Find all Ring homomorphisms of $f: mathbb Qto mathbb Q$
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Could you please tell me whether my proof is correct or not,
$f(x)=f(x.1)=f(x).f(1)$ (Since we are finding ring homomorphism and unity belongs to set of Rationals, assuming f as ring homomorphism)
from above it is clear that $f(1)$ is unity element.
also $f(x) = f(1+1+1+1+.....+$x times) $=f(1)+f(1)+f(1)+.......+f(1) = x . f(1)$
or
$f(x) . f(1)= x . f(1) $
$f(1)(f(x)-x)=0$ (since Q is intergral domain, we get two cases)
$f(1) = 0$ corresponds to zero homomorphism
and $f(x) = x$ corresponds to Identity homomorphism
Thus we get only two ring homomorphisms from $f: mathbb Qto mathbb Q$
abstract-algebra ring-theory
asked Nov 24 at 16:39
Rohit Bharadwaj
518
add a comment |
up vote
3
down vote
favorite
Could you please tell me whether my proof is correct or not,
$f(x)=f(x.1)=f(x).f(1)$ (Since we are finding ring homomorphism and unity belongs to set of Rationals, assuming f as ring homomorphism)
from above it is clear that $f(1)$ is unity element.
also $f(x) = f(1+1+1+1+.....+$x times) $=f(1)+f(1)+f(1)+.......+f(1) = x . f(1)$
or
$f(x) . f(1)= x . f(1) $
$f(1)(f(x)-x)=0$ (since Q is intergral domain, we get two cases)
$f(1) = 0$ corresponds to zero homomorphism
and $f(x) = x$ corresponds to Identity homomorphism
Thus we get only two ring homomorphisms from $f: mathbb Qto mathbb Q$
abstract-algebra ring-theory
asked Nov 24 at 16:39
Rohit Bharadwaj
518
$f(x) = f(1+1+cdots+1)$ only holds for $x in mathbb N^*$.
– xbh
Nov 24 at 16:54
so I can't use that to prove the above result for Rational Numbers? but just one doubt, why am i still getting the required results using $f(x) = f(1+1+...+1)$
– Rohit Bharadwaj
Nov 24 at 17:13
@RohitBharadwaj: $forall : minmathbb{Q}quad f(x)=fleft(mcdotdfrac xmright)=mcdot fleft(dfrac xmright)=f(m)cdot fleft(dfrac xmright)implies fleft(dfrac xmright)=0 $ or $f(m)=m$.
– Yadati Kiran
Nov 24 at 17:14
@RohitBharadwaj No, that's not what I meant. I mean your reasoning just shows the possible conclusions for $x in mathbb N^*$, not necessarily applicable to all $x in mathbb Q$.
– xbh
Nov 24 at 17:18
@RohitBharadwaj: The two ring homomorphism are $f(x)=x$ and $f(x)=0$.
– Yadati Kiran
Nov 24 at 17:20
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Could you please tell me whether my proof is correct or not,
$f(x)=f(x.1)=f(x).f(1)$ (Since we are finding ring homomorphism and unity belongs to set of Rationals, assuming f as ring homomorphism)
from above it is clear that $f(1)$ is unity element.
also $f(x) = f(1+1+1+1+.....+$x times) $=f(1)+f(1)+f(1)+.......+f(1) = x . f(1)$
or
$f(x) . f(1)= x . f(1) $
$f(1)(f(x)-x)=0$ (since Q is intergral domain, we get two cases)
$f(1) = 0$ corresponds to zero homomorphism
and $f(x) = x$ corresponds to Identity homomorphism
Thus we get only two ring homomorphisms from $f: mathbb Qto mathbb Q$
abstract-algebra ring-theory
asked Nov 24 at 16:39
Rohit Bharadwaj
518
Could you please tell me whether my proof is correct or not,
$f(x)=f(x.1)=f(x).f(1)$ (Since we are finding ring homomorphism and unity belongs to set of Rationals, assuming f as ring homomorphism)
from above it is clear that $f(1)$ is unity element.
also $f(x) = f(1+1+1+1+.....+$x times) $=f(1)+f(1)+f(1)+.......+f(1) = x . f(1)$
or
$f(x) . f(1)= x . f(1) $
$f(1)(f(x)-x)=0$ (since Q is intergral domain, we get two cases)
$f(1) = 0$ corresponds to zero homomorphism
and $f(x) = x$ corresponds to Identity homomorphism
Thus we get only two ring homomorphisms from $f: mathbb Qto mathbb Q$
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Nov 24 at 16:39
Rohit Bharadwaj
518
asked Nov 24 at 16:39
Rohit Bharadwaj
518
asked Nov 24 at 16:39
Rohit Bharadwaj
518
asked Nov 24 at 16:39
Rohit Bharadwaj
518
asked Nov 24 at 16:39
Rohit Bharadwaj
518
518
$f(x) = f(1+1+cdots+1)$ only holds for $x in mathbb N^*$.
– xbh
Nov 24 at 16:54
so I can't use that to prove the above result for Rational Numbers? but just one doubt, why am i still getting the required results using $f(x) = f(1+1+...+1)$
– Rohit Bharadwaj
Nov 24 at 17:13
@RohitBharadwaj: $forall : minmathbb{Q}quad f(x)=fleft(mcdotdfrac xmright)=mcdot fleft(dfrac xmright)=f(m)cdot fleft(dfrac xmright)implies fleft(dfrac xmright)=0 $ or $f(m)=m$.
– Yadati Kiran
Nov 24 at 17:14
@RohitBharadwaj No, that's not what I meant. I mean your reasoning just shows the possible conclusions for $x in mathbb N^*$, not necessarily applicable to all $x in mathbb Q$.
– xbh
Nov 24 at 17:18
@RohitBharadwaj: The two ring homomorphism are $f(x)=x$ and $f(x)=0$.
– Yadati Kiran
Nov 24 at 17:20
add a comment |
$f(x) = f(1+1+cdots+1)$ only holds for $x in mathbb N^*$.
– xbh
Nov 24 at 16:54
so I can't use that to prove the above result for Rational Numbers? but just one doubt, why am i still getting the required results using $f(x) = f(1+1+...+1)$
– Rohit Bharadwaj
Nov 24 at 17:13
@RohitBharadwaj: $forall : minmathbb{Q}quad f(x)=fleft(mcdotdfrac xmright)=mcdot fleft(dfrac xmright)=f(m)cdot fleft(dfrac xmright)implies fleft(dfrac xmright)=0 $ or $f(m)=m$.
– Yadati Kiran
Nov 24 at 17:14
@RohitBharadwaj No, that's not what I meant. I mean your reasoning just shows the possible conclusions for $x in mathbb N^*$, not necessarily applicable to all $x in mathbb Q$.
– xbh
Nov 24 at 17:18
@RohitBharadwaj: The two ring homomorphism are $f(x)=x$ and $f(x)=0$.
– Yadati Kiran
Nov 24 at 17:20
$f(x) = f(1+1+cdots+1)$ only holds for $x in mathbb N^*$.
– xbh
Nov 24 at 16:54
$f(x) = f(1+1+cdots+1)$ only holds for $x in mathbb N^*$.
– xbh
Nov 24 at 16:54
so I can't use that to prove the above result for Rational Numbers? but just one doubt, why am i still getting the required results using $f(x) = f(1+1+...+1)$
– Rohit Bharadwaj
Nov 24 at 17:13
so I can't use that to prove the above result for Rational Numbers? but just one doubt, why am i still getting the required results using $f(x) = f(1+1+...+1)$
– Rohit Bharadwaj
Nov 24 at 17:13
@RohitBharadwaj: $forall : minmathbb{Q}quad f(x)=fleft(mcdotdfrac xmright)=mcdot fleft(dfrac xmright)=f(m)cdot fleft(dfrac xmright)implies fleft(dfrac xmright)=0 $ or $f(m)=m$.
– Yadati Kiran
Nov 24 at 17:14
@RohitBharadwaj: $forall : minmathbb{Q}quad f(x)=fleft(mcdotdfrac xmright)=mcdot fleft(dfrac xmright)=f(m)cdot fleft(dfrac xmright)implies fleft(dfrac xmright)=0 $ or $f(m)=m$.
– Yadati Kiran
Nov 24 at 17:14
@RohitBharadwaj No, that's not what I meant. I mean your reasoning just shows the possible conclusions for $x in mathbb N^*$, not necessarily applicable to all $x in mathbb Q$.
– xbh
Nov 24 at 17:18
@RohitBharadwaj No, that's not what I meant. I mean your reasoning just shows the possible conclusions for $x in mathbb N^*$, not necessarily applicable to all $x in mathbb Q$.
– xbh
Nov 24 at 17:18
@RohitBharadwaj: The two ring homomorphism are $f(x)=x$ and $f(x)=0$.
– Yadati Kiran
Nov 24 at 17:20
@RohitBharadwaj: The two ring homomorphism are $f(x)=x$ and $f(x)=0$.
– Yadati Kiran
Nov 24 at 17:20
add a comment |
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$f(x) = f(1+1+cdots+1)$ only holds for $x in mathbb N^*$.
– xbh
Nov 24 at 16:54
so I can't use that to prove the above result for Rational Numbers? but just one doubt, why am i still getting the required results using $f(x) = f(1+1+...+1)$
– Rohit Bharadwaj
Nov 24 at 17:13
@RohitBharadwaj: $forall : minmathbb{Q}quad f(x)=fleft(mcdotdfrac xmright)=mcdot fleft(dfrac xmright)=f(m)cdot fleft(dfrac xmright)implies fleft(dfrac xmright)=0 $ or $f(m)=m$.
– Yadati Kiran
Nov 24 at 17:14
@RohitBharadwaj No, that's not what I meant. I mean your reasoning just shows the possible conclusions for $x in mathbb N^*$, not necessarily applicable to all $x in mathbb Q$.
– xbh
Nov 24 at 17:18
@RohitBharadwaj: The two ring homomorphism are $f(x)=x$ and $f(x)=0$.
– Yadati Kiran
Nov 24 at 17:20