Is $F=-1, 0, 1$ a subfield of $mathbb C$?

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Is $F=-1, 0, 1$ a subfield of $mathbb C$?



I have just started reading Linear Algebra text by Kenneth Hoffman and Ray Kunze.
They wrote that "any subfield of $mathbb C$ must contain every rational number", this statement implies that $F$ is not a subfield of $mathbb C$ even though the operations of addition and multiplication on $F$ are defined in $F$.



Am I overlooking some fact?




field-theory




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  • What's $1+1$? $$
    – John Ma
    Jul 16 at 8:49






  • 1




    Then how is addition defined in $-1,0,1subseteq mathbb C$?
    – drhab
    Jul 16 at 8:50















up vote
0
down vote

favorite












Is $F=-1, 0, 1$ a subfield of $mathbb C$?



I have just started reading Linear Algebra text by Kenneth Hoffman and Ray Kunze.
They wrote that "any subfield of $mathbb C$ must contain every rational number", this statement implies that $F$ is not a subfield of $mathbb C$ even though the operations of addition and multiplication on $F$ are defined in $F$.



Am I overlooking some fact?




field-theory




share|cite|improve this question





















  • What's $1+1$? $$
    – John Ma
    Jul 16 at 8:49






  • 1




    Then how is addition defined in $-1,0,1subseteq mathbb C$?
    – drhab
    Jul 16 at 8:50













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Is $F=-1, 0, 1$ a subfield of $mathbb C$?



I have just started reading Linear Algebra text by Kenneth Hoffman and Ray Kunze.
They wrote that "any subfield of $mathbb C$ must contain every rational number", this statement implies that $F$ is not a subfield of $mathbb C$ even though the operations of addition and multiplication on $F$ are defined in $F$.



Am I overlooking some fact?




field-theory




share|cite|improve this question













Is $F=-1, 0, 1$ a subfield of $mathbb C$?



I have just started reading Linear Algebra text by Kenneth Hoffman and Ray Kunze.
They wrote that "any subfield of $mathbb C$ must contain every rational number", this statement implies that $F$ is not a subfield of $mathbb C$ even though the operations of addition and multiplication on $F$ are defined in $F$.



Am I overlooking some fact?





field-theory





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share|cite|improve this question




share|cite|improve this question








edited Jul 16 at 8:51









John Ma

37.5k93669




37.5k93669









asked Jul 16 at 8:48









Satishchandra Chitrapu

41




41











  • What's $1+1$? $$
    – John Ma
    Jul 16 at 8:49






  • 1




    Then how is addition defined in $-1,0,1subseteq mathbb C$?
    – drhab
    Jul 16 at 8:50

















  • What's $1+1$? $$
    – John Ma
    Jul 16 at 8:49






  • 1




    Then how is addition defined in $-1,0,1subseteq mathbb C$?
    – drhab
    Jul 16 at 8:50
















What's $1+1$? $$
– John Ma
Jul 16 at 8:49




What's $1+1$? $$
– John Ma
Jul 16 at 8:49




1




1




Then how is addition defined in $-1,0,1subseteq mathbb C$?
– drhab
Jul 16 at 8:50





Then how is addition defined in $-1,0,1subseteq mathbb C$?
– drhab
Jul 16 at 8:50











1 Answer
1






active

oldest

votes

















up vote
3
down vote













$F$ is not a subfield of $mathbbC$; it is not closed under addition.



Example: $1+1=2notin F$. Although $1$ is an element of $F$, $2$ is not.






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  • If we were to adhere to the common usage of the phrase, the point is that addition and multiplication aren't defined on $F$, for the reason you've said.
    – Saucy O'Path
    Jul 16 at 9:18











  • @SaucyO'Path thank you for pointing that out, I've corrected it.
    – csch2
    Jul 16 at 9:19










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active

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up vote
3
down vote













$F$ is not a subfield of $mathbbC$; it is not closed under addition.



Example: $1+1=2notin F$. Although $1$ is an element of $F$, $2$ is not.






share|cite|improve this answer























  • If we were to adhere to the common usage of the phrase, the point is that addition and multiplication aren't defined on $F$, for the reason you've said.
    – Saucy O'Path
    Jul 16 at 9:18











  • @SaucyO'Path thank you for pointing that out, I've corrected it.
    – csch2
    Jul 16 at 9:19














up vote
3
down vote













$F$ is not a subfield of $mathbbC$; it is not closed under addition.



Example: $1+1=2notin F$. Although $1$ is an element of $F$, $2$ is not.






share|cite|improve this answer























  • If we were to adhere to the common usage of the phrase, the point is that addition and multiplication aren't defined on $F$, for the reason you've said.
    – Saucy O'Path
    Jul 16 at 9:18











  • @SaucyO'Path thank you for pointing that out, I've corrected it.
    – csch2
    Jul 16 at 9:19












up vote
3
down vote










up vote
3
down vote









$F$ is not a subfield of $mathbbC$; it is not closed under addition.



Example: $1+1=2notin F$. Although $1$ is an element of $F$, $2$ is not.






share|cite|improve this answer















$F$ is not a subfield of $mathbbC$; it is not closed under addition.



Example: $1+1=2notin F$. Although $1$ is an element of $F$, $2$ is not.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 16 at 9:19


























answered Jul 16 at 8:57









csch2

220211




220211











  • If we were to adhere to the common usage of the phrase, the point is that addition and multiplication aren't defined on $F$, for the reason you've said.
    – Saucy O'Path
    Jul 16 at 9:18











  • @SaucyO'Path thank you for pointing that out, I've corrected it.
    – csch2
    Jul 16 at 9:19
















  • If we were to adhere to the common usage of the phrase, the point is that addition and multiplication aren't defined on $F$, for the reason you've said.
    – Saucy O'Path
    Jul 16 at 9:18











  • @SaucyO'Path thank you for pointing that out, I've corrected it.
    – csch2
    Jul 16 at 9:19















If we were to adhere to the common usage of the phrase, the point is that addition and multiplication aren't defined on $F$, for the reason you've said.
– Saucy O'Path
Jul 16 at 9:18





If we were to adhere to the common usage of the phrase, the point is that addition and multiplication aren't defined on $F$, for the reason you've said.
– Saucy O'Path
Jul 16 at 9:18













@SaucyO'Path thank you for pointing that out, I've corrected it.
– csch2
Jul 16 at 9:19




@SaucyO'Path thank you for pointing that out, I've corrected it.
– csch2
Jul 16 at 9:19












 

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