Answer to Question #179494 in Abstract Algebra for swarnajeet kumar

Question

Which of the following statements are true, and which are false? Give reasons for your answers. i) For any ring R and . a, b R, (a b) a 2ab b 2 2 2 ∈ + = + + ii) Every ring has at least two elements. iii) If R is a ring with identity and I is an ideal of R, then the identity of R I is the same as the identity of R. iv) If f : R → S is a ring homomorphism, then it is a group homomorphism from (R, +) to (S, +). v) If R is a ring, then any ring homomorphism from R ×R into R is surjective

Answer 1

(i) True: Suppose R is a commutative ring. Then $ab=ba$ for every $a, b ∈ R, \text{ so } −ab + ba = 0,$

so by the first equation above, $(a + b)(a − b) = a^2 + 0 − b^2 = a^2 − b^2$ .

(ii) False: There is a unique ring with one element called the zero ring. It is a really bad idea to adopt the convention that the zero ring isn't a ring; without it the category of rings fails to have a terminal object and the tensor product of two rings (which is the coproduct for commutative rings) fails to exist in general.

(iii) True:As with subgroups, we call ideal I proper in R if I = 0 and I= R. Noticethat if R has identity (or “unity”) R and if R is in ideal I then I = R. In fact,if u is a unit in R and u ∈ I, then R = u−1 ∈ I and so I = R. If R ∈ thenI = R. So we can say that a left (or right) nonzero ideal I of ring R with unity Ris proper if and only if I contains no units of R. So a division ring (in which everynonzero element is a unit) has no proper left (or right) ideals.

(iv) False:Let ϕ_(R) = s. Observe that $s = ϕ(R) = ϕ(R.R) = ϕ(R) · ϕ(R) = s^2.$

Then we have $s = s^2$ , and equivalently $s(1 − s) = 0$ . Note that s − 1 = 0 by construction.

Now suppose that $ϕ(R) = 0$ , then, for any r in R $ϕ(r) = ϕ(r · R) = ϕ(r)ϕ(R) = ϕ(r) · 0 = 0$

and thus, $ϕ$ is the zero homomorphism, a contradiction. We conclude that $s(s − 1) = 0$

holds for s = 0. Hence R is not surjective.