A two-dimensional, time-dependent analysis of the mammalian retina is presented. A computer simulation of experimental data considers ganglion-cell activity as a two-dimensional process. The properties of the simulation model were examined with several input stimuli. Circular receptive fields enhance not only borders, but other spatial complexities, such as ends of lines and angles. The enhancement is remarkably insensitive to the shape of receptive fields. The processing of the input picture by on-center and off-center receptive fields is complementary but the division of labor varies with the task.
In a recent paper [2] one of the authors has introduced the concept of module type of a ring, for rings with unit. The object of this paper is to generalize this concept to arbitrary rings, without assuming the existence of a unit. This is easily accomplished for rings with one-sided unit, and we shall define the type of such a ring. Theorem 2.5 gives a relation between this type and the module type of [2], and permits the immediate extension of all results in [2] to rings with one-sided unit.
A ring R is said to be a P-ancestral ring if all proper non-zero sub-rings of R have property P. If/ 1 is the property that every proper non-zero sub-ring of R is a (two-sided) ideal then the ring Z of rational integers furnishes an example of a P-ancestral ring.If 5 is a sub-ring of R we define the left-idealizer of S, written I{S), by I(S) = {xeR: xs e S for s e S}. Clearly I(S) is the largest sub-ring of R in which S is a left ideal and I(S) = R if and only if S is a left ideal of R. With obvious changes we may consider right-idealizer and (two-sided) idealizer. We assume R has a unit denoted by 1.Our theorems relate conditions of P-ancestral types to conditions of leftidealizers.Let 5 and T be sub-rings of a ring R. Then the following results are immediate:
0) 16/(5), (ii) S£J(S), (iii) 7(5) s 7(7(5)), (iv) I(S)nl(T)£I(SnT), (v) I(TuS)£l(T)uI(S), (vi) 7(S)ci7(S 2 ).Let D be the ring of all two by two matrices over Z and let
H(oo) -
T = U o y)-x >Then K<= S and I(K) <= I(S) properly. Now S 2 £ S always and by (vi) I(S) £7 (5 2 ). These observations show that knowing the relation between the sub-rings we may still not conclude the direction in which the inclusion relation will go for the left-idealizers. Also in D, 7(r)n7(/<:)c7(rn^) properly and properly. This shows that (iv) and (v) are the best possible results.
Lemma. Let S be a non-zero sub-ring of R. Then I(S) = S if and only if leS.
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