Definition 26.12.1. Let $X$ be a scheme. We say $X$ is *reduced* if every local ring $\mathcal{O}_{X, x}$ is reduced.

## 26.12 Reduced schemes

Lemma 26.12.2. A scheme $X$ is reduced if and only if $\mathcal{O}_ X(U)$ is a reduced ring for all $U \subset X$ open.

**Proof.**
Assume that $X$ is reduced. Let $f \in \mathcal{O}_ X(U)$ be a section such that $f^ n = 0$. Then the image of $f$ in $\mathcal{O}_{U, u}$ is zero for all $u \in U$. Hence $f$ is zero, see Sheaves, Lemma 6.11.1. Conversely, assume that $\mathcal{O}_ X(U)$ is reduced for all opens $U$. Pick any nonzero element $f \in \mathcal{O}_{X, x}$. Any representative $(U, f \in \mathcal{O}(U))$ of $f$ is nonzero and hence not nilpotent. Hence $f$ is not nilpotent in $\mathcal{O}_{X, x}$.
$\square$

Lemma 26.12.3. An affine scheme $\mathop{\mathrm{Spec}}(R)$ is reduced if and only if $R$ is reduced.

**Proof.**
The direct implication follows immediately from Lemma 26.12.2 above. In the other direction it follows since any localization of a reduced ring is reduced, and in particular the local rings of a reduced ring are reduced.
$\square$

Lemma 26.12.4. Let $X$ be a scheme. Let $T \subset X$ be a closed subset. There exists a unique closed subscheme $Z \subset X$ with the following properties: (a) the underlying topological space of $Z$ is equal to $T$, and (b) $Z$ is reduced.

**Proof.**
Let $\mathcal{I} \subset \mathcal{O}_ X$ be the sub presheaf defined by the rule

Here we use $f(t)$ to indicate the image of $f$ in the residue field $\kappa (t)$ of $X$ at $t$. Because of the local nature of the condition it is clear that $\mathcal{I}$ is a sheaf of ideals. Moreover, let $U = \mathop{\mathrm{Spec}}(R)$ be an affine open. We may write $T \cap U = V(I)$ for a unique radical ideal $I \subset R$. Given a prime $\mathfrak p \in V(I)$ corresponding to $t \in T \cap U$ and an element $f \in R$ we have $f(t) = 0 \Leftrightarrow f \in \mathfrak p$. Hence $\mathcal{I}(U) = \bigcap _{\mathfrak p \in V(I)} \mathfrak p = I$ by Algebra, Lemma 10.17.2. Moreover, for any standard open $D(g) \subset \mathop{\mathrm{Spec}}(R) = U$ we have $\mathcal{I}(D(g)) = I_ g$ by the same reasoning. Thus $\widetilde I$ and $\mathcal{I}|_ U$ agree (as ideals) on a basis of opens and hence are equal. Therefore $\mathcal{I}$ is a quasi-coherent sheaf of ideals.

At this point we may define $Z$ as the closed subspace associated to the sheaf of ideals $\mathcal{I}$. For every affine open $U = \mathop{\mathrm{Spec}}(R)$ of $X$ we see that $Z \cap U = \mathop{\mathrm{Spec}}(R/I)$ where $I$ is a radical ideal and hence $Z$ is reduced (by Lemma 26.12.3 above). By construction the underlying closed subset of $Z$ is $T$. Hence we have found a closed subscheme with properties (a) and (b).

Let $Z' \subset X$ be a second closed subscheme with properties (a) and (b). For every affine open $U = \mathop{\mathrm{Spec}}(R)$ of $X$ we see that $Z' \cap U = \mathop{\mathrm{Spec}}(R/I')$ for some ideal $I' \subset R$. By Lemma 26.12.3 the ring $R/I'$ is reduced and hence $I'$ is radical. Since $V(I') = T \cap U = V(I)$ we deduced that $I = I'$ by Algebra, Lemma 10.17.2. Hence $Z'$ and $Z$ are defined by the same sheaf of ideals and hence are equal. $\square$

Definition 26.12.5. Let $X$ be a scheme. Let $Z \subset X$ be a closed subset. A *scheme structure on $Z$* is given by a closed subscheme $Z'$ of $X$ whose underlying set is equal to $Z$. We often say “let $(Z, \mathcal{O}_ Z)$ be a scheme structure on $Z$” to indicate this. The *reduced induced scheme structure* on $Z$ is the one constructed in Lemma 26.12.4. The *reduction $X_{red}$ of $X$* is the reduced induced scheme structure on $X$ itself.

Often when we say “let $Z \subset X$ be an irreducible component of $X$” we think of $Z$ as a reduced closed subscheme of $X$ using the reduced induced scheme structure.

Remark 26.12.6. Let $X$ be a scheme. Let $T \subset X$ be a locally closed subset. In this situation we sometimes also use the phrase “reduced induced scheme structure on $T$”. It refers to the reduced induced scheme structure from Definition 26.12.5 when we view $T$ as a closed subset of the open subscheme $X \setminus \partial T$ of $X$. Here $\partial T = \overline{T} \setminus T$ is the “boundary” of $T$ in the topological space of $X$.

Lemma 26.12.7. Let $X$ be a scheme. Let $Z \subset X$ be a closed subscheme. Let $Y$ be a reduced scheme. A morphism $f : Y \to X$ factors through $Z$ if and only if $f(Y) \subset Z$ (set theoretically). In particular, any morphism $Y \to X$ factors as $Y \to X_{red} \to X$.

**Proof.**
Assume $f(Y) \subset Z$ (set theoretically). Let $\mathcal{I} \subset \mathcal{O}_ X$ be the ideal sheaf of $Z$. For any affine opens $U \subset X$, $\mathop{\mathrm{Spec}}(B) = V \subset Y$ with $f(V) \subset U$ and any $g \in \mathcal{I}(U)$ the pullback $b = f^\sharp (g) \in \Gamma (V, \mathcal{O}_ Y) = B$ maps to zero in the residue field of any $y \in V$. In other words $b \in \bigcap _{\mathfrak p \subset B} \mathfrak p$. This implies $b = 0$ as $B$ is reduced (Lemma 26.12.2, and Algebra, Lemma 10.17.2). Hence $f$ factors through $Z$ by Lemma 26.4.6.
$\square$

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