Inverse of a ring isomorphism is also an isomorphism?
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1
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Let $f:R rightarrow S$ be an isomorphism of rings.
Let $g:S rightarrow R$ be the inverse function of $f$.
Show that $g$ is also an isomorphism.
I know I can show and isomorphism by first showing $g$ is injective, surjective, and operation preserving. I am unclear on how to do that exactly. Any help would be greatly appreciated.
abstract-algebra ring-theory
asked Nov 19 at 1:14
Raul Quintanilla Jr.
552
add a comment |
up vote
1
down vote
favorite
Let $f:R rightarrow S$ be an isomorphism of rings.
Let $g:S rightarrow R$ be the inverse function of $f$.
Show that $g$ is also an isomorphism.
I know I can show and isomorphism by first showing $g$ is injective, surjective, and operation preserving. I am unclear on how to do that exactly. Any help would be greatly appreciated.
abstract-algebra ring-theory
asked Nov 19 at 1:14
Raul Quintanilla Jr.
552
The only problem here, assuming you already know the basic set theory stuff for inverse functions, is to show the inverse preserves the operations. So suppose $;s,s'in S;$ . You must prove $;g(ss')=g(s)g(s');,;;g(s+s')=g(s)+g(s');$ ...Use bijectivity of $;f;$ for this
– DonAntonio
Nov 19 at 1:19
Related: math.stackexchange.com/questions/2039702/…
– Ethan Bolker
Nov 19 at 1:42
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f:R rightarrow S$ be an isomorphism of rings.
Let $g:S rightarrow R$ be the inverse function of $f$.
Show that $g$ is also an isomorphism.
I know I can show and isomorphism by first showing $g$ is injective, surjective, and operation preserving. I am unclear on how to do that exactly. Any help would be greatly appreciated.
abstract-algebra ring-theory
asked Nov 19 at 1:14
Raul Quintanilla Jr.
552
Let $f:R rightarrow S$ be an isomorphism of rings.
Let $g:S rightarrow R$ be the inverse function of $f$.
Show that $g$ is also an isomorphism.
I know I can show and isomorphism by first showing $g$ is injective, surjective, and operation preserving. I am unclear on how to do that exactly. Any help would be greatly appreciated.
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Nov 19 at 1:14
Raul Quintanilla Jr.
552
asked Nov 19 at 1:14
Raul Quintanilla Jr.
552
asked Nov 19 at 1:14
Raul Quintanilla Jr.
552
asked Nov 19 at 1:14
Raul Quintanilla Jr.
552
asked Nov 19 at 1:14
Raul Quintanilla Jr.
552
552
The only problem here, assuming you already know the basic set theory stuff for inverse functions, is to show the inverse preserves the operations. So suppose $;s,s'in S;$ . You must prove $;g(ss')=g(s)g(s');,;;g(s+s')=g(s)+g(s');$ ...Use bijectivity of $;f;$ for this
– DonAntonio
Nov 19 at 1:19
Related: math.stackexchange.com/questions/2039702/…
– Ethan Bolker
Nov 19 at 1:42
add a comment |
The only problem here, assuming you already know the basic set theory stuff for inverse functions, is to show the inverse preserves the operations. So suppose $;s,s'in S;$ . You must prove $;g(ss')=g(s)g(s');,;;g(s+s')=g(s)+g(s');$ ...Use bijectivity of $;f;$ for this
– DonAntonio
Nov 19 at 1:19
Related: math.stackexchange.com/questions/2039702/…
– Ethan Bolker
Nov 19 at 1:42
The only problem here, assuming you already know the basic set theory stuff for inverse functions, is to show the inverse preserves the operations. So suppose $;s,s'in S;$ . You must prove $;g(ss')=g(s)g(s');,;;g(s+s')=g(s)+g(s');$ ...Use bijectivity of $;f;$ for this
– DonAntonio
Nov 19 at 1:19
The only problem here, assuming you already know the basic set theory stuff for inverse functions, is to show the inverse preserves the operations. So suppose $;s,s'in S;$ . You must prove $;g(ss')=g(s)g(s');,;;g(s+s')=g(s)+g(s');$ ...Use bijectivity of $;f;$ for this
– DonAntonio
Nov 19 at 1:19
Related: math.stackexchange.com/questions/2039702/…
– Ethan Bolker
Nov 19 at 1:42
Related: math.stackexchange.com/questions/2039702/…
– Ethan Bolker
Nov 19 at 1:42
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
because $f$ is a bijection, there are $r,r' in R$ such that $f(r)=s$ and $f(r')=s'$.
So now ,
$g(ss')=g(f(r)f(r'))=g(f(rr'))=rr'=g(s)g(s')$
also
$g(s+s')=g(f(r)+f(r'))=g(f(r+r'))=r+r'=g(s)+g(s')$.
finally,
$g(1)=g(f(1))=1$
Hence, $g$ is a homomorphism and since inverse of a bijective function $f$ is bijective, $g$ is also a bijective hence an isomorphism.
edited Nov 19 at 1:38
Q the Platypus
2,754933
answered Nov 19 at 1:33
mathnoob
1,507320
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
because $f$ is a bijection, there are $r,r' in R$ such that $f(r)=s$ and $f(r')=s'$.
So now ,
$g(ss')=g(f(r)f(r'))=g(f(rr'))=rr'=g(s)g(s')$
also
$g(s+s')=g(f(r)+f(r'))=g(f(r+r'))=r+r'=g(s)+g(s')$.
finally,
$g(1)=g(f(1))=1$
Hence, $g$ is a homomorphism and since inverse of a bijective function $f$ is bijective, $g$ is also a bijective hence an isomorphism.
edited Nov 19 at 1:38
Q the Platypus
2,754933
answered Nov 19 at 1:33
mathnoob
1,507320
add a comment |
up vote
1
down vote
accepted
because $f$ is a bijection, there are $r,r' in R$ such that $f(r)=s$ and $f(r')=s'$.
So now ,
$g(ss')=g(f(r)f(r'))=g(f(rr'))=rr'=g(s)g(s')$
also
$g(s+s')=g(f(r)+f(r'))=g(f(r+r'))=r+r'=g(s)+g(s')$.
finally,
$g(1)=g(f(1))=1$
Hence, $g$ is a homomorphism and since inverse of a bijective function $f$ is bijective, $g$ is also a bijective hence an isomorphism.
edited Nov 19 at 1:38
Q the Platypus
2,754933
answered Nov 19 at 1:33
mathnoob
1,507320
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
because $f$ is a bijection, there are $r,r' in R$ such that $f(r)=s$ and $f(r')=s'$.
So now ,
$g(ss')=g(f(r)f(r'))=g(f(rr'))=rr'=g(s)g(s')$
also
$g(s+s')=g(f(r)+f(r'))=g(f(r+r'))=r+r'=g(s)+g(s')$.
finally,
$g(1)=g(f(1))=1$
Hence, $g$ is a homomorphism and since inverse of a bijective function $f$ is bijective, $g$ is also a bijective hence an isomorphism.
edited Nov 19 at 1:38
Q the Platypus
2,754933
answered Nov 19 at 1:33
mathnoob
1,507320
because $f$ is a bijection, there are $r,r' in R$ such that $f(r)=s$ and $f(r')=s'$.
So now ,
$g(ss')=g(f(r)f(r'))=g(f(rr'))=rr'=g(s)g(s')$
also
$g(s+s')=g(f(r)+f(r'))=g(f(r+r'))=r+r'=g(s)+g(s')$.
finally,
$g(1)=g(f(1))=1$
Hence, $g$ is a homomorphism and since inverse of a bijective function $f$ is bijective, $g$ is also a bijective hence an isomorphism.
edited Nov 19 at 1:38
Q the Platypus
2,754933
answered Nov 19 at 1:33
mathnoob
1,507320
edited Nov 19 at 1:38
Q the Platypus
2,754933
edited Nov 19 at 1:38
Q the Platypus
2,754933
edited Nov 19 at 1:38
Q the Platypus
2,754933
2,754933
answered Nov 19 at 1:33
mathnoob
1,507320
answered Nov 19 at 1:33
mathnoob
1,507320
answered Nov 19 at 1:33
mathnoob
1,507320
1,507320
add a comment |
add a comment |
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The only problem here, assuming you already know the basic set theory stuff for inverse functions, is to show the inverse preserves the operations. So suppose $;s,s'in S;$ . You must prove $;g(ss')=g(s)g(s');,;;g(s+s')=g(s)+g(s');$ ...Use bijectivity of $;f;$ for this
– DonAntonio
Nov 19 at 1:19
Related: math.stackexchange.com/questions/2039702/…
– Ethan Bolker
Nov 19 at 1:42