# Homomorphism between commutative rings

1. Oct 23, 2008

### infinityQ

When I read a ring homomophism of wiki, I found a sentence in the properties section.

Let f:R->S be a ring homomorphism (Assuming R and S have a mulitplicative identity).
"If R and S are commutative, S is a field, and f is surgective, then ker(f) is a maximal ideal of R".

I was trying to prove the above sentence, but I could not finish proving it.

My proof so far is

Step 1: showing ker(f) is a subring of R.
f(0) = 0, f(1) = 1.
Let r belongs to ker(f), called I, which is a subset of R.
Then f(r)=0. f(r + (-r))=0. Thus r has an additive inverse.
f(x+(y+z))=f(x)+f(y+z)=f(x)+f(y)+f(z)=f(x+y)+f(z)=f((x+y)+z)=0 for x,y,z in I, which shows the associative of addition.
f(a(bc)) = f(a)f(bc)=f(a)f(b)f(c)=f(ab)f(c)=f((ab)c)=0, for a,b,c in I.
f(a(b+c)) = f(a)f(b+c)=f(a)(f(b)+f(c))=f(ab)+f(ac)=f(ab+ac)=0

Step 2:showing I(=ker(f)) is an ideal of R.
Let x belongs to R and r belongs to I.
We claim that xr belongs to I. Suppose on the contrary that xr does not belong to I.
Then f(xr)=f(x)f(r) $$\neq$$0, which means both f(x) and f(r) should not be zero, contradicting that r belongs to ker(f). Thus xr belongs to I.

Step 3:Using f is surgective to S which is a field, show that ker(f) is a maximal ideal of R.

Please correct me if something is wrong in step 1&2, and give me some hints for step 3.
Thanks.

Last edited: Oct 23, 2008
2. Oct 24, 2008

### morphism

By the first isomorphism theorem R/kerf is a field. So there won't be a proper ideal in R larger than kerf, because field don't have proper ideals.