Find all Ring homomorphisms of $f: mathbb Qto mathbb Q$












3












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    $f(x) = f(1+1+cdots+1)$ only holds for $x in mathbb N^*$.
    $endgroup$
    – xbh
    Nov 24 '18 at 16:54












  • $begingroup$
    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 $f(x) = f(1+1+...+1)$
    $endgroup$
    – Rohit Bharadwaj
    Nov 24 '18 at 17:13










  • $begingroup$
    @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 $f(m)=m$.
    $endgroup$
    – Yadati Kiran
    Nov 24 '18 at 17:14












  • $begingroup$
    @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 $x in mathbb Q$.
    $endgroup$
    – xbh
    Nov 24 '18 at 17:18












  • $begingroup$
    @RohitBharadwaj: The two ring homomorphism are $f(x)=x$ and $f(x)=0$.
    $endgroup$
    – Yadati Kiran
    Nov 24 '18 at 17:20


















3












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    $f(x) = f(1+1+cdots+1)$ only holds for $x in mathbb N^*$.
    $endgroup$
    – xbh
    Nov 24 '18 at 16:54












  • $begingroup$
    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 $f(x) = f(1+1+...+1)$
    $endgroup$
    – Rohit Bharadwaj
    Nov 24 '18 at 17:13










  • $begingroup$
    @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 $f(m)=m$.
    $endgroup$
    – Yadati Kiran
    Nov 24 '18 at 17:14












  • $begingroup$
    @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 $x in mathbb Q$.
    $endgroup$
    – xbh
    Nov 24 '18 at 17:18












  • $begingroup$
    @RohitBharadwaj: The two ring homomorphism are $f(x)=x$ and $f(x)=0$.
    $endgroup$
    – Yadati Kiran
    Nov 24 '18 at 17:20
















3












3








3





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 24 '18 at 16:39









Rohit BharadwajRohit Bharadwaj

518




518












  • $begingroup$
    $f(x) = f(1+1+cdots+1)$ only holds for $x in mathbb N^*$.
    $endgroup$
    – xbh
    Nov 24 '18 at 16:54












  • $begingroup$
    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 $f(x) = f(1+1+...+1)$
    $endgroup$
    – Rohit Bharadwaj
    Nov 24 '18 at 17:13










  • $begingroup$
    @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 $f(m)=m$.
    $endgroup$
    – Yadati Kiran
    Nov 24 '18 at 17:14












  • $begingroup$
    @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 $x in mathbb Q$.
    $endgroup$
    – xbh
    Nov 24 '18 at 17:18












  • $begingroup$
    @RohitBharadwaj: The two ring homomorphism are $f(x)=x$ and $f(x)=0$.
    $endgroup$
    – Yadati Kiran
    Nov 24 '18 at 17:20




















  • $begingroup$
    $f(x) = f(1+1+cdots+1)$ only holds for $x in mathbb N^*$.
    $endgroup$
    – xbh
    Nov 24 '18 at 16:54












  • $begingroup$
    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 $f(x) = f(1+1+...+1)$
    $endgroup$
    – Rohit Bharadwaj
    Nov 24 '18 at 17:13










  • $begingroup$
    @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 $f(m)=m$.
    $endgroup$
    – Yadati Kiran
    Nov 24 '18 at 17:14












  • $begingroup$
    @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 $x in mathbb Q$.
    $endgroup$
    – xbh
    Nov 24 '18 at 17:18












  • $begingroup$
    @RohitBharadwaj: The two ring homomorphism are $f(x)=x$ and $f(x)=0$.
    $endgroup$
    – Yadati Kiran
    Nov 24 '18 at 17:20


















$begingroup$
$f(x) = f(1+1+cdots+1)$ only holds for $x in mathbb N^*$.
$endgroup$
– xbh
Nov 24 '18 at 16:54






$begingroup$
$f(x) = f(1+1+cdots+1)$ only holds for $x in mathbb N^*$.
$endgroup$
– xbh
Nov 24 '18 at 16:54














$begingroup$
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 $f(x) = f(1+1+...+1)$
$endgroup$
– Rohit Bharadwaj
Nov 24 '18 at 17:13




$begingroup$
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 $f(x) = f(1+1+...+1)$
$endgroup$
– Rohit Bharadwaj
Nov 24 '18 at 17:13












$begingroup$
@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 $f(m)=m$.
$endgroup$
– Yadati Kiran
Nov 24 '18 at 17:14






$begingroup$
@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 $f(m)=m$.
$endgroup$
– Yadati Kiran
Nov 24 '18 at 17:14














$begingroup$
@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 $x in mathbb Q$.
$endgroup$
– xbh
Nov 24 '18 at 17:18






$begingroup$
@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 $x in mathbb Q$.
$endgroup$
– xbh
Nov 24 '18 at 17:18














$begingroup$
@RohitBharadwaj: The two ring homomorphism are $f(x)=x$ and $f(x)=0$.
$endgroup$
– Yadati Kiran
Nov 24 '18 at 17:20






$begingroup$
@RohitBharadwaj: The two ring homomorphism are $f(x)=x$ and $f(x)=0$.
$endgroup$
– Yadati Kiran
Nov 24 '18 at 17:20












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