description/proof of that 'rings - homomorphisms' isomorphism is restricted to multiplicative 'groups - homomorphisms' isomorphism
Topics
About: ring
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of ring.
- The reader knows a definition of %category name% isomorphism.
- The reader admits the proposition that for any ring, the subset of the multiplicatively-invertible elements forms a multiplicative group.
- The reader admits the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.
- The reader admits the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that any 'rings - homomorphisms' isomorphism is restricted to the multiplicative 'groups - 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:
\(R_1\): \(\in \{\text{ the rings }\}\)
\(R_2\): \(\in \{\text{ the rings }\}\)
\(f\): \(: R_1 \to R_2\), \(\in \{\text{ the 'rings - homomorphisms' isomorphisms }\}\)
\(G_1\): \(= \{r_1 \in R_1 \vert \exists {r_1}^{- 1} \in R_1\}\) with the induced multiplication, \(\in \{\text{ the groups }\}\)
\(G_2\): \(= \{r_2 \in R_2 \vert \exists {r_2}^{- 1} \in R_2\}\) with the induced multiplication, \(\in \{\text{ the groups }\}\)
\(f^`\): \(: G_1 \to G_2\), \(= \text{ the domain codomain restriction of } f\)
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Statements:
\(f^` \in \{\text{ the 'groups - homomorphisms' isomorphisms }\}\)
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2: Note
\(f\) itself is an additive 'groups - homomorphisms' isomorphism, because being a 'rings - homomorphisms' isomorphism requires it.
3: Proof
Whole Strategy: Step 1: see that \(f^`\) is valid; Step 2: see that \(f^`\) is a group homomorphism; Step 3: see that \(f^`\) is a bijection; Step 4: conclude the proposition.
Step 1:
Let us see that \(f^`\) is valid.
\(G_1\) and \(G_2\) are indeed some groups, by the proposition that for any ring, the subset of the multiplicatively-invertible elements forms a multiplicative group.
Is \(f^`\) indeed into \(G_2\)?
Let \(g_1 \in G_1\) be any.
\(f^` (g_1)^{- 1} = f (g_1)^{- 1} = f (g_1^{- 1})\), because \(f (g_1^{- 1}) f (g_1) = f (g_1^{- 1} g_1)\), because \(f\) is a ring homomorphism, \(= f (1) = 1\), because \(f\) is a ring homomorphism, and \(f (g_1) f (g_1^{- 1}) = f (g_1 g_1^{- 1})\), because \(f\) is a ring homomorphism, \(= f (1) = 1\), because \(f\) is a ring homomorphism.
So, \(f^` (g_1) \in G_2\).
So, \(f^`\) is into \(G_2\).
So, \(f^`\) is valid.
Step 2:
Let us see that \(f^`\) is a group homomorphism.
For each \(g_1, g'_1 \in G_1\), \(f^` (g_1 g'_1) = f (g_1 g'_1) = f (g_1) f (g'_1)\), because \(f\) is a ring homomorphism, \(= f^` (g_1) f^` (g'_1)\).
So, \(f^`\) is a group homomorphism, by the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.
Step 3:
\(f^`\) is an injection, because \(f\) is so.
\(f^`\) is a surjection, because for each \(g_2 \in G_2\), there is an \(r_1 \in R_1\) such that \(f (r_1) = g_2\), because \(f\) is a surjection, but \({g_2}^{- 1} = f (r'_1)\), because \(f\) is a surjection, and \(f (r_1 r'_1) = f (r_1) f (r'_1)\), because \(f\) is a ring homomorphism, \(= g_2 {g_2}^{- 1} = 1\), which implies that \(r_1 r'_1 = 1\), because \(f\) is a 'rings - homomorphisms' isomorphism, and \(f (r'_1 r_1) = f (r'_1) f (r_1)\), because \(f\) is a ring homomorphism, \(= {g_2}^{- 1} g_2 = 1\), which implies that \(r'_1 r_1 = 1\), because \(f\) is a 'rings - homomorphisms' isomorphism, so, \(r'_1 = {r_1}^{- 1}\), so, \(r_1 \in G_1\).
So, \(f^`\) is a bijection.
Step 4:
So, \(f^`\) is a 'groups - homomorphisms' isomorphism, by the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism Bijective Group Homomorphism Is 'Groups - Homomorphisms' Isomorphism
