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I have the following commutative algebra question. It came up at the time of reading local rings of schemes and their tangent spaces.

Let $R$ be a local ring with unique maximal ideal $m$. Then is it true that $R/m^2 \simeq R/m \oplus m/m^2$?

Thank you.

KAK
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    Really in a case like this you ought to say what kind of isomorphism you're talking about. It could be a ring isomorphism or vector space isomorphism, or module isomorphism or group isomorphism... Depending on what you say, the responses could be completely different. Please keep it in mind for future posts and try to update this one. – rschwieb Jan 20 '24 at 13:01
  • You should never expect an $R$-module isomorphism between these two modules. In particular, $R/m^2$ is a principally generated $R$-module, with generating set $1+m^2$. However, $R/m\oplus m/m^2$ cannot ever be generated by a single element. – walkar Jan 21 '24 at 02:22

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No. Let $\Bbb{Z}_{(p)}$ be the localization of $\Bbb{Z}$ at the prime ideal $(p)=p\Bbb{Z}$. $$ \begin{split} \Bbb{Z}_{(p)}/p^2\Bbb{Z}_{(p)}\simeq \Bbb{Z}/p^2\Bbb{Z}\\ \Bbb{Z}_{(p)}/p\Bbb{Z}_{(p)}\simeq \Bbb{Z}/p\Bbb{Z}\\ (\Bbb{Z}_{(p)}/p\Bbb{Z}_{(p)})/(\Bbb{Z}_{(p)}/p^2\Bbb{Z}_{(p)})\simeq \Bbb{Z}/p\Bbb{Z}\\ \Bbb{Z}/p^2\Bbb{Z}\not\simeq\Bbb{Z}/p\Bbb{Z}\oplus\Bbb{Z}/p\Bbb{Z} \end{split} $$

Chad K
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  • Thank you for your example. Can you tell how the local ring splits in this (https://math.stackexchange.com/a/339180/1088447) case? – KAK Jan 20 '24 at 09:34
  • @KAK: $k[x]/(x^2)$ is two-dimensional as a vector space over $k$. They're saying $k[x]/(x^2)\simeq k\oplus k$ as a vector space over $k$. But the isomorphism is not an isomorphism as a ring. $k[x]/(x^2)$ has a nilpotent element $x$, which the direct sum $k\oplus k$ of fields does not. – Chad K Jan 20 '24 at 09:47
  • Ok thanks. But why $O_x/m_x^2 $ $\simeq $ $O_x/m_x \oplus m_x/m_x^2$ as $k$ vector space? – KAK Jan 20 '24 at 09:55
  • @KAK: There's a surjection $\tau\colon O\to O/m\simeq k$. Since $\tau(m^2)={0}$, the map factors $\hat{\tau}\colon O/m^2\to O/m\simeq k$. $\ker(\hat{\tau})=m/m^2$. As a map of $k$-modules, $\hat{\tau}$ has a right-inverse $\sigma$ and then $O/m^2\simeq \ker(\hat{\tau})\oplus \text{im}(\sigma)\simeq m/m^2\oplus k$. – Chad K Jan 20 '24 at 10:13