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Let $R := k[x_1, \cdots, x_n, \cdots]/(x_1^1, \cdots , x^n_n, \cdots),$ where $k$ is a field. Set $I:=(x_1, \cdots, x_n, \cdots)$. the questions are:

Is $\inf\{i\in \mathbb N \cup \{0\}\cup \{\infty\}\mid Ext^i_R(R/I,R)\neq 0\}=0?$

In the case of $S := k[x_1, \cdots, x_n]/(x_1^1, \cdots , x^n_n),$ and $J:=(x_1, \cdots, x_n)$, there is, (If I'm not mistaken), an affirmative answer because of Rees' theorem (B-H,1.2.5). But I've not any idea about the questions above.

What about: $\inf\{i\in \mathbb N \cup \{0\}\cup\{\infty\}\mid \lim_{\to_n} Ext^i_R(R/I^n,R)\neq 0\}=0?$

thank you.

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I think the answers to both questions are yes.

Put a grading on $R$ so that it is connected (zero in negative degrees, $k$ in degree 0): for example, put $x_n$ in degree $n$. (It also seems safest to grade it so that each homogeneous piece is finite-dimensional, although this may not be necessary.) This graded version of $R$ is what Margolis calls a "P-algebra" (H. R. Margolis, Spectra and the Steenrod Algebra, Chapter 13). In the category of bounded below graded modules, $R$ is then injective (Ch. 13, Thm. 12). Since $R/I$ and $R/I^n$ should both be graded modules in this setting, and they are bounded below, so $Ext^i_R(R/I,R)=0=Ext^i_R(R/I^n,R)$ for all positive $i$.

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    $\begingroup$ I don't think he or she wants to put a grading on R. Actually, direct computation shows that the answer is no since there are no module maps from the quotient to the ring. $\endgroup$
    – Fernando Muro
    Commented Apr 30, 2015 at 23:10
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    $\begingroup$ Graded Ext is not necessarily the same as ungraded Ext! $\endgroup$
    – Fred Rohrer
    Commented May 1, 2015 at 5:18
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    $\begingroup$ Fernando: you're right, I was misinterpreting the question as asking whether Ext^i was zero for positive i. $\endgroup$
    – John Palmieri
    Commented May 2, 2015 at 15:11
  • $\begingroup$ Fred: It could be in this case, though. Take a graded free resolution of $R/I$; this is still free in the ungraded case. I think the only issue is that there will be extra ungraded maps from a graded free module to $R$. I wonder if those can be dealt with in some nice way. $\endgroup$
    – John Palmieri
    Commented May 2, 2015 at 15:17
  • $\begingroup$ Fernando Muro is right, but this answer is helpful. $\endgroup$
    – user 1
    Commented May 12, 2015 at 15:48

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