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Let S be a subring of the ring R. Thus, by definition (S, +) is an abelian group. Let 0s denote the identity element of this group, and write 0R for the usual "0" of R. Show that 0s = 0R. (See also Exercise 3.54 in Chapter 3 ahead.)

Let S be a subring of the ring R. Thus, by definition (S, +) is an abelian group. Let 0s denote the identity element of this group, and write 0R for the usual "0" of R. Show that 0s = 0R. (See also Exercise 3.54 in Chapter 3 ahead.)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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**Exercise 2.29** Let \( S \) be a subring of the ring \( R \). Thus, by definition \((S, +)\) is an abelian group. Let \( 0_S \) denote the identity element of this group, and write \( 0_R \) for the usual “0” of \( R \). Show that \( 0_S = 0_R \). (See also Exercise 3.54 in Chapter 3 ahead.)
Transcribed Image Text:**Exercise 2.29** Let \( S \) be a subring of the ring \( R \). Thus, by definition \((S, +)\) is an abelian group. Let \( 0_S \) denote the identity element of this group, and write \( 0_R \) for the usual “0” of \( R \). Show that \( 0_S = 0_R \). (See also Exercise 3.54 in Chapter 3 ahead.)
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