]> A ringlet modulo an ideal

A ringlet modulo an ideal

Given a ringlet R and an ideal of that ringlet, we can construct an equivalence on the ringlet that identifies values precisely if adding members of the ideal to each gives the same result; E = (R| r←s: r+i = s+j for some i, j in I |R). As long as I is non-empty, this is reflexive on R (pick any i in the ideal and r+i = r+i for all r in R); it is symmetric by specification; and the additive closure of I ensures it is transitive.

Now suppose E relates R to r and S to s; so we have some i, j, h, k in I for which R+i = r+j, S+h = s+k; thus R.S +(R.h +i.S +i.h) = (R+i).(S+h) = (r+j).(s+k) = r.s +(r.k +j.s +j.k) and, as each term in either end's parentheses has a member of I among its factors, the parenthetical tems of the ends are in I, so we can infer that E relates R.S to r.s; our equivalence respects our mutliplication. Likewise, R +S +(i +h) = r +s +(j +k) shows that the equivalence respects addition. Consequently, when we consider R modulo E, we get a ringlet; I shall denote this R/I, formally the ringlet induced from R by equivalence modulo differences in I.

For i, j in I, as addition is abelian, we have i+j = j+i whence E identifies i with j. Furthermore, as I is closed under additive completions within R, for any r in R that E does identify with any i in I, we have r+h = i+k with h and i+k in I, hence r in I; so everything E identifies with any member of I is in fact in I. For any i in I and r in R, we have i+i also in i and associativity of our addition gives r +(i+i) = (r+i) +i from which, with i+i and i in I, we infer that E identifies r+i with r; thus the members of I (all equivalent to one another and only to one another) are an identity for the addition in R/I (which considers them all to be the same one value). Thus R/I necessarily does have an additive identity.

When a member r of R is considered modulo E, i.e. as a member of R/I, it is usual to denote it r+I, emphasising that we identify it with every r+i for i in I; in the same spirit, the additive identity in R/I may be denoted I rather than 0. (Orthodoxy has R/I's members be the equivalence classes of E, each of which, at least when R is a ring, is {r+i: i in I} for some r in R, which orthodoxy likewise naturally denotes r+I; in this case, I genuinely is one of those equivalence classes.)

Conversely, suppose we have an equivalence E on a ringlet R that respects the addition and multiplication on R; we can then obtain the ringlet R/E in the usual way. In so far as E identifies any distinct values in R for which there is an additive completion, E identifies all such additive completions with one another and they consitute an identity in R/E; whenever E relates r to r+i for any i, R/E treats i as the identity. As E respects R's multiplication, for each s in R it must then also relate r.s to r.s +r.i and R/E treats r.i as an identity; as E respects R's addition, if it also relates some t to t+j then it must relate r+t to r+t +i+j and relate r+t +i to r+t +j, ensuring that r's identitis in R/E are closed under its addition and under completions of that addition. Thus the identities in R/E do form an ideal in R; although, absent additive completion in R, I'm not sure I can show that ideal generates (all of) E; it might relates some r to s for which R provides no additive completion (although I cannot think of a counter-example, so suspect I'm merely failing to see why this can't happen).

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