Fibre Products Base Change and Scheme Theoretic Fibres: Examples

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The fibres of xy=t

Example

Over any field $k$ , consider $X = Spec k [x, y, t] / (x y - t) \to Spec k [t]$ . At $t = a \in k$ the fibre is $Spec k [x, y] / (x y - a)$ . At zero it is the union of two reduced axes; at $a \neq 0$ it is $G_{m, k} = Spec k [x, x^{- 1}]$ . The generic fibre is $Spec k (t) [x, x^{- 1}]$ .

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

Let $A \to B$ be a ring map, $p \in Spec A$ , and $M = A ∖ p$ , acting on $B$ through the ring map. The fibre over $p$ is canonically $Spec (B \otimes_{A} κ (p)) ≅ Spec (M^{- 1} B / p M^{- 1} B) .$ The residue field is $κ (p) = A_{p} / p A_{p}$ . No reduction of the tensor ring is taken. (Coordinate ring of an affine fibre)

[F2]

Let $A \to C$ be a unital ring map. For any set of variables $(t_{i})$ and any ideal $I \subseteq A [t_{i}]$ , $(A [t_{i}] / I) \otimes_{A} C ≅ C [t_{i}] / I C [t_{i}] .$ Here the extended ideal is generated by the coefficient images of all elements of $I$ . For a multiplicative subset $M \subseteq A$ , $(M^{- 1} A) \otimes_{A} C ≅ {\overline{M}}^{- 1} C .$ These are ring isomorphisms; no flatness, finite-generation or nonzero-ring hypothesis is required. (Presentations and localization under base extension)

Verification

1.1givenF1F2

F1 computes a fibre by tensoring with its residue field. F2 substitutes $t = a$ to give $k [x, y] / (x y - a)$ , and at the generic point it extends coefficients to $k (t)$ with relation $x y = t$ .

2.1step 1.1algebra

At $a = 0$ , $(x y) = (x) \cap (y)$ : a polynomial divisible by both $x$ and $y$ has every monomial divisible by $x y$ . The ideals $(x), (y)$ are prime, so their intersection is radical and the quotient is reduced. Every prime containing $x y$ contains $x$ or $y$ ; these two incomparable minimal primes give exactly the two axes, meeting at $(x, y)$ .

3.1step 1.1algebra∎

At $a \neq 0$ , the inverse maps send $x \mapsto x$ , $y \mapsto a x^{- 1}$ in one direction and $x^{- 1} \mapsto a^{- 1} y$ in the other. They identify the ring with $k [x, x^{- 1}]$ . The same formulas with the nonzero unit $t \in k (t)$ give the generic fibre. In particular $a = 1$ has the same form; all three rings are nonzero, so none of these fibres is empty.

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Quadratic fibres over rational points and the generic point

Example

For any field $k$ , the family $Spec k [x, t] / (x^{2} - t) \to Spec k [t]$ has fibre $Spec k [x] / (x^{2})$ at zero, nonreduced in every characteristic. Over $Q$ , the fibre rings at $t = 1, 0, - 1$ and at the generic point are respectively $Q \times Q, Q [ϵ] / (ϵ^{2}), Q (i), Q (x),$ where the generic base field $Q (t)$ embeds in $Q (x)$ via $t = x^{2}$ , with degree two.

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

Let $A \to B$ be a ring map, $p \in Spec A$ , and $M = A ∖ p$ , acting on $B$ through the ring map. The fibre over $p$ is canonically $Spec (B \otimes_{A} κ (p)) ≅ Spec (M^{- 1} B / p M^{- 1} B) .$ The residue field is $κ (p) = A_{p} / p A_{p}$ . No reduction of the tensor ring is taken. (Coordinate ring of an affine fibre)

[F2]

Let $A \to C$ be a unital ring map. For any set of variables $(t_{i})$ and any ideal $I \subseteq A [t_{i}]$ , $(A [t_{i}] / I) \otimes_{A} C ≅ C [t_{i}] / I C [t_{i}] .$ Here the extended ideal is generated by the coefficient images of all elements of $I$ . For a multiplicative subset $M \subseteq A$ , $(M^{- 1} A) \otimes_{A} C ≅ {\overline{M}}^{- 1} C .$ These are ring isomorphisms; no flatness, finite-generation or nonzero-ring hypothesis is required. (Presentations and localization under base extension)

[F3]

Let $R$ be a commutative ring and let $I_{1}, \dots, I_{r}$ be pairwise comaximal ideals, where $r \geq 1$ . Then the canonical map $R ⟶ \prod_{i = 1}^{r} R / I_{i}, x ⟼ (x + I_{1}, \dots, x + I_{r})$ is surjective, its kernel is $⋂_{i = 1}^{r} I_{i}$ , and $⋂_{i = 1}^{r} I_{i} = \prod_{i = 1}^{r} I_{i} .$ Equivalently, $R / \prod_{i = 1}^{r} I_{i} ≅ \prod_{i = 1}^{r} R / I_{i} .$ (Chinese remainder theorem for pairwise comaximal ideals)

Verification

1.1givenF1F2

F1 and F2 give $k [x] / (x^{2} - a)$ at $t = a$ . At $a = 0$ , the classes of $1, x$ are a basis, so $x$ is nonzero with square zero, in every characteristic. The ideal $(x)$ is the only prime and the residue field is $k$ .

1.2F1F2algebra

The generic fibre is $Q (t) [x] / (x^{2} - t)$ , equivalently $Q [x]$ with all nonzero polynomials in $x^{2}$ inverted. For nonzero $g (x)$ , the product $g (x) g (- x)$ is a nonzero even polynomial, hence a polynomial in $x^{2}$ . Thus $g (x)^{- 1} = g (- x) / (g (x) g (- x))$ belongs to this ring. It is exactly $Q (x)$ . Its dimension over $Q (t)$ is two, because division by the monic polynomial $x^{2} - t$ leaves unique remainders of degree less than two.

2.1F3step 1.1algebra

Over $Q$ , at $a = 1$ the factors $x - 1, x + 1$ generate comaximal ideals since their difference is 2. F3 therefore gives $Q \times Q$ . At $a = - 1$ , $x^{2} + 1$ has no rational root, hence is irreducible of degree two and its quotient is the field $Q (i)$ .

3.1F3step 1.1step 2.1step 1.2∎

The displayed fibres are nonempty: each coordinate ring contains 1 nonzero. Over an algebraic closure of any residue field in this rational family, $x^{2} - a$ splits into two distinct factors when $a \neq 0$ , since the characteristic is zero and a root then has nonzero derivative $2 x$ . At $a = 0$ it remains a doubled point. CRT proves the two-point assertion just as at $a = 1$ .

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A real conic acquires complex points

Example

The real scheme $X = Spec R [x, y] / (x^{2} + y^{2} + 1)$ has no $R$ -valued points over $R$ . Its complex base change is $Spec C [x, y] / (x^{2} + y^{2} + 1)$ and has the complex point $(i, 0)$ . Absence of real-valued points does not mean $X$ is empty.

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

For every field $K$ and scheme $X$ , morphisms $Spec K \to X$ correspond bijectively to pairs $(x, ι)$ with $x \in X$ and a field embedding $ι : κ (x) \to K$ . The identity embedding gives a canonical morphism $Spec κ (x) \to X$ , compatible with all scheme morphisms. More generally, for a nonzero local ring $(R, m)$ , morphisms $Spec R \to X$ correspond to pairs $(x, φ)$ with a local homomorphism $φ : O_{X, x} \to R$ . Assuming Choice, two field-valued points have the same image in $X$ if and only if they are dominated by a common field-valued point, by compatible embeddings of their fields into a third field. (Field-valued points and local-ring points)

[F2]

For a field extension $K / k$ and a $k$ -scheme $X$ , the inverse image of every affine open $U = Spec A$ in $X_{K}$ is $Spec (A \otimes_{k} K)$ . These affine charts cover $X_{K}$ and are compatible on overlaps and with coefficient localizations. (Affine charts after extension of the ground field)

[F3]

Let $A \to C$ be a unital ring map. For any set of variables $(t_{i})$ and any ideal $I \subseteq A [t_{i}]$ , $(A [t_{i}] / I) \otimes_{A} C ≅ C [t_{i}] / I C [t_{i}] .$ Here the extended ideal is generated by the coefficient images of all elements of $I$ . For a multiplicative subset $M \subseteq A$ , $(M^{- 1} A) \otimes_{A} C ≅ {\overline{M}}^{- 1} C .$ These are ring isomorphisms; no flatness, finite-generation or nonzero-ring hypothesis is required. (Presentations and localization under base extension)

Verification

1.1givenF1algebra

A real-valued point over $R$ gives a real algebra map to $R$ , hence images $a, b \in R$ satisfying $a^{2} + b^{2} + 1 = 0$ . This is impossible since the left side is at least 1. This agrees with the residue-field description of F1; the requirement that the morphism be over $R$ is retained.

2.1F2F3algebra∎

F2 and F3 give the displayed complex ring. Evaluation $x \mapsto i, y \mapsto 0$ kills its defining polynomial and sends 1 to 1, so it is a complex point. Its composition with the projection gives a point of $X$ , proving $X$ is nonempty. The same evaluation directly shows neither coordinate ring is zero.

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An integral real scheme splits over the complex numbers

Statement refuted

False claims: an integral real scheme must stay irreducible or connected after extension to $C$ ; an injective morphism of schemes must remain injective after arbitrary base change. The morphism $Spec C \to Spec R$ refutes all these assertions.

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

For a field extension $K / k$ and a $k$ -scheme $X$ , the inverse image of every affine open $U = Spec A$ in $X_{K}$ is $Spec (A \otimes_{k} K)$ . These affine charts cover $X_{K}$ and are compatible on overlaps and with coefficient localizations. (Affine charts after extension of the ground field)

[F2]

Let $A \to C$ be a unital ring map. For any set of variables $(t_{i})$ and any ideal $I \subseteq A [t_{i}]$ , $(A [t_{i}] / I) \otimes_{A} C ≅ C [t_{i}] / I C [t_{i}] .$ Here the extended ideal is generated by the coefficient images of all elements of $I$ . For a multiplicative subset $M \subseteq A$ , $(M^{- 1} A) \otimes_{A} C ≅ {\overline{M}}^{- 1} C .$ These are ring isomorphisms; no flatness, finite-generation or nonzero-ring hypothesis is required. (Presentations and localization under base extension)

[F3]

Let $R$ be a commutative ring and let $I_{1}, \dots, I_{r}$ be pairwise comaximal ideals, where $r \geq 1$ . Then the canonical map $R ⟶ \prod_{i = 1}^{r} R / I_{i}, x ⟼ (x + I_{1}, \dots, x + I_{r})$ is surjective, its kernel is $⋂_{i = 1}^{r} I_{i}$ , and $⋂_{i = 1}^{r} I_{i} = \prod_{i = 1}^{r} I_{i} .$ Equivalently, $R / \prod_{i = 1}^{r} I_{i} ≅ \prod_{i = 1}^{r} R / I_{i} .$ (Chinese remainder theorem for pairwise comaximal ideals)

Counterexample

1.1givenF1F2

The source has one point with field local ring $C$ , so it is integral and connected, and the map to the one-point spectrum of $R$ is injective. By F1 and F2 its complex base change has ring $C \otimes_{R} C ≅ C [z] / (z^{2} + 1)$ .

2.1F3step 1.1algebra∎

The factors $z - i, z + i$ are comaximal because $2 i$ is a unit. F3 gives $C [z] / (z^{2} + 1) ≅ C \times C$ . The two prime ideals are $C \times 0$ and $0 \times C$ ; the complementary idempotents exhibit them as two disjoint nonempty clopen points. Thus the base change is reduced but disconnected and reducible, and its map to $Spec C$ is not injective.

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The affine plane and the pair of generic points

Example

For any field $k$ , $A_{k}^{1} \times_{k} A_{k}^{1} ≅ A_{k}^{2}$ functorially. Nevertheless the product has distinct points over the pair of generic points: in $k [x, y]$ , both $(0)$ and $(y - x)$ contract to $(0)$ in each of $k [x]$ and $k [y]$ .

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

Let $A \to B$ and $A \to C$ be maps of commutative unital rings, allowing the zero ring. In the category of all schemes, $Spec B \times_{Spec A} Spec C ≅ Spec (B \otimes_{A} C) .$ The projections correspond to $b \mapsto b \otimes 1$ and $c \mapsto 1 \otimes c$ . (Affine fibre products are spectra of tensor products)

[F2]

For scheme morphisms $f : X \to S$ and $g : Y \to S$ , points of $P = X \times_{S} Y$ are in bijection with quadruples $(x, y, s, r)$ where $f (x) = g (y) = s$ and $r \in Spec (κ (x) \otimes_{κ (s)} κ (y)) .$ The residue field at the corresponding point of $P$ is canonically $κ (r)$ . (Points of a fibre product via residue-field tensors)

Verification

1.1givenF1algebra

F1 identifies the product ring with $k [x] \otimes_{k} k [y] ≅ k [x, y]$ : the maps send $x \otimes 1$ to $x$ and $1 \otimes y$ to $y$ , with inverse on every polynomial specified by these images. Consequently morphisms from every $k$ -scheme $T$ into the plane are compatible pairs of morphisms into the two lines. Empty test schemes are allowed.

2.1F2step 1.1algebra∎

The ring $k [x, y]$ is a domain, so $(0)$ is prime. The quotient by $(y - x)$ is $k [x]$ , also a domain, so this is another prime, distinct because $y - x \neq 0$ . Substitution $y = x$ is injective on either single-variable subring, giving the asserted contractions. F2 explains these as distinct residue-tensor primes over the same point pair. This works over every field, including characteristic two.

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A closed-immersion fibre is one residue point or empty

Example

For an ideal $I \subseteq A$ and $p \in Spec A$ , the fibre of $Spec (A / I) \to Spec A$ at $p$ is $Spec κ (p)$ if $I \subseteq p$ , and is empty otherwise.

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

Let $A \to B$ be a ring map, $p \in Spec A$ , and $M = A ∖ p$ , acting on $B$ through the ring map. The fibre over $p$ is canonically $Spec (B \otimes_{A} κ (p)) ≅ Spec (M^{- 1} B / p M^{- 1} B) .$ The residue field is $κ (p) = A_{p} / p A_{p}$ . No reduction of the tensor ring is taken. (Coordinate ring of an affine fibre)

[F2]

Let $A \to C$ be a unital ring map. For any set of variables $(t_{i})$ and any ideal $I \subseteq A [t_{i}]$ , $(A [t_{i}] / I) \otimes_{A} C ≅ C [t_{i}] / I C [t_{i}] .$ Here the extended ideal is generated by the coefficient images of all elements of $I$ . For a multiplicative subset $M \subseteq A$ , $(M^{- 1} A) \otimes_{A} C ≅ {\overline{M}}^{- 1} C .$ These are ring isomorphisms; no flatness, finite-generation or nonzero-ring hypothesis is required. (Presentations and localization under base extension)

Verification

1.1givenF1F2

F1 and F2 give the fibre ring $(A / I) \otimes_{A} κ (p) ≅ κ (p) / I κ (p)$ .

2.1step 1.1algebra∎

If $I \subseteq p$ , every element maps to zero, so the quotient is the residue field. Otherwise some element of $I$ maps to a nonzero element of the field, hence a unit, so the quotient is zero and its spectrum empty. These cases exhaust all ideals, including $I = 0$ and $I = A$ . The argument applies to any prime, not just a maximal one.

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The ideal of a polynomial graph

Example

Let $k$ be a field, $n, m \geq 0$ , and let $u : A_{k}^{n} \to A_{k}^{m}$ be given by polynomials $f_{1}, \dots, f_{m} \in k [x_{1}, \dots, x_{n}]$ . Its graph is the closed subscheme of $A_{k}^{n + m}$ with ideal $(y_{1} - f_{1} (x), \dots, y_{m} - f_{m} (x)) .$

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

For an $S$ -morphism $u : X \to Y$ (as in def-scheme-over-base), the graph morphism is $Γ_{u} = ({id}_{X}, u) : X \to X \times_{S} Y$ , supplied by thm-fibre-products-of-schemes-exist. Its first projection is the identity and its second projection is $u$ . The definition alone does not assert that its image is closed. (The graph morphism over a base)

[F2]

For an $S$ -morphism $u : X \to Y$ , put $H = (u {pr}_{X}, {pr}_{Y}) : X \times_{S} Y \to Y \times_{S} Y$ . The square with top arrow $Γ_{u} : X \to X \times_{S} Y$ , bottom arrow $Δ_{Y / S} : Y \to Y \times_{S} Y$ , left arrow $u$ , and right arrow $H$ is Cartesian. (The graph is a pullback of the diagonal)

[F3]

For a ring $A$ , closed immersions $Z \to Spec A$ are, up to unique isomorphism over $Spec A$ , precisely the morphisms $Spec (A / I) \to Spec A$ for ideals $I \subseteq A$ . (Closed immersions into affine schemes are quotient spectra)

[F4]

Let $A \to C$ be a unital ring map. For any set of variables $(t_{i})$ and any ideal $I \subseteq A [t_{i}]$ , $(A [t_{i}] / I) \otimes_{A} C ≅ C [t_{i}] / I C [t_{i}] .$ Here the extended ideal is generated by the coefficient images of all elements of $I$ . For a multiplicative subset $M \subseteq A$ , $(M^{- 1} A) \otimes_{A} C ≅ {\overline{M}}^{- 1} C .$ These are ring isomorphisms; no flatness, finite-generation or nonzero-ring hypothesis is required. (Presentations and localization under base extension)

Verification

1.1givenF1F3algebra

By F1 the graph has coordinate map sending $x_{i}$ to $x_{i}$ and $y_{j}$ to $f_{j} (x)$ . This is a surjection to $k [x_{1}, \dots, x_{n}]$ . In the quotient by the displayed ideal, successive substitution replaces every $y_{j}$ by $f_{j} (x)$ , giving inverse ring maps with $k [x]$ . Thus the displayed ideal is exactly the kernel; F3 makes the morphism a closed immersion.

2.1F2F4step 1.1∎

The diagonal in $A_{k}^{m} \times_{k} A_{k}^{m}$ has ideal $(z_{j} - y_{j})$ by the same substitution calculation. Pulling it back along $(u {pr}_{X}, {pr}_{Y})$ sends those generators to $f_{j} (x) - y_{j}$ . By F2 this pullback is the graph; F4 computes its quotient ideal and gives the identical presentation. If $m = 0$ the list is empty and the graph is the identity; if $n = 0$ it is the point given by the constants $f_{j}$ .

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Equal fibre points can have different multiplicities

Statement refuted

False claim: the underlying set of a scheme fibre together with its pointwise residue fields determines that fibre up to scheme isomorphism. For any field $k$ , the zero fibres of $x^{2} = t$ and $x^{3} = t$ have the same one-point underlying space and residue field $k$ , but are not isomorphic as schemes. Their coordinate rings have $k$ -dimensions two and three.

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

Let $A \to B$ be a ring map, $p \in Spec A$ , and $M = A ∖ p$ , acting on $B$ through the ring map. The fibre over $p$ is canonically $Spec (B \otimes_{A} κ (p)) ≅ Spec (M^{- 1} B / p M^{- 1} B) .$ The residue field is $κ (p) = A_{p} / p A_{p}$ . No reduction of the tensor ring is taken. (Coordinate ring of an affine fibre)

[F2]

For commutative unital rings $A, B$ , the assignment $φ \mapsto Spec (φ)$ gives a natural bijection ${Hom}_{C R i n g} (A, B) ≅ {Hom}_{L R S} (Spec B, Spec A) .$ Consequently $A \mapsto Spec A$ is a contravariant equivalence from commutative rings to affine schemes, with quasi-inverse global sections. (Affine schemes are contravariantly equivalent to commutative rings)

Counterexample

1.1givenF1algebra

F1 identifies the two fibre rings as $R_{2} = k [x] / (x^{2})$ and $R_{3} = k [x] / (x^{3})$ . Each has exactly one prime, $(x)$ , since every prime contains the nilpotent class of $x$ and the quotient is the field $k$ . Hence their underlying spaces and pointwise residue fields agree.

2.1F2step 1.1algebra∎

The monomial classes $1, x, \dots, x^{r - 1}$ form a $k$ -basis of $R_{r}$ by division by $x^{r}$ . Thus their dimensions are two and three, excluding a $k$ -algebra isomorphism. To exclude even an abstract ring isomorphism, note that the nilradical of $R_{2}$ has square zero, whereas the nilradical of $R_{3}$ has nonzero square generated by $x^{2}$ . Every ring isomorphism preserves the nilradical and its powers. F2 therefore excludes any scheme isomorphism. These rings are nonzero and the argument works in every characteristic.

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An empty fibre from a zero tensor ring

Example

For the open immersion $Spec k [t, t^{- 1}] \to Spec k [t]$ , the fibre at $t = 0$ is empty: its tensor coordinate ring is the zero ring.

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

Let $A \to B$ be a ring map, $p \in Spec A$ , and $M = A ∖ p$ , acting on $B$ through the ring map. The fibre over $p$ is canonically $Spec (B \otimes_{A} κ (p)) ≅ Spec (M^{- 1} B / p M^{- 1} B) .$ The residue field is $κ (p) = A_{p} / p A_{p}$ . No reduction of the tensor ring is taken. (Coordinate ring of an affine fibre)

[F2]

Let $A \to C$ be a unital ring map. For any set of variables $(t_{i})$ and any ideal $I \subseteq A [t_{i}]$ , $(A [t_{i}] / I) \otimes_{A} C ≅ C [t_{i}] / I C [t_{i}] .$ Here the extended ideal is generated by the coefficient images of all elements of $I$ . For a multiplicative subset $M \subseteq A$ , $(M^{- 1} A) \otimes_{A} C ≅ {\overline{M}}^{- 1} C .$ These are ring isomorphisms; no flatness, finite-generation or nonzero-ring hypothesis is required. (Presentations and localization under base extension)

Verification

1.1givenF1F2

F1 computes its ring as $k [t, t^{- 1}] \otimes_{k [t]} k$ , with $t$ acting as zero on $k$ . By F2 this is the localization of $k$ at the image of the powers of $t$ .

2.1step 1.1algebra∎

In this ring $t$ is simultaneously zero and invertible, forcing $1 = t t^{- 1} = 0$ . Thus the ring is zero and has no prime ideals, so the fibre is empty. This agrees with the absence of $(t)$ from $D (t)$ . At $t = 1$ the same formula instead gives $k$ , since 1 is already invertible; the emptiness depends on the specified point.

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A reduced field spectrum becomes nonreduced

Statement refuted

False claim: reducedness survives algebraic extension of the ground field. Let $p$ be prime, $k = F_{p} (v)$ with $v$ transcendental, and $L = k [u] / (u^{p} - v)$ . Then $L$ is a field, but $L \otimes_{k} L ≅ L [ϵ] / (ϵ^{p}),$ so the base change of the reduced $k$ -scheme $Spec L$ along the finite algebraic extension $L / k$ is nonreduced.

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

For a field extension $K / k$ and a $k$ -scheme $X$ , the inverse image of every affine open $U = Spec A$ in $X_{K}$ is $Spec (A \otimes_{k} K)$ . These affine charts cover $X_{K}$ and are compatible on overlaps and with coefficient localizations. (Affine charts after extension of the ground field)

[F2]

Let $A \to C$ be a unital ring map. For any set of variables $(t_{i})$ and any ideal $I \subseteq A [t_{i}]$ , $(A [t_{i}] / I) \otimes_{A} C ≅ C [t_{i}] / I C [t_{i}] .$ Here the extended ideal is generated by the coefficient images of all elements of $I$ . For a multiplicative subset $M \subseteq A$ , $(M^{- 1} A) \otimes_{A} C ≅ {\overline{M}}^{- 1} C .$ These are ring isomorphisms; no flatness, finite-generation or nonzero-ring hypothesis is required. (Presentations and localization under base extension)

[F3]

Let $F$ have characteristic $p > 0$ , let $a \in F$ not be a $p$ th power in $F$ , and let $n \geq 1$ . Then $x^{p^{n}} - a$ is irreducible in $F [x]$ . (If $a$ is not a $p$ th power in a characteristic- $p$ field, then $x^{p^{n}} - a$ is irreducible for every $n \geq 1$ )

[F4]

For every field $F$ , the polynomial ring $F [x]$ is a unique factorisation domain. (For every field $F$ , $F [x]$ is a unique factorisation domain)

Counterexample

1.1givenF3F4

By F4, $F_{p} [v]$ is a UFD. If $v = (a / b)^{p}$ for nonzero polynomials $a, b$ , comparison of the exponent of the prime polynomial $v$ in $v b^{p} = a^{p}$ gives $1 + p {ord}_{v} (b) = p {ord}_{v} (a)$ , impossible modulo $p$ . Thus $v$ is not a $p$ th power in $k$ , and F3 with exponent parameter 1 proves $u^{p} - v$ irreducible. Its quotient $L$ is a field of degree $p$ over $k$ .

2.1F1F2step 1.1algebra

By F1 and F2 the base-change ring is $L [z] / (z^{p} - v)$ . In $L$ one has $v = u^{p}$ , and the characteristic- $p$ binomial formula gives $z^{p} - u^{p} = (z - u)^{p}$ . Substitution $ϵ = z - u$ supplies the claimed ring isomorphism, with inverse $z = u + ϵ$ .

3.1step 1.1step 2.1algebra∎

The classes $1, ϵ, \dots, ϵ^{p - 1}$ form a basis by division by the monic polynomial $ϵ^{p}$ . Since $p \geq 2$ , $ϵ$ is nonzero but nilpotent. Thus this ring, whose unique prime is $(ϵ)$ , is nonreduced whereas $L$ is reduced. The argument includes $p = 2$ ; no integer-only Eisenstein criterion is used.

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The self fibre product of the quadratic cover

Example

For $A_{k}^{1} \to A_{k}^{1}$ given by $t \mapsto x^{2}$ , the self fibre product is $Spec k [x, y] / (x^{2} - y^{2}) .$ If $char k \neq 2$ it is a reduced union of two distinct line components; if $char k = 2$ it is the doubled diagonal.

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

Let $A \to B$ and $A \to C$ be maps of commutative unital rings, allowing the zero ring. In the category of all schemes, $Spec B \times_{Spec A} Spec C ≅ Spec (B \otimes_{A} C) .$ The projections correspond to $b \mapsto b \otimes 1$ and $c \mapsto 1 \otimes c$ . (Affine fibre products are spectra of tensor products)

[F2]

Let $A \to C$ be a unital ring map. For any set of variables $(t_{i})$ and any ideal $I \subseteq A [t_{i}]$ , $(A [t_{i}] / I) \otimes_{A} C ≅ C [t_{i}] / I C [t_{i}] .$ Here the extended ideal is generated by the coefficient images of all elements of $I$ . For a multiplicative subset $M \subseteq A$ , $(M^{- 1} A) \otimes_{A} C ≅ {\overline{M}}^{- 1} C .$ These are ring isomorphisms; no flatness, finite-generation or nonzero-ring hypothesis is required. (Presentations and localization under base extension)

Verification

1.1givenF1F2

Use F1 with the two maps from $k [t]$ . F2 eliminates $t$ from the presentation $k [x, y, t] / (t - x^{2}, t - y^{2})$ and obtains $k [x, y] / (x^{2} - y^{2})$ , with the two required projections.

2.1step 1.1algebra

If $2 \neq 0$ in $k$ , the invertible linear substitution $u = x - y, v = x + y$ turns the ring into $k [u, v] / (u v)$ . Every prime over $(u v)$ contains $u$ or $v$ . The ideals $(u), (v)$ are incomparable primes and their intersection is $(u v)$ by monomial comparison. Thus the ring is reduced with exactly two line components meeting at the origin. They are not comaximal: their sum is $(u, v)$ , so this is not a product-ring decomposition.

3.1step 1.1algebra∎

If $char k = 2$ , then $x^{2} - y^{2} = (x - y)^{2}$ . Put $ϵ = x - y$ to obtain $k [y, ϵ] / (ϵ^{2})$ . Its class $ϵ$ is nonzero with square zero, and its reduced support is the diagonal. Both characteristic cases give nonzero rings and exhaust all fields.

Fibre Products Base Change and Scheme Theoretic Fibres — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The fibres of xy=t

Example

Facts & Assumptions

Verification

Quadratic fibres over rational points and the generic point

Example

Facts & Assumptions

Verification

A real conic acquires complex points

Example

Facts & Assumptions

Verification

An integral real scheme splits over the complex numbers

Statement refuted

Facts & Assumptions

Counterexample

The affine plane and the pair of generic points

Example

Facts & Assumptions

Verification

A closed-immersion fibre is one residue point or empty

Example

Facts & Assumptions

Verification

The ideal of a polynomial graph

Example

Facts & Assumptions

Verification

Equal fibre points can have different multiplicities

Statement refuted

Facts & Assumptions

Counterexample

An empty fibre from a zero tensor ring

Example

Facts & Assumptions

Verification

A reduced field spectrum becomes nonreduced

Statement refuted

Facts & Assumptions

Counterexample

The self fibre product of the quadratic cover

Example

Facts & Assumptions

Verification

Sources