it is V, and so all $\phi|_{U_i}$ factor through R. Hence so does ϕ.

Proof. Fix $g\in G$ and $x\in X$. We consider two bases of E_gx: we have the basis $\{e_i(gx)\}$ and the basis $\{g e_i(x)\}$. Because A is an irreducible (hence connected) zero-sum-free semiring and E_gx is a free A-module, the basis is unique up to reordering and scalar multiplication (see remark A.5). Thus, there is some permutation σ and some units $a_i\in A^\times$ such that $g e_i(x) = a_i e_{\sigma(i)}(gx)$. The desired map ϕ is the map that sends (g, x) to this $\sigma \in S_n$. Note that ϕ is unique because a basis vector cannot be a multiple of another element of the basis.

Proof. For any irreducible idempotent K-algebra A, we consider the group G(A) of A-valued points of G and its action on the sets X(A) and E(A). Pick a basis e_i for E and observe that this induces a basis $x^*(e_i)$ of E_x for each $x\in X(A)$. Note that this is essentially due to the fact that $x^*$ and $x_*$ are adjoint functors as in the case for rings (lemma 3.8), implying that if $E=\mathcal{O}_X^n$, then $x^*E=\mathcal{O}_{\operatorname{Spec} A}^n$ since a left adjoint functor preserves colimits. It follows that $E_x=A^n$, and hence we can identify E(A) with $X(A) \times A^n$.

Now, equivariance of the bundle E means that G(A) acts on E(A) in a manner that is linear on fibres (definition 4.1 (3)) and that it makes the projection $\pi_A:E(A)\rightarrow X(A)$ equivariant (definition 4.1 (2)). Apply lemma 5.16 to the G(A)-action on $E(A)\rightarrow X(A)$. We obtain for each A, a map $\phi_A: G(A) \times X(A) \rightarrow S_n$ such that $g e_i(x)$ is a scalar multiple of $e_{\phi_A(g, x)(i)}(gx)$ for all $(g, x)\in G(A) \times X(A)$.

First, we want to show this map is natural. Fix a morphism $f: \operatorname{Spec} B \rightarrow \operatorname{Spec} A$. The induced map $f^*: X(A)\rightarrow X(B)$ sends a point $x:\operatorname{Spec} A\rightarrow X$ to $x\circ f$, and the induced map $f^*: G(A)\rightarrow G(B)$ is described similarly.

Let $\alpha:G \times_K X \to X$ be a G-action on X. Then for each K-algebra A, one has

\begin{equation*} \alpha_A:G(A) \times X(A) \to X(A). \end{equation*}

Since the action α is a morphism of schemes, it induces a natural transformation of functors of points, i.e., for $g \in G(A)$ and $x \in X(A)$, one has

(13)\begin{equation} f^*(gx) = f^*(\alpha_A(g, x)) = \alpha_B(f^*(g,x)) = \alpha_B(f^*(g), f^*(x)) = f^*(g)f^*(x). \end{equation}

The induced map $f^*: E(A)\rightarrow E(B)$ sends $e_i(x)$ to

(14)\begin{equation} f^*(e_i(x))= f^* x^*(e_i) = e_i(x\circ f) = e_i(f^*(x)). \end{equation}

By (1) of definition 4.1, similar to (13), for any $e\in E(A)$ and $g\in G(A)$, we have

\begin{equation*} f^*(g) f^*(e) = f^*(ge). \end{equation*}

Observe that

(15)\begin{equation} f^*(g) f^*(e_i(x)) = f^*(g e_i(x)) \end{equation}

is a scalar multiple of

(16)\begin{equation} f^*(e_{\phi_A(g, x)(i)}(gx)) = e_{\phi_A(g, x)(i)}(f^*(gx)) = e_{\phi_A(g, x)(i)}(f^*(g)f^*(x)) \end{equation}

because $ge_i(x)$ is a multiple of $e_{\phi_A(g, x)(i)}(gx)$. We claim the following:

(17)\begin{equation} \phi_B(f^*(g), f^*(x)) = \phi_A(g, x), \end{equation}

which is precisely the naturality result we want to prove. Note that $f^*$ does not occur on the right side because S_n is a constant functor, so $f^*$ is the identity here.

To see (17), define for $g' \in G(B)$ and $x' \in X(B)$,

(18)\begin{equation} \psi(g',x')=\begin{cases} \phi_B(g',x') & \textrm{if } (g', x') \neq (f^*(g), f^*(x)),\\ \phi_A(g, x) & \textrm{if } (g', x') = (f^*(g), f^*(x)). \end{cases} \end{equation}

Then for any $g' \in G(B)$ and $x' \in X(B)$, we see that $g' e_i(x')$ is a multiple of $e_{\psi(g', x')(i)}(g'x')$; when $(g', x') = (f^*(g), f^*(x))$ this is because (15) is a multiple of (16), and otherwise this holds because it is true for ϕ_B. Now the uniqueness in the defining property of ϕ_B (in lemma 5.16) implies $\phi_B = \psi$, so in particular $\phi_B(f^*(g), f^*(x)) = \phi_A(g, x)$, as claimed.

We thus have constructed a natural map $(G \times_K X)(A) \rightarrow S_n = S_n(A)$. Moreover, it follows from proposition 5.13 that $G \times_K X$ is irreducible. Therefore, from lemma 5.3, this natural transformation comes from a morphism of schemes $\phi: G \times_K X \rightarrow S_n$. Since $G\times_K X$ is connected, and S_n is the spectrum of a direct sum of $n!$ copies of K, ϕ is constant, i.e., there is some $\sigma \in S_n$ such that ϕ factors through the inclusion $\operatorname{Spec} K\rightarrow S_n$ corresponding to the point σ. The same must be true of the corresponding maps on functors of points. Thus, $\phi_A(g, x) = \sigma$ for all $(g, x)\in G(A) \times X(A)$. So we obtain that for any $(g, x)\in G(A) \times X(A)$, $g e_i(x)$ is a scalar multiple of $e_{\sigma (i)}(gx)$. By applying this to g = 1, we must have σ = 1, so $g e_i(x)$ is a scalar multiple of $e_i(gx)$.

Let L_i be the line bundle contained in E given as a sheaf by $L_i(U) = \mathrm{span}(e_i|_U)$. Its functor of points is such that $L_i(A)$ is the set of pairs (x, e) such that $x\in X(A)$, and $e\in \mathrm{span}(e_i(x))\subseteq E_x$. We saw $g e_i(x)$ is a scalar multiple of $e_i(gx)$ so belongs to $L_i(A)$, and the same is true for ge where $e\in L_i(A)$ since any such e is a multiple of $e_i(x)$ for some x. Since $ge\in L_i(A)$ for $g\in G(A)$ and $e\in L_i(A)$, it follows that the G(A) action on E(A) descends to an action on $L_i(A)$, making $L_i(A)$ equivariant. Even without equivariance, we of course have $E = \bigoplus L_i$, and since L_i is equivariant, this implies the result.

Proof. We prove the case for semirings. The case for monoids is similar.

Note that we have a canonical global section $t \in \Gamma(GL_1, \mathcal{O}_{\text{GL}_1}^\times)$, so $f: X\rightarrow \text{GL}_1$ induces a global section $f^*t \in \Gamma(X, \mathcal{O}_X^\times)$. It is natural, because given $g: Y\rightarrow \text{GL}_1$, the induced map on $\mathrm{Hom}_{\operatorname{Spec} K}(-, \text{GL}_1)$ sends f to $f\circ g$, while the induced map on $\Gamma(-, \mathcal{O}_-^\times)$ sends $f^* t$ to $g^* f^* t = (f\circ g)^* t$.

To show it is an isomorphism, we begin with the affine case, where $X = \operatorname{Spec} A$ for some semiring A and so $\Gamma(X, \mathcal{O}_X^\times) = A^\times$. The universal property of $K[t, t^{-1}]$ asserts that morphisms

\begin{equation*} f: K[t, t^{-1}]\rightarrow A \end{equation*}

are in bijection with units of A, and that the corresponding element of $A^\times$ is $f(t) = f^* t$. So this bijection is the natural map from before.

Now consider the general case. Fix an open affine cover $\{U_i \}$ and let $\iota_i: U_i\rightarrow X$ be the inclusion. For each $i, j$, fix an open affine cover $\{V_{ijk} \}$ of $U_i \cap U_j$ with $V_{ijk} = V_{jik}$, and let $\phi_{ijk}: V_{ijk} \rightarrow U_i$ be the inclusion. Given $u \in \Gamma(X, \mathcal{O}_X^\times)$, we obtain elements $u_i = u\mid_{U_i} \in \Gamma(U_i, \mathcal{O}_{U_i}^\times)$ which satisfy the following condition:

\begin{equation*} \phi_{ikj}^* u_i = \phi_{jik}^* u_j. \end{equation*}

We let f_i be the morphism corresponding to u_i; note that the restrictions of f_i and f_j to each V_ijk agree. Our goal is to show there is a unique $f: X\rightarrow GL_1$ satisfying $u = f^* t$.

Given a morphism $f: X\rightarrow GL_1$, we have

\begin{equation*} (f^* t)\mid_{U_i} = \iota_i^* f^* t = (f\mid_{U_i})^* t. \end{equation*}

Thus $f^* t = u$ if and only if $(f\mid_{U_i})^* t = u_i$. By the affine case of this result, this holds if and only if $f\mid_{U_i}$ is the morphism corresponding to u_i, i.e., if and only if $f\mid_{U_i} = f_i$. But since the f_i agree on overlaps they glue together to give a unique morphism satisfying this property.

Proof. We prove the case for semirings. The case for monoids is similar.

We will construct u via its functor of points. Let v be a basis vector for global sections for L. Since v_x is a basis vector for the free module L_x, gv_x is a basis vector for L_gx. But v_gx is also a basis vector, so there is a unique $a\in GL_1(A)$ such that $a v_{gx} = g v_x$. We let $u_A(g,x)=a$.

We have thus constructed a unique map u_A such that $u_A(g, x) v_{gx} = g v_x$ holds. To obtain the map u, we must show u_A is natural in A. Fix a morphism $f:\operatorname{Spec} B \to \operatorname{Spec} A$. As in the proof of theorem 5.17, we let the induced maps be $f^*:X(A) \to X(B)$, $f^*:G(A) \to G(B)$, and $f^*:L(A) \to L(B)$. Since $f^*(gx)=f^*(g)f^*(x)$, we have

\begin{equation*} u_B(f^* g, f^* x) v_{f^*(gx)} = (f^*g) v_{f^* x}. \end{equation*}

Recall that $v_{f^* x}$ is the result of pulling v back along $f^* x : \operatorname{Spec} B \rightarrow X$, while v_x is the pullback along $x: \operatorname{Spec} A \rightarrow X$. Thus $v_{f^* x}$ is the pullback of v_x along $f: \operatorname{Spec} B \rightarrow \operatorname{Spec} A$. In other words, $v_{f^*x} = f^* v_x$, where the latter $f^*$ denotes the induced map $f^*:L(A)\rightarrow L(B)$.

Thus we have

\begin{equation*} u_B(f^* g, f^* x) f^*v_{gx} = (f^*g) f^* v_{x}. \end{equation*}

But applying $f^*$ to both sides of

\begin{equation*} u_A(g, x) v_{gx} = g v_x \end{equation*}

gives

\begin{equation*} u_A(g, x) f^* v_{gx} = (f^* g) f^* v_x. \end{equation*}

Thus $u_B(f^* g, f^* x) = u_A(g, x)$, which is the desired naturality.

Proof. To show the first part of the statement, observe that

\begin{equation*} u_A(gh, x) v_{ghx} = gh v_{x} = u_A(h, x) g v_{hx} = u_A(h, x) u_A(g, hx) v_{ghx}. \end{equation*}

Since v_ghx is a basis vector, the first part of the statement follows. Similarly, the second half of the statement follows from $u_A(1, x) v_x = 1 v_x = v_x$.

Proof. We choose any trivial line bundle L and basis vector v. To construct the equivariant structure, we define the action $\rho_A:G(A) \times L(A) \rightarrow L(A)$ by

\begin{equation*} g(av_x) = a u_A(g, x) v_{gx}. \end{equation*}

By construction this action is linear and satisfies $u_A(g, x) v_{gx} = g v_x$. We must check that it is actually a group action, that it is compatible with the action on X, and that it is natural in A.

To see that it is a group action, observe that

\begin{equation*} g(h(a v_x)) = g(a u_A(h, x) v_{hx}) = a u_A(g, hx) u_A(h, x) v_{ghx} = a u_A(gh, x) v_{ghx} = (gh)(a v_x), \end{equation*}

which is the desired homomorphism property. Also 1 acts via $1(a v_x) = u_A(1, x) a v_x = a v_x$. The action is compatible with the action on X because $g(a v_x) = a u_A(g, x) v_{gx}$ lies over gx.

To show that the action is natural in A, let $f:\operatorname{Spec} B \to \operatorname{Spec} A$, and $f^*$ be as in the proof of lemma 6.4. We need to show that the following diagram commutes:

Now for $(g,av_x) \in G(A) \times L(A)$, we have

\begin{align*} f^*\rho_A(g,av_x)&=f^*(au_A(g,x)v_{gx})=au_A(g,x)f^*(v_{gx})\\ &=au_A(g,x)v_{f^*({gx})}=au_A(g,x)v_{f^*gf^*x}. \end{align*}

On the other hand,

\begin{equation*} \rho_B(f^*g,f^*(av_x))=\rho_B(f^*g,af^*(v_x))=\rho_B(f^*g,av_{f^*x})=au_B(f^*g,f^*x)v_{f^*gf^*x}. \end{equation*}

From the functoriallity of u, we have

\begin{equation*} u_B(f^*g,f^*x)=u_A(g,x), \end{equation*}

implying that

\begin{equation*} f^*\rho_A(g,av_x)=\rho_B(f^*g,f^*(av_x))=\rho_Bf^*(g,av_x). \end{equation*}

Proof of proposition 6.8

We have already proved that equivariant structures on $\mathcal{O}_X$ uniquely give rise to principal morphisms u (lemmas 6.4 and 6.5), and that every such u arises in this way (lemma 6.7).

Next we prove the uniqueness. Consider two actions $\alpha, \beta: G \times_K \mathcal{O}_X \rightarrow \mathcal{O}_X$ which give rise to the same morphism u. Any element of L(A) can be written $a 1_x$ for some $a\in A$ and $x\in X(A)$. Under both actions, linearity and the definition of u give

\begin{equation*} g(a 1_x) = a g(1_x) = a u_A(g, x) 1_x, \end{equation*}

and hence so both actions are the same. In other words, such u arises from a unique equivariant structure on $\mathcal{O}_X$.

Finally, we show this bijective correspondence is a homomorphism. Let $L, L'$ be equivariant line bundles with basis sections v and w, respectively. Let $u, u', \phi: G \times_K X \rightarrow \mathrm{GL}_1$ correspond to equivariant structures of $L, L', L\otimes L'$, respectively. Then by definition of ϕ and of the action on the tensor product,

(21)\begin{equation}\phi_A(g, x)(v_{gx} \otimes w_{gx}) = g(v_x \otimes w_x) = gv_x \otimes gw_x. \end{equation}

On the other hand, by definition of u and u ^′,

(22)\begin{equation}u_A(g, x)u'_A(g, x)(v_{gx} \otimes w_{gx}) = gv_x \otimes gw_x. \end{equation}

So $\phi = uu'$, which implies the correspondence in question is a group homomorphism.

Proof. We first consider the case when X and G are affine. Let $X=\operatorname{Spec} M$ and $G=\operatorname{Spec} N$. Then,

\begin{equation*} G\times X = \operatorname{Spec} (N\otimes_{\mathbb{F}_1}M), \quad G_K=\operatorname{Spec} K[N], \quad X_K=\operatorname{Spec} K[M], \end{equation*}

and since tensor products preserve colimits

\begin{equation*} G_K \times X_K= (G\times X)_K. \end{equation*}

Hence, we have

\begin{equation*} \Gamma(G_K\times X_K, \mathcal{O}_{G_K\times X_K}^\times) = (K[N\otimes_{\mathbb{F}_1}M])^\times. \end{equation*}

It follows from [Reference Jun, Mincheva and Tolliver17, Proposition 3.4] that

\begin{equation*} (K[N\otimes_{\mathbb{F}_1}M])^\times = K^\times \times (N\otimes_{\mathbb{F}_1}M)^\times. \end{equation*}

On the other hand, one can easily check that

\begin{equation*} \Gamma (G\times X, K^\times)=K^\times, \quad \Gamma(G\times X, \mathcal{O}_{G\times X}^\times) = (N\otimes_{\mathbb{F}_1}M)^\times, \end{equation*}

showing the claimed identity for affine G and X.

For general G and X, we cover $G_K \times X_K$ by affine open cover $\mathcal{U}_K$ obtained by affine open subsets of $\mathcal{U}=\{U_i\times V_j\}$, where U_i (resp. V_j) are affine open subsets of G (resp. X). Note that this is possible by [Reference Jun, Mincheva and Tolliver17, Lemma 3.1]. Then, as in [Reference Jun, Mincheva and Tolliver17, Theorem 3.8], by using the affine case, one can obtain isomorphisms of Čech cochains

\begin{equation*} \mathbf{C}^k(\mathcal{U}_K,\mathcal{O}_{G_K\times X_K}^\times) = \mathbf{C}^k(\mathcal{U},K^\times \times \mathcal{O}_{G\times X}^\times), \end{equation*}

showing the desired identity (23).

Proof. Let $u \in \Gamma(G_K\times X_K, \mathcal{O}_{G_K\times X_K}^\times)$. We write $u=(u_1,u_2)$ under the isomorphism in lemma 6.11. Under a connectedness assumption, u ₁ is constant, and hence the identity

\begin{equation*} u(gh, x) = u(g, hx) u(h, x) \end{equation*}

together with the fact that $K^\times$ is a group (so that $x^2 = x$ implies x = 1) implies that u ₁ is 1. It follows that u actually lies in $\Gamma(G\times X, \mathcal{O}_{G\times X}^\times)$. Now, the desired group isomorphism follows from corollary 6.10.

Proof. We consider first the case where $X = \operatorname{Spec} A$ and $Y = \operatorname{Spec} B$ are affine. As a monoid, $A \otimes_{\mathbb{F}_1} B$ is obtained from the (unpointed) monoid A × B by identifying $(0, b)$ and $(a, 0)$ with $(0, 0)$ for all $a \in A$ and $b \in B$; this is an $\mathbb{F}_1$-algebra with zero element $(0, 0)$. The quotient map $\pi:A \times B \to (A\otimes_{\mathbb{F}_1}B)$ gives rise to a map

\begin{equation*} \theta: A^\times \times B^\times = (A \times B)^\times \rightarrow (A \otimes_{\mathbb{F}_1} B)^\times. \end{equation*}

We claim that θ is bijective. In fact, θ is injective because the only fibre of the quotient map π containing more than one element is $A\times \{0\} \cup \{0\} \times B$, and this fibre contains no units. Surjectivity is similar: if $x \in (A \otimes_{\mathbb{F}_1} B)^\times$ then we can let y be the unique element of $\pi^{-1}(x)$ and z be the unique element of $\pi^{-1}(x^{-1})$ and observe that $\pi(yz) = 1$. Since the fibre over 1 contains a single element, yz = 1 so $x = \pi(y)$ for some unit y and $x \in \mathrm{im} \theta$. Thus, θ is bijective. Since we can view $\pi: A \times B \rightarrow (A \otimes_{\mathbb{F}_1} B)$ as the multiplication map, θ is the multiplication map on units, which establishes the result in the affine case.

Next, suppose Y is such that the result holds for any pair (Z, Y) with Z affine. Suppose X is irreducible. Let $U \subseteq X$ be open affine. Then by the affine case, for $u \in \Gamma(X\times Y, \mathcal{O}_{X\times Y}^\times)$, there exists unique elements $\alpha_U \in \Gamma(U, \mathcal{O}_X^\times)$ and $\beta_U \in \Gamma(Y, \mathcal{O}_Y^\times)$ such that

\begin{equation*} u \mid_{U \times Y} = \alpha_U \beta_U. \end{equation*}

Let $U_1, U_2$ be open affine subsets of X; they are nondisjoint by irreducibility. Let $V \subseteq U_1 \cap U_2$ be a nonempty open affine subset. Then restricting the factorization of $u\mid_{U_1\times Y}$ to V × Y gives

\begin{equation*} u\mid_{V\times Y} = \alpha_{U_1}\mid_V \beta_{U_1}. \end{equation*}

By applying the argument to U ₂ and using uniqueness of $\alpha_V, \beta_V$, we obtain

\begin{equation*} \alpha_{U_1}\mid_V = \alpha_V = \alpha_{U_2}\mid_V \textrm{and } \beta_{U_1} = \beta_V = \beta_{U_2}. \end{equation*}

Since β_U does depend on U, we denote it β. Fix the open cover $\{U_i \}$ of X containing all open affine sets. We have seen that the sections $\alpha_{U_i}\in \Gamma(U_i, \mathcal{O}_X^\times)$ agree on overlaps, so give rise to $\alpha \in \Gamma(X, \mathcal{O}_X^\times)$.