description/proof of that range of ring homomorphism is subring of codomain
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of ring.
- The reader knows a definition of %structure kind name% homomorphism.
- The reader admits the proposition that for any group homomorphism, the range of the homomorphism is a subgroup of the codomain.
Target Context
- The reader will have a description and a proof of the proposition that the range of any ring homomorphism is a subring of the codomain.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R_1\): \(\in \{\text{ the rings }\}\)
\(R_2\): \(\in \{\text{ the rings }\}\)
\(f\): \(:R_1 \to R_2\), \(\in \{\text{ the ring homomorphisms }\}\)
//
Statements:
\(f (R_1) \in \{\text{ the rings }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(f (R_1)\) is an Abelian group under addition; Step 2: see that \(f (R_1)\) is a monoid under multiplication; Step 3: see that multiplication is distributive with respect to addition.
Step 1:
Let us see that \(f (R_1)\) is an Abelian group under addition.
\(R_1\) and \(R_2\) are some groups under additions and \(f\) is a group homomorphism with respect to the additive groups.
By the proposition that for any group homomorphism, the range of the homomorphism is a subgroup of the codomain, \(f (R_1)\) is an additive subgroup of \(R_2\).
\(f (R_1)\) is Abelian under addition, because the addition is inherited from ambient \(R_2\), which is Abelian under addition.
Step 2:
Let us see that \(f (R_1)\) is a monoid under multiplication.
\(f (R_1)\) is closed under the multiplication: for each \(f (r_1), f (r'_1) \in f (R_1)\), \(f (r_1) f (r'_1) = f (r_1 r'_1) \in f (R_1)\).
The multiplication is associative, because it is inherited from the ambient \(R_2\), whose multiplication is associative.
\(f (R_1)\) has the identity element, because \(f (1) = 1 \in f (R_1)\).
Step 3:
The multiplication is distributive with respect to addition, because it is so in the ambient \(R_2\).
