Mixing
Doubt about proof the smallest subfield that contains a ring
Proposition. Let $F$ a field and $Rne{0}$ a subring such that $1_Fin R$. We place $$F'=bigg{ab^{-1};|;ain R, bin Rsetminus{{0}}bigg},$$
the $F'$ is the smalles subfield of $F$ which contains $R$.
The I did not understand just one thing in the proof: because if $1_Fin R$, then $Rsubseteq F'$. Would anyone be kind enough to explain it?
Thanks!
abstract-algebra proof-verification
asked Nov 20 '18 at 17:26
Jack J.
4421419
add a comment |
Proposition. Let $F$ a field and $Rne{0}$ a subring such that $1_Fin R$. We place $$F'=bigg{ab^{-1};|;ain R, bin Rsetminus{{0}}bigg},$$
the $F'$ is the smalles subfield of $F$ which contains $R$.
The I did not understand just one thing in the proof: because if $1_Fin R$, then $Rsubseteq F'$. Would anyone be kind enough to explain it?
Thanks!
abstract-algebra proof-verification
asked Nov 20 '18 at 17:26
Jack J.
4421419
1
The ring must be much more than just that: it must be an integral domain...And you also must define the operations in that $;F';$ that you defined.
– DonAntonio
Nov 20 '18 at 17:54
add a comment |
Proposition. Let $F$ a field and $Rne{0}$ a subring such that $1_Fin R$. We place $$F'=bigg{ab^{-1};|;ain R, bin Rsetminus{{0}}bigg},$$
the $F'$ is the smalles subfield of $F$ which contains $R$.
The I did not understand just one thing in the proof: because if $1_Fin R$, then $Rsubseteq F'$. Would anyone be kind enough to explain it?
Thanks!
abstract-algebra proof-verification
asked Nov 20 '18 at 17:26
Jack J.
4421419
Proposition. Let $F$ a field and $Rne{0}$ a subring such that $1_Fin R$. We place $$F'=bigg{ab^{-1};|;ain R, bin Rsetminus{{0}}bigg},$$
the $F'$ is the smalles subfield of $F$ which contains $R$.
The I did not understand just one thing in the proof: because if $1_Fin R$, then $Rsubseteq F'$. Would anyone be kind enough to explain it?
Thanks!
abstract-algebra proof-verification
abstract-algebra proof-verification
asked Nov 20 '18 at 17:26
Jack J.
4421419
asked Nov 20 '18 at 17:26
Jack J.
4421419
asked Nov 20 '18 at 17:26
Jack J.
4421419
asked Nov 20 '18 at 17:26
Jack J.
4421419
asked Nov 20 '18 at 17:26
Jack J.
4421419
4421419
1
The ring must be much more than just that: it must be an integral domain...And you also must define the operations in that $;F';$ that you defined.
– DonAntonio
Nov 20 '18 at 17:54
add a comment |
1
The ring must be much more than just that: it must be an integral domain...And you also must define the operations in that $;F';$ that you defined.
– DonAntonio
Nov 20 '18 at 17:54
1
1
The ring must be much more than just that: it must be an integral domain...And you also must define the operations in that $;F';$ that you defined.
– DonAntonio
Nov 20 '18 at 17:54
The ring must be much more than just that: it must be an integral domain...And you also must define the operations in that $;F';$ that you defined.
– DonAntonio
Nov 20 '18 at 17:54
add a comment |
1 Answer
1
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oldest
votes
Take any $ain R$ and $b=1_F$. Then $a=ab^{-1}in F'$. Hence $Rsubseteq F'$.
answered Nov 20 '18 at 17:28
user1551
71.5k566125
add a comment |
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Take any $ain R$ and $b=1_F$. Then $a=ab^{-1}in F'$. Hence $Rsubseteq F'$.
answered Nov 20 '18 at 17:28
user1551
71.5k566125
add a comment |
Take any $ain R$ and $b=1_F$. Then $a=ab^{-1}in F'$. Hence $Rsubseteq F'$.
answered Nov 20 '18 at 17:28
user1551
71.5k566125
add a comment |
Take any $ain R$ and $b=1_F$. Then $a=ab^{-1}in F'$. Hence $Rsubseteq F'$.
answered Nov 20 '18 at 17:28
user1551
71.5k566125
Take any $ain R$ and $b=1_F$. Then $a=ab^{-1}in F'$. Hence $Rsubseteq F'$.
answered Nov 20 '18 at 17:28
user1551
71.5k566125
answered Nov 20 '18 at 17:28
user1551
71.5k566125
answered Nov 20 '18 at 17:28
user1551
71.5k566125
answered Nov 20 '18 at 17:28
user1551
71.5k566125
71.5k566125
add a comment |
add a comment |
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1
The ring must be much more than just that: it must be an integral domain...And you also must define the operations in that $;F';$ that you defined.
– DonAntonio
Nov 20 '18 at 17:54