According to this reference an elliptic curve is $F$-pure if and only if the $F$-pure threshold of its defining ideal is $1$. Does there exist an $F$-pure local ring $R=A/\mathfrak{a}$ such that $\text{fpt}(\mathfrak{a})\neq 1$?
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1$\begingroup$ The question isn't worded particularly well. What is A? The $F$-pure threshold takes as input the data of a pair $(R, \mathfrak{a})$ so I assume you are using $(A, \mathfrak{a})$ when you write $fpt(\mathfrak{a})$. There are examples of regular local rings $A$ with $F$-pure quotient $A/\mathfrak{a}$ and yet the $F$-pure threshold of the pair $(A, \mathfrak{a}) \neq 1$; an easy one that comes to mind is taking $A$ to be a polynomial ring in $n^2$ variables and $\mathfrak{a}$ the ideal defined by minors of the matrix of variables. $\endgroup$– lemillerCommented Oct 7, 2016 at 15:29
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$\begingroup$ @lemiller, Thanks. Would you please mention its reference? The question can be modifies as, whether $F$-pure ideals have a particular $F$-pure threshold, something like a function of the dimension of the ring or the codimension of the defining ideal $\endgroup$– AuroraCommented Oct 7, 2016 at 15:32
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1$\begingroup$ This paper has the calculation for the $F$-pure threshold which can easily fail to be $1$. I'm sure there are much easier examples however. $\endgroup$– lemillerCommented Oct 7, 2016 at 15:37
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