Proving f(1) = 1 for a Surjective Ring Function

[mattmns]

I have this question, just curious if what I did is correct.
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Let A and R be rings, and let f:A->R be a function satisfying:

f(a+b) = f(a)+f(b) and f(ab) = f(a)f(b).

Prove that if f is surjective, then f(1) = 1.
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Note: I will use * for multiplication for clarity where things would have just otherwise been next to each other.
Also, note that A and R are commutative rings, both of which have a multiplicative identity 1.

First, R is a ring, so 1 is in R. Also, because this function is surjective, there must be some a in A, that satisfies f(a) = 1. So, we can rewrite f(a) = 1 as: 1 = f(a*1) = f(a)f(1) = 1*f(1) [because f(a) = 1] = f(1) = 1.

That looks sufficient to me, but I don't like it when I don't use something I could have (ie, the + part of the function). Does this look correct? Thanks.

 

[AKG]

Yes, but you have 1 = ... = f(1) = 1. You shouldn't have that underlined part.

 

[mattmns]

Yes, but you have 1 = ... = f(1) = 1. You shouldn't have that underlined part.

Thanks. You are right, I cannot say that f(1) = 1 in that sense.

 

[AKG]

Yes, but of course you don't need to. You have 1 = ... = f(1), and that's all you needed.