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A noetherian local ring A is said to be analytically unramified if the complete local ring $\hat{A}$ is reduced.

I don't see why it makes sense to call such a ring analytically unramified. The only possibility is, I guess, that $A \to \hat{A}$ is unramified ; meaning $\Omega_{\hat{A}/A} = 0$. But I don't see why $\hat{A}$ reduced would be equivalent to $\Omega_{\hat{A}/A} = 0$.

EDIT : if $A$ is already complete and non-reduced then the map $A \to \hat{A}$ is the identity so it is unramified.

So please, let me know why we use that definition...

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Is $A$ reduced ? –  user18119 Feb 4 '13 at 16:01
    
$A \to \widehat{A}$ is injective. So if $\widehat{A}$ is reduced, then so is $A$. –  Damien L Feb 4 '13 at 16:02
    
Yes I know. But then a complet non-reduced local ring is not analytically unramified. I checked this in Matsumura and I find the definition a little strange. Anyway, in this situation, the $\Omega$ vanishes, but $A$ is not analytically unramified. –  user18119 Feb 4 '13 at 16:11
    
Ho nice ! If $A$ is complete non-reduced then $A \to \widehat{A}$ is the identity so it is unramified. Ok. So then, when this definition ? –  Damien L Feb 4 '13 at 16:14
    
If $A$ is a discrete valuation ring, then $A$ is analytically unramified, but I suspect that $\Omega_{\hat{A}/A}$ is non-zero if $A$ is not excellent as there will be inseparable elements in $\hat{A}$. –  user18119 Feb 4 '13 at 16:20
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