Lemma 15.45.5. Let $R$ be a local ring. Let $nil(R)$ denote the ideal of nilpotent elements of $R$. Then $nil(R)R^ h = nil(R^ h)$ and $nil(R)R^{sh} = nil(R^{sh})$.

**Proof.**
Note that $nil(R)$ is the biggest ideal consisting of nilpotent elements such that the quotient $R/nil(R)$ is reduced. Note that $nil(R)R^ h$ consists of nilpotent elements by Algebra, Lemma 10.31.3. Also, note that $R^ h/nil(R) R^ h$ is the henselization of $R/nil(R)$ by Algebra, Lemma 10.154.10. Hence $R^ h/nil(R)R^ h$ is reduced by Lemma 15.45.4. We conclude that $nil(R) R^ h = nil(R^ h)$ as desired. Similarly for the strict henselization but using Algebra, Lemma 10.154.16.
$\square$

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