The Stacks project

26.12 Reduced schemes

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

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

\[ \mathcal{I}(U) = \{ f \in \mathcal{O}_ X(U) \mid f(t) = 0\text{ for all }t \in T\cap U\} \]

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01IZ. Beware of the difference between the letter 'O' and the digit '0'.