Geometric irreducibility, reducedness, and integrality after base change

Choose an embedding \[ k_s\subseteq\bar k. \] The projection \[ X_{\bar k}\longrightarrow X_{k_s} \] is surjective because \(\Spec\bar k\to\Spec k_s\) is faithfully flat and hence surjective, and surjectivity is preserved by base change.

The continuous image of an irreducible topological space is irreducible. Hence irreducibility of \(X_{\bar k}\) implies irreducibility of \(X_{k_s}\).

This is the standard separable-closure criterion for geometric irreducibility. The algebraic content is the following pair of facts.

First, over a separably closed field, an irreducible scheme remains irreducible after every field extension: affinely, tensoring a ring with a unique minimal prime over a separably closed field with a field extension again has a unique minimal prime.

Second, given \(K/k\) there is a common overfield \(\Omega\) containing both \(K\) and \(k_s\) over \(k\). The first fact gives that \[ X_\Omega=(X_{k_s})_\Omega \] is irreducible. The projection \[ X_\Omega\longrightarrow X_K \] is surjective, so \(X_K\) is irreducible.

This is precisely the implication in the separable-closure test for geometric irreducibility.

Step 1 proves \((1)\Rightarrow(2)\), and 2 proves \((2)\Rightarrow(3)\). The implication \((3)\Rightarrow(1)\) is immediate by taking \(K=\bar k\).

Embed the perfect closure \(k_p\) into \(\bar k\). On an affine open \(U=\Spec A\subseteq X\), the ring map \[ A\otimes_k k_p \longrightarrow A\otimes_k\bar k \] is injective because \(k_p\to\bar k\) is an extension of fields and tensoring a \(k_p\)-vector space with \(\bar k\) is faithful.

If \(A\otimes_k k_p\) had a nonzero nilpotent, its image would be a nonzero nilpotent in \(A\otimes_k\bar k\), contradiction. Hence every affine chart of \(X_{k_p}\) is reduced.

The field \(k_p\) is perfect. A reduced finite-type scheme over a perfect field is geometrically reduced: affinely, a reduced finitely generated algebra over a perfect field remains reduced after every field extension.

Choose a common overfield \(\Omega\) of \(K\) and \(k_p\). The preceding fact gives that \[ X_\Omega=(X_{k_p})_\Omega \] is reduced.

On every affine chart of \(X_K\), the base-change map to the corresponding affine chart of \(X_\Omega\) is injective because \(K\to\Omega\) is a field extension. A nilpotent in the source would therefore give a nilpotent in the reduced target. Hence \(X_K\) is reduced.

Equivalently, this is the perfect-closure criterion for geometric reducedness: finite purely inseparable extensions are exactly the obstruction, and they all embed in \(k_p\).

Step 4 proves \((1)\Rightarrow(2)\), and 5 proves \((2)\Rightarrow(3)\). The implication \((3)\Rightarrow(1)\) is immediate by taking \(K=\bar k\).

Take \[ k=\mathbb R, \qquad X=\Spec\mathbb C \] with its natural structure as an \(\mathbb R\)-scheme. Since \(\mathbb C\) is a field, \(X\) is integral.

After base change to \(\mathbb C\), \[ \mathbb C\otimes_{\mathbb R}\mathbb C \cong \mathbb C\times\mathbb C. \] Thus \[ X_\mathbb C \cong \Spec\mathbb C\amalg\Spec\mathbb C \] is reducible. Hence \(X\) is not geometrically irreducible.

Let \[ k=\mathbb F_p(u) \] and set \[ X=\Spec k[x]/(x^p-u). \] The element \(u\) is not a \(p\)th power in \(k\): its valuation at the prime \(u\) is \(1\), whereas a \(p\)th power has valuation divisible by \(p\). Hence \[ x^p-u \] is irreducible over \(k\), so the quotient is a field and \(X\) is integral.

After adjoining \(u^{1/p}\), \[ x^p-u=(x-u^{1/p})^p. \] Therefore \[ X_{k(u^{1/p})} \cong \Spec k(u^{1/p})[x]/(x-u^{1/p})^p \] is nonreduced. Thus \(X\) is not geometrically reduced.

Let \[ k=\mathbb F_p(u,v), \] and let \(w\) satisfy the Artin–Schreier equation \[ w^p-w=v. \] This polynomial is separable because its derivative is \(-1\). It has no root in \(k\): using the valuation at infinity in the variable \(v\), normalized by \[ \operatorname{ord}_\infty(v)=-1, \] one has \[ \operatorname{ord}_\infty(r^p-r) \] either nonnegative, or a negative integer divisible by \(p\), and therefore never \(-1\). An Artin–Schreier polynomial \(T^p-T-v\) with no root is irreducible. Thus \[ L=k(w) \] is a finite separable extension of degree \(p\).

Now put \[ E=L(u^{1/p}), \qquad X=\Spec E. \] The element \(u^{1/p}\) cannot lie in \(L\): otherwise the nontrivial purely inseparable extension \[ k(u^{1/p})/k \] would be a subextension of the separable extension \(L/k\). Hence \[ [E:L]=p. \] Hence \(E\) is a field and \(X\) is integral.

Over an algebraic closure \(\bar k\), the separable extension \(L/k\) splits into \(p\) distinct embeddings, while the purely inseparable equation becomes a \(p\)th power. Consequently \[ E\otimes_k\bar k \cong \prod_{j=1}^{p} \bar k[\epsilon]/(\epsilon^p). \] This ring has more than one minimal prime and has nonzero nilpotents. Therefore \(X_{\bar k}\) is both reducible and nonreduced, so \(X\) is neither geometrically irreducible nor geometrically reduced.

Step 3 proves part (a), step 6 proves part (b), and steps <1>7–<1>9 give the examples requested in part (c).