Let $R$ be a commutative ring, and let $f \in R[\![X]\!]$ be a formal power series. Sometimes (and for example, this will always be possible if $R$ is Noetherian), one may write $f$ in the form $$ f = \sum_{i=1}^n a_i u_i X^i, $$ where $a_i \in R$ and $u_i$ is a unit in $R[\![X]\!]$. I call the smallest such $n$ for which such an expression is possible the pseudodegree of $f$.

At least, that's what I and a coauthor called it in a recent preprint. It seemed reasonable, since if one makes the same definition in the polynomial ring $R[X]$, this concept coincides with the usual notion of the degree of a polynomial, at least when $R$ is reduced. Also, like the degree of a polynomial, the pseudodegree of $f$ is at least as big as one less than the minimal number of generators of the ideal of $R$ that is generated by the coefficients of $f$.

But this must have a name somewhere, right? Can anyone point to an article, book, preprint or the like where this concept has been used? What else can be said about pseudodegree?

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    $\begingroup$ A trivial comment. Pseudodegree may not coincide with the usual degree in polynomial rings if the base ring had nilpotents. $\endgroup$ – Mohan Feb 15 '14 at 17:03
  • $\begingroup$ Oh, good point; I'll make the corresponding edit. $\endgroup$ – Neil Epstein Feb 15 '14 at 19:31
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    $\begingroup$ I see. I figured there was something I was missing. $\endgroup$ – Ben Webster Feb 16 '14 at 19:20
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    $\begingroup$ If $f=\sum_{i=0}^\infty a_i X^i$, and $I_n=(a_0,\dots,a_n)$, then this is the $n$ at which the increasing sequence of ideals $I_n$ stabilizes, right? $\endgroup$ – Will Sawin Feb 18 '14 at 3:08
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    $\begingroup$ Question: is the psuedo degree multiplicative? is there a counter-example? $\endgroup$ – user82090 Oct 30 '15 at 9:57

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