958: Range of Field Homomorphism Is Subfield of Codomain

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

description/proof of that range of field homomorphism is subfield of codomain

---

Topics

About: field

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of field.
• The reader knows a definition of %structure kind name% homomorphism.
• The reader admits the proposition that the range of any ring homomorphism is a subring of the codomain.

---

Target Context

• The reader will have a description and a proof of the proposition that the range of any field homomorphism is a subfield 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:
$F_1$: $\in \{\text{ the fields }\}$
$F_2$: $\in \{\text{ the fields }\}$
$f$: $:F_1\to F_2$, $\in \{\text{ the field homomorphisms }\}$
//

Statements:
$f(F_1)\in \{\text{ the fields }\}$
//

---

2: Proof

Whole Strategy: Step 1: see that $f(F_1)$ is a subring of $F_2$; Step 2: see that $f(F_1)$ is commutative under the multiplication; Step 3: see that each element of $f(F_1)$ has an inverse.

Step 1:

$F_1$ and $F_2$ are some rings and $f$ is a ring homomorphism.

By the proposition that the range of any ring homomorphism is a subring of the codomain, $f(F_1)$ is a subring of $F_2$.

Step 2:

$f(F_1)$ is commutative under the multiplication, because the multiplication is inherited from ambient $F_2$, which is commutative under the multiplication.

Step 3:

Let us see that each element of $f(F_1)$ has an inverse.

Let $f(r_1)\in f(F_1)$ be any.

$f({r_1}^{-1})\in f(F_1)$ is an inverse of $f(r_1)$: $f(r_1)f({r_1}^{-1})=f(r_1{r_1}^{-1})=f(1)=1$ and $f({r_1}^{-1})f(r_1)=f({r_1}^{-1}{r}_1)=f(1)=1$.

---

References

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