Question

18. Let o: R+ S be a ring homomorphism. Prove each of the following statements. (a) If R is a commutative ring, then (R) is a commutative ring. (b) (0)=0. (c) Let 18 and 1s be the identities for R and S, respectively. If o is onto, then (1r) = 1s. (d) If R is a field and $(R) +0, then (R) is a field.

Answer 1

solution - Given that, let pir“ be a ring homomorphism. if R is that . Now - we have prove to then Ø(R) is a a commutative ring commutative ring proof - R is a commutative oling 86,82 ER » 01. r2 = 82.8 for each 81,82 ER Then there exists S1, S2. ES such that op (81) = s, and $(12) = S2 Now consider, S1: 52 = (81). $(82.) := $(8,12)

. ; = $(x281) $ is homomorphism = $(52). $(81) - S2.si '. I . 52 = 52.511 to Porove NOW we have that (0) = 0 Consider 0 (0) +00) = $(0+X) = $(8) we get, $ (0) +$() = $(V) -> $(01 +*(X). - (V) => $(0) = $(0) - $ (8) : 110860 = 07 (0) = 0

- Given, Ir and is are identities of k and S. Trespectively. O is onto Wole we have to porove that $C!R) = 15 let e- $(IR) » EES Suppose that SES " is onto There exists an Ø (81) = si element VIER such that consider esi = $ CIR) 0 (01) CIR Bu - $ (81) - Si esi = Si and . eSIE SI = Sie te E.S. . CES

→ - e is the multiplicative identity of the ring s. is are identity of 5 Multiplicative identity is unique we have .: - that is e - 1s = * (IR) © if R is a field and $(R) +0 , then OCR is a field. has * → OCR} #0, #XER R is a field, then every element r a multiplicative invenise 9 dity & R torr = IR Consider, =IR

(8N) = (IR) → ☆ .. (81). 008) = 15 ☆ Cou's=1s («) +0. SES. → Ø CVI) is the inverse Of Thus, se S has a multiplicative inverse foor By each the definition of field, we conclude that Is is a field