961: 'Rings - Homomorphisms' Isomorphism Between Fields Is 'Fields - Homomorphisms' Isomorphism

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description/proof of that 'rings - homomorphisms' isomorphism between fields is 'fields - homomorphisms' isomorphism

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Topics

About: field

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

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Target Context

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

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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:
$F_1$: $\in \{\text{ the fields }\}$
$F_2$: $\in \{\text{ the fields }\}$
$f$: $F_1\to F_2$, $\in \{\text{ the 'rings - homomorphisms' isomorphisms }\}$
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Statements:
$f\in \{\text{ the 'fields - homomorphisms' isomorphisms }\}$
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2: Proof

Whole Strategy: Step 1: see that all what needs to be checked is that for each $r_1\in F_1$, $f(r_1^{-1})=f(r_1{)}^{-1}$; Step 2: see that $f(r_1^{-1})=f(r_1{)}^{-1}$.

Step 1:

As any field is a ring, calling $f$ "'rings - homomorphisms' isomorphism" makes sense.

While only the differences of 'field' from 'ring' are being multiplicative commutative and each element's having the inverse, the commutativity is just about inside $F_1$ or $F_2$, not about $f$, and the only concern about $f$ is whether for each $r_1\in F_1$, $f(r_1^{-1})=f(r_1{)}^{-1}$ and whether for each $r_2\in F_2$, $f^{-1}(r_2^{-1})=f^{-1}(r_2{)}^{-1}$.

If $f(r_1^{-1})=f(r_1{)}^{-1}$ holds, $f^{-1}(r_2^{-1})=f^{-1}(r_2{)}^{-1}$ will hold likewise, because the situation is symmetric.

So, let us check that $f(r_1^{-1})=f(r_1{)}^{-1}$.

Step 2:

$r_1{r}_1^{-1}=r_1^{-1}{r}_1=1$. $f(r_1{r}_1^{-1})=f(r_1^{-1}{r}_1)=f(1)=1$. But $f(r_1{r}_1^{-1})=f(r_1)f(r_1^{-1})$ and $f(r_1^{-1}{r}_1)=f(r_1^{-1})f(r_1)$. So, $f(r_1)f(r_1^{-1})=f(r_1^{-1})f(r_1)=1$, which means that $f(r_1{)}^{-1}=f(r_1^{-1})$.

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References

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