In the category of varieties, the Zariski topology on a product is not the product topology. In the category of schemes, the underlying point set of a product is not even the product set.
Let \(k\) be a field and let \(\AA_k^1 = \Spec k[x]\) be the affine line over \(k\). Show that \(\fiberprod{\AA_k^1}{\Spec k}{\AA_k^1} \cong \AA_k^2\), and show that the underlying point set of the product is not the product of the underlying point sets of the factors, even if \(k\) is algebraically closed.
Let \(k\) be a field and let \(s\) and \(t\) be indeterminates over \(k\). Then \(\Spec k(s)\), \(\Spec k(t)\), and \(\Spec k\) are all one-point spaces. Describe the product scheme \(\fiberprod{\Spec k(s)}{\Spec k}{\Spec k(t)}\).
Solution
1. The product of the two affine lines over \(k\) is\[ \boxed{ \mathbb A^1_k\times_{\Spec k}\mathbb A^1_k \cong \Spec k[x,y] =\mathbb A^2_k.} \]
For affine schemes, fibre products are computed by tensor products of coordinate rings. Thus \[ \Spec k[x] \times_{\Spec k} \Spec k[y] \cong \Spec\bigl(k[x]\otimes_k k[y]\bigr). \] The natural map \[ k[x]\otimes_k k[y]\longrightarrow k[x,y], \qquad p(x)\otimes q(y)\longmapsto p(x)q(y), \] is an isomorphism, giving the result.
2. Even when \(k\) is algebraically closed, the underlying point set of \(\mathbb A^2_k\) is not the Cartesian product of the underlying point sets of the two copies of \(\mathbb A^1_k\).
Let \(\eta\in\Spec k[x]\) and \(\eta'\in\Spec k[y]\) be the generic points, corresponding to the zero ideals.
Under the two projections \[ \Spec k[x,y]\longrightarrow\Spec k[x], \qquad \Spec k[x,y]\longrightarrow\Spec k[y], \] both of the distinct primes \[ (0) \qquad\text{and}\qquad (x-y) \] map to \[ (\eta,\eta'). \] Indeed, \[ (x-y)\cap k[x]=(0), \qquad (x-y)\cap k[y]=(0). \] Thus the natural map of point sets \[ |\mathbb A^2_k| \longrightarrow |\mathbb A^1_k|\times|\mathbb A^1_k| \] is not injective. Hence the point set of the scheme product is not the set-theoretic product, regardless of algebraic closedness of \(k\).
3. For part (b),\[ \boxed{ \Spec k(s)\times_{\Spec k}\Spec k(t) \cong \Spec\bigl(k(s)\otimes_k k(t)\bigr).} \]
This is again the affine fibre-product formula.
4. The tensor product in 3 is the localization\[ R=k(s)\otimes_k k(t) \cong S^{-1}k[s,t], \]where \(S\) is the multiplicative set generated by all nonzero polynomials in \(k[s]\) and all nonzero polynomials in \(k[t]\).
Write \[ k(s)=\bigl(k[s]\bigr)_{k[s]\setminus\{0\}}, \qquad k(t)=\bigl(k[t]\bigr)_{k[t]\setminus\{0\}}. \] Tensor product commutes with localization, so localizing the two factors before tensoring is the same as first forming \[ k[s]\otimes_k k[t]\cong k[s,t] \] and then inverting the images of all nonzero elements from the two one-variable polynomial rings. This is exactly \(S^{-1}k[s,t]\).
5. The primes of \(R\) correspond to the prime ideals\[ \mathfrak p\subseteq k[s,t] \]such that\[ \mathfrak p\cap k[s]=(0), \qquad \mathfrak p\cap k[t]=(0). \]
Prime ideals of a localization \(S^{-1}A\) correspond to prime ideals of \(A\) disjoint from \(S\). Here being disjoint from \(S\) means containing no nonzero polynomial in \(s\) alone and no nonzero polynomial in \(t\) alone, which is precisely the displayed pair of contraction conditions.
6. The primes in 5 are exactly\[ (0) \]and the height-one primes\[ (h(s,t)), \]where \(h\in k[s,t]\) is irreducible and depends nontrivially on both \(s\) and \(t\).
The zero ideal clearly satisfies the contraction conditions.
Let \(\mathfrak p\ne(0)\) satisfy them. Since \(k[s,t]\) has dimension \(2\), the prime has height \(1\) or \(2\).
It cannot have height \(2\): then it is maximal, and the field \[ k[s,t]/\mathfrak p \] is a finitely generated \(k\)-algebra. By Zariski’s lemma it is a finite algebraic extension of \(k\). The image of \(s\) would therefore be algebraic over \(k\), so some nonzero polynomial in \(k[s]\) would lie in \(\mathfrak p\), contradicting the contraction condition.
Hence \(\mathfrak p\) has height \(1\). Since \(k[s,t]\) is a unique factorization domain, every height-one prime is generated by an irreducible polynomial: \[ \mathfrak p=(h). \] If \(h\) depended only on \(s\), or only on \(t\), the corresponding contraction would be nonzero. Thus \(h\) must involve both variables.
Conversely, let \(h\) be irreducible and involve both variables. If a nonzero \(p(s)\in k[s]\) lay in \((h)\), then \(h\) would divide \(p(s)\) in the UFD \(k[s,t]\), forcing \(h\) to be associated to a polynomial in \(s\) alone, contradiction. Thus \[ (h)\cap k[s]=(0), \] and similarly for \(k[t]\). Hence \((h)\) survives the localization.
7. Therefore the product in part (b) is a one-dimensional scheme with one generic point and closed points indexed by irreducible mixed polynomials \(h(s,t)\), up to multiplication by a nonzero scalar.
By 6, the zero prime is the unique generic point. Every surviving nonzero prime has height one, and there are no larger surviving primes, so it becomes maximal in \(R\) and hence is a closed point.
There are many such points. For example, \[ t-s^n, \qquad n\ge1, \] are irreducible mixed polynomials and give distinct closed points.
8. At the closed point defined by an irreducible mixed polynomial \(h(s,t)\), the residue field is\[ \boxed{ \kappa(h) \cong \operatorname{Frac}\bigl(k[s,t]/(h)\bigr).} \]
Because \(h\) involves \(t\), it is primitive as a polynomial in \(t\) over \(k[s]\) after multiplication by a scalar, and irreducibility in \(k[s,t]\) implies irreducibility over \(k(s)[t]\) by Gauss’s lemma. Localizing \[ k[s,t]/(h) \] at all nonzero elements of \(k[s]\) therefore gives the field \[ k(s)[t]/(h). \] This is the fraction field of \(k[s,t]/(h)\). Further localization by nonzero elements from \(k[t]\) does not change the field. Hence it is the residue field of the corresponding maximal ideal of \(R\).
9. In particular, although each of\[ \Spec k(s),\qquad\Spec k(t),\qquad\Spec k \]has one point, their fibre product has infinitely many points.
Step 7 exhibits the generic point and infinitely many distinct closed points \((t-s^n)\).
10. Q.E.D.
Steps <1>1–<1>2 prove part (a), and steps <1>3–<1>9 describe the product in part (b).