959: Bijective Ring Homomorphism Is 'Rings - Homomorphisms' Isomorphism

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that bijective ring homomorphism is 'rings - homomorphisms' isomorphism

---

Topics

About: ring

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of ring.
• The reader knows a definition of bijection.
• The reader knows a definition of %structure kind name% homomorphism.
• The reader knows a definition of %category name% isomorphism.

---

Target Context

• The reader will have a description and a proof of the proposition that any bijective ring homomorphism is a 'rings - homomorphisms' isomorphism.

---

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:
$R_1$: $\in \{\text{ the rings }\}$
$R_2$: $\in \{\text{ the rings }\}$
$f$: $R_1\to R_2$, $\in \{\text{ the bijections }\}\cap \{\text{ the ring homomorphisms }\}$
//

Statements:
$f\in \{\text{ the 'rings - homomorphisms' isomorphisms }\}$
//

---

2: Note

Sometimes, the definition that dictates any bijective ring homomorphism to be a 'rings - homomorphisms' isomorphism is seen, but that does not seem a good practice: 'isomorphism' is a general concept defined in the category theory and requires the inverse to be a homomorphism in the category.

In general, a bijective morphism of a category is not necessarily any %category name% isomorphism. For example, a bijective continuous map, which is a morphism of the 'topological spaces - continuous maps' category, is not necessarily a homeomorphism, which is a 'topological spaces - continuous maps' isomorphism.

As this proposition holds, some people think that that definition that requires only bijective-ness is valid, but just because this proposition holds does not mean that the general definition made in the category theory should be deformed for the ring case.

---

3: Proof

Whole Strategy: Step 1: take the inverse, $f^{-1}$; Step 2: see that $f^{-1}$ is a ring homomorphism.

Step 1:

As $f$ is bijective, there is the inverse, $f^{-1}:R_2\to R_1$.

Step 2:

Let us see that $f^{-1}$ is necessarily ring homomorphic.

$R_1$ and $R_2$ are some additive groups and $f$ is a bijective group homomorphism between the additive groups.

By the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism, $f$ is a 'groups - homomorphisms' isomorphism between the additive groups. So, $f^{-1}$ is a group homomorphism between the additive groups.

$f^{-1}(1)=1$, because $f(1)=1$ and $1=f^{-1}(f(1))=f^{-1}(1)$.

For each $f(r_1),f(r_1^{\prime })\in R_2$, $f^{-1}(f(r_1)f(r_1^{\prime }))=f^{-1}(f(r_1))f^{-1}(f(r_1^{\prime }))$, because $f(r_1{r}_1^{\prime })=f(r_1)f(r_1^{\prime })$ and $f^{-1}(f(r_1))f^{-1}(f(r_1^{\prime }))=r_1{r}_1^{\prime }=f^{-1}(f(r_1{r}_1^{\prime }))=f^{-1}(f(r_1)f(r_1^{\prime }))$.

So, $f^{-1}$ is a ring homomorphism.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>