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Tag 0AE9

Chapter 79: Resolution of Surfaces Revisited > Section 79.9: Examples

Example 79.9.1. Let $k$ be a field. The ring $A = k[x, y, z]/(x^r + y^s + z^t)$ is a UFD for $r, s, t$ pairwise coprime integers. Namely, since $x^r + y^s + z^t$ is irreducible $A$ is a domain. The element $z$ is a prime element, i.e., generates a prime ideal in $A$. On the other hand, if $t = 1 + ers$ for some $e$, then $$ A[1/z] \cong k[x', y', 1/z] $$ where $x' = x/z^{es}$, $y' = y/z^{et}$ and $z = (x')^r + (y')^s$. Thus $A[1/z]$ is a localization of a polynomial ring and hence a UFD. It follows from an argument of Nagata that $A$ is a UFD. See Algebra, Lemma 10.119.7. A similar argument can be given if $t$ is not congruent to $1$ modulo $rs$.

    The code snippet corresponding to this tag is a part of the file spaces-resolve.tex and is located in lines 969–988 (see updates for more information).

    \begin{example}
    \label{example-factorial}
    \begin{reference}
    \cite[4(c)]{Samuel-UFD}
    \end{reference}
    Let $k$ be a field. The ring $A = k[x, y, z]/(x^r + y^s + z^t)$
    is a UFD for $r, s, t$ pairwise coprime integers. Namely, since
    $x^r + y^s + z^t$ is irreducible $A$ is a domain. The element $z$
    is a prime element, i.e., generates a prime ideal in $A$.
    On the other hand, if $t = 1 + ers$ for some $e$, then
    $$
    A[1/z] \cong k[x', y', 1/z]
    $$
    where $x' = x/z^{es}$, $y' = y/z^{et}$ and $z = (x')^r + (y')^s$.
    Thus $A[1/z]$ is a localization of a polynomial ring and hence
    a UFD. It follows from an argument of Nagata that $A$ is a UFD.
    See Algebra, Lemma \ref{algebra-lemma-invert-prime-elements}.
    A similar argument can be given if $t$ is not congruent to $1$
    modulo $rs$.
    \end{example}

    References

    [Samuel-UFD, 4(c)]

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