957: Range of Ring Homomorphism Is Subring of Codomain

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description/proof of that range of ring homomorphism is subring of codomain

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Topics

About: ring

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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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.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$R_1$: $\in \{\text{ the rings }\}$
$R_2$: $\in \{\text{ the rings }\}$
$f$: $:R_1\to R_2$, $\in \{\text{ the ring homomorphisms }\}$
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Statements:
$f(R_1)\in \{\text{ the rings }\}$
//

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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 $f(r_1),f(r_1^{\prime })\in f(R_1)$, $f(r_1)f(r_1^{\prime })=f(r_1{r}_1^{\prime })\in f(R_1)$.

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 $R_2$.

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References

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