Homomorphisms between two isomorphic rings ?

  • #1

Homework Statement


True or False?
Let R and S be two isomorphic commutative rings (S=/={0}). Then any ring homomorphism from R to S is an isomorphism.

Homework Equations



R being a commutative ring means it's an abelian group under addition, and has the following additional properties:

i) a*(b+c)=a*b+a*c
ii) ab=ba
iii) a*(b*c)=(a*b)*c
iv) there exists an element eR s.t. a*eR=a for all a in R.



A "ring homomorphism" from R to S is a function f from R to S such that
i) f(a)*f(b)=f(a*b)
ii) f(a+b)=f(a)+f(b)
iii) f(eS)=eR

The Attempt at a Solution


BAck of the book says false

I thought to make f(a)=0S for all a in S which would have worked as a counterexample but but it implies f(eR)=0S which by property (iii) of ring homomorphisms implies eR=0S which means a=a*eS=a*0S=0 so a=0 for all a in S but that means S={0} which is a contradiction.


Thanks for reading
 

Answers and Replies

  • #2
jbunniii
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The Attempt at a Solution


BAck of the book says false

I thought to make f(a)=0S for all a in S which would have worked as a counterexample but but it implies f(eR)=0S which by property (iii) of ring homomorphisms implies eR=0S which means a=a*eS=a*0S=0 so a=0 for all a in S but that means S={0} which is a contradiction.
Consider for example a polynomial ring, such as ##\mathbb{R}[x]##. Clearly it is isomorphic to itself. But there are many homomorphisms from ##\mathbb{R}[x] \to \mathbb{R}[x]## which are not isomorphisms. Can you find one?
 
  • #3
Consider for example a polynomial ring, such as ##\mathbb{R}[x]##. Clearly it is isomorphic to itself. But there are many homomorphisms from ##\mathbb{R}[x] \to \mathbb{R}[x]## which are not isomorphisms. Can you find one?


A homomorphism could me mapping every polynomial to its constant term (term without an x).
Say ##P_{1}(x)=c_{n}x^n+...+c_{1}x+c_{0}## and
##P_{2}(x)=k_{m}x^n+...+k_{1}x+k_{0}##

So ##f(P_{1}(x)P_{2}(x))=c_{0}k_{0}=f(P_{1}(x)f(P_{2}(x))##
Also, ##f(P_{1}(x)+P_{2}(x))=c_{0}+k_{0}=f(P_{1}(x)+f(P_{2}(x))##
And lastly, ##f(1)=1##.

So indeed this is a ring homomorphism, but obviously it is not surjective or injective.


Is this right?
 
  • #4
jbunniii
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A homomorphism could me mapping every polynomial to its constant term (term without an x).
Say ##P_{1}(x)=c_{n}x^n+...+c_{1}x+c_{0}## and
##P_{2}(x)=k_{m}x^n+...+k_{1}x+k_{0}##

So ##f(P_{1}(x)P_{2}(x))=c_{0}k_{0}=f(P_{1}(x)f(P_{2}(x))##
Also, ##f(P_{1}(x)+P_{2}(x))=c_{0}+k_{0}=f(P_{1}(x)+f(P_{2}(x))##
And lastly, ##f(1)=1##.

So indeed this is a ring homomorphism, but obviously it is not surjective or injective.


Is this right?
Yes, that's the example I had in mind.
 

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