The Stacks Project


Tag 033J

Chapter 27: Properties of Schemes > Section 27.7: Normal schemes

Lemma 27.7.2. Let $X$ be a scheme. The following are equivalent:

  1. The scheme $X$ is normal.
  2. For every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$ is normal.
  3. There exists an affine open covering $X = \bigcup U_i$ such that each $\mathcal{O}_X(U_i)$ is normal.
  4. There exists an open covering $X = \bigcup X_j$ such that each open subscheme $X_j$ is normal.

Moreover, if $X$ is normal then every open subscheme is normal.

Proof. This is clear from the definitions. $\square$

    The code snippet corresponding to this tag is a part of the file properties.tex and is located in lines 772–786 (see updates for more information).

    \begin{lemma}
    \label{lemma-locally-normal}
    Let $X$ be a scheme. The following are equivalent:
    \begin{enumerate}
    \item The scheme $X$ is normal.
    \item For every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$
    is normal.
    \item There exists an affine open covering $X = \bigcup U_i$ such that
    each $\mathcal{O}_X(U_i)$ is normal.
    \item There exists an open covering $X = \bigcup X_j$
    such that each open subscheme $X_j$ is normal.
    \end{enumerate}
    Moreover, if $X$ is normal then every open subscheme
    is normal.
    \end{lemma}
    
    \begin{proof}
    This is clear from the definitions.
    \end{proof}

    Comments (0)

    There are no comments yet for this tag.

    There is also 1 comment on Section 27.7: Properties of Schemes.

    Add a comment on tag 033J

    Your email address will not be published.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 lower-right corner).

    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 box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?