Suppose $R$ is a unitary ring and $S$ has no zero divisor. Prove that if $f$ is a homomorphic function from...












1















Suppose $R$ is a unitary ring and $S$ has no zero divisor. Prove that if $f$ is a homomorphic function from $R$ to $S$ then $S$ is unitary.




My attempt:
I think if S is going to be unitary, then it's unit element should be the image of the unit element of R.
but no idea for starting...






ring-theory







share|cite|improve this question




















  • 1




    You have to assume that $f$ is non-trivial. With that in mind, you might not know yet that $f(1) = 1$, but what do you know about $f(1)$?
    – Arthur
    yesterday












  • f(1) isnt zero divisor @arthur
    – Arman_jr
    yesterday












  • Sure, because there are none in $S$. You're told that $f$ is a homomorphism. Does that tell you anything else about $f(1)$ than it not being a zero divisor?
    – Arthur
    yesterday










  • @arthur if S be unitary then if we look at s in S which is not zero then, f(1).s=s so [f(1) - 1].s=0, because S has no left divisor and s in nonzero so f(1)=1
    – Arman_jr
    yesterday












  • You can't assume $S$ is unitary (because that's what we're asked to prove), and you can't assume $f(1)cdot s = s$ for arbitrary $s$ because, again, we don't know yet that $f(1) = 1$. What about $(f(1))^2$?
    – Arthur
    yesterday


















1















Suppose $R$ is a unitary ring and $S$ has no zero divisor. Prove that if $f$ is a homomorphic function from $R$ to $S$ then $S$ is unitary.




My attempt:
I think if S is going to be unitary, then it's unit element should be the image of the unit element of R.
but no idea for starting...






ring-theory







share|cite|improve this question




















  • 1




    You have to assume that $f$ is non-trivial. With that in mind, you might not know yet that $f(1) = 1$, but what do you know about $f(1)$?
    – Arthur
    yesterday












  • f(1) isnt zero divisor @arthur
    – Arman_jr
    yesterday












  • Sure, because there are none in $S$. You're told that $f$ is a homomorphism. Does that tell you anything else about $f(1)$ than it not being a zero divisor?
    – Arthur
    yesterday










  • @arthur if S be unitary then if we look at s in S which is not zero then, f(1).s=s so [f(1) - 1].s=0, because S has no left divisor and s in nonzero so f(1)=1
    – Arman_jr
    yesterday












  • You can't assume $S$ is unitary (because that's what we're asked to prove), and you can't assume $f(1)cdot s = s$ for arbitrary $s$ because, again, we don't know yet that $f(1) = 1$. What about $(f(1))^2$?
    – Arthur
    yesterday
















1












1








1








Suppose $R$ is a unitary ring and $S$ has no zero divisor. Prove that if $f$ is a homomorphic function from $R$ to $S$ then $S$ is unitary.




My attempt:
I think if S is going to be unitary, then it's unit element should be the image of the unit element of R.
but no idea for starting...






ring-theory







share|cite|improve this question
















Suppose $R$ is a unitary ring and $S$ has no zero divisor. Prove that if $f$ is a homomorphic function from $R$ to $S$ then $S$ is unitary.




My attempt:
I think if S is going to be unitary, then it's unit element should be the image of the unit element of R.
but no idea for starting...






ring-theory




ring-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









rschwieb

105k1299244




105k1299244










asked yesterday









Arman_jr

284




284








  • 1




    You have to assume that $f$ is non-trivial. With that in mind, you might not know yet that $f(1) = 1$, but what do you know about $f(1)$?
    – Arthur
    yesterday












  • f(1) isnt zero divisor @arthur
    – Arman_jr
    yesterday












  • Sure, because there are none in $S$. You're told that $f$ is a homomorphism. Does that tell you anything else about $f(1)$ than it not being a zero divisor?
    – Arthur
    yesterday










  • @arthur if S be unitary then if we look at s in S which is not zero then, f(1).s=s so [f(1) - 1].s=0, because S has no left divisor and s in nonzero so f(1)=1
    – Arman_jr
    yesterday












  • You can't assume $S$ is unitary (because that's what we're asked to prove), and you can't assume $f(1)cdot s = s$ for arbitrary $s$ because, again, we don't know yet that $f(1) = 1$. What about $(f(1))^2$?
    – Arthur
    yesterday
















  • 1




    You have to assume that $f$ is non-trivial. With that in mind, you might not know yet that $f(1) = 1$, but what do you know about $f(1)$?
    – Arthur
    yesterday












  • f(1) isnt zero divisor @arthur
    – Arman_jr
    yesterday












  • Sure, because there are none in $S$. You're told that $f$ is a homomorphism. Does that tell you anything else about $f(1)$ than it not being a zero divisor?
    – Arthur
    yesterday










  • @arthur if S be unitary then if we look at s in S which is not zero then, f(1).s=s so [f(1) - 1].s=0, because S has no left divisor and s in nonzero so f(1)=1
    – Arman_jr
    yesterday












  • You can't assume $S$ is unitary (because that's what we're asked to prove), and you can't assume $f(1)cdot s = s$ for arbitrary $s$ because, again, we don't know yet that $f(1) = 1$. What about $(f(1))^2$?
    – Arthur
    yesterday










1




1




You have to assume that $f$ is non-trivial. With that in mind, you might not know yet that $f(1) = 1$, but what do you know about $f(1)$?
– Arthur
yesterday






You have to assume that $f$ is non-trivial. With that in mind, you might not know yet that $f(1) = 1$, but what do you know about $f(1)$?
– Arthur
yesterday














f(1) isnt zero divisor @arthur
– Arman_jr
yesterday






f(1) isnt zero divisor @arthur
– Arman_jr
yesterday














Sure, because there are none in $S$. You're told that $f$ is a homomorphism. Does that tell you anything else about $f(1)$ than it not being a zero divisor?
– Arthur
yesterday




Sure, because there are none in $S$. You're told that $f$ is a homomorphism. Does that tell you anything else about $f(1)$ than it not being a zero divisor?
– Arthur
yesterday












@arthur if S be unitary then if we look at s in S which is not zero then, f(1).s=s so [f(1) - 1].s=0, because S has no left divisor and s in nonzero so f(1)=1
– Arman_jr
yesterday






@arthur if S be unitary then if we look at s in S which is not zero then, f(1).s=s so [f(1) - 1].s=0, because S has no left divisor and s in nonzero so f(1)=1
– Arman_jr
yesterday














You can't assume $S$ is unitary (because that's what we're asked to prove), and you can't assume $f(1)cdot s = s$ for arbitrary $s$ because, again, we don't know yet that $f(1) = 1$. What about $(f(1))^2$?
– Arthur
yesterday






You can't assume $S$ is unitary (because that's what we're asked to prove), and you can't assume $f(1)cdot s = s$ for arbitrary $s$ because, again, we don't know yet that $f(1) = 1$. What about $(f(1))^2$?
– Arthur
yesterday












2 Answers
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If $f$ is the zero homomorphism, this is false. Let's suppose otherwise from here on out. It turns out that you don't even need to know $S$ has an identity up front:




its unit element should be the image of the unit element of R




This is a good idea. Here is a hint to make it work: verify $f(1)$ is an idempotent of $S$. That reduces the problem to this question.






share|cite|improve this answer





























    0














    Since $f(1)$ is the unit element of $im(f)$ and we assume $f$ to be non-zero, it follows that $f(1) neq 0$.
    Let $a in S$. Then $f(1)^2 a = f(1) a$ and therefore $f(1) a = a$, since $f(1)$ is not a zero divisor. In the same way it follows that $a f(1) = a$, so $f(1) = 1_S$.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      0














      If $f$ is the zero homomorphism, this is false. Let's suppose otherwise from here on out. It turns out that you don't even need to know $S$ has an identity up front:




      its unit element should be the image of the unit element of R




      This is a good idea. Here is a hint to make it work: verify $f(1)$ is an idempotent of $S$. That reduces the problem to this question.






      share|cite|improve this answer


























        0














        If $f$ is the zero homomorphism, this is false. Let's suppose otherwise from here on out. It turns out that you don't even need to know $S$ has an identity up front:




        its unit element should be the image of the unit element of R




        This is a good idea. Here is a hint to make it work: verify $f(1)$ is an idempotent of $S$. That reduces the problem to this question.






        share|cite|improve this answer
























          0












          0








          0






          If $f$ is the zero homomorphism, this is false. Let's suppose otherwise from here on out. It turns out that you don't even need to know $S$ has an identity up front:




          its unit element should be the image of the unit element of R




          This is a good idea. Here is a hint to make it work: verify $f(1)$ is an idempotent of $S$. That reduces the problem to this question.






          share|cite|improve this answer












          If $f$ is the zero homomorphism, this is false. Let's suppose otherwise from here on out. It turns out that you don't even need to know $S$ has an identity up front:




          its unit element should be the image of the unit element of R




          This is a good idea. Here is a hint to make it work: verify $f(1)$ is an idempotent of $S$. That reduces the problem to this question.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          rschwieb

          105k1299244




          105k1299244























              0














              Since $f(1)$ is the unit element of $im(f)$ and we assume $f$ to be non-zero, it follows that $f(1) neq 0$.
              Let $a in S$. Then $f(1)^2 a = f(1) a$ and therefore $f(1) a = a$, since $f(1)$ is not a zero divisor. In the same way it follows that $a f(1) = a$, so $f(1) = 1_S$.






              share|cite|improve this answer


























                0














                Since $f(1)$ is the unit element of $im(f)$ and we assume $f$ to be non-zero, it follows that $f(1) neq 0$.
                Let $a in S$. Then $f(1)^2 a = f(1) a$ and therefore $f(1) a = a$, since $f(1)$ is not a zero divisor. In the same way it follows that $a f(1) = a$, so $f(1) = 1_S$.






                share|cite|improve this answer
























                  0












                  0








                  0






                  Since $f(1)$ is the unit element of $im(f)$ and we assume $f$ to be non-zero, it follows that $f(1) neq 0$.
                  Let $a in S$. Then $f(1)^2 a = f(1) a$ and therefore $f(1) a = a$, since $f(1)$ is not a zero divisor. In the same way it follows that $a f(1) = a$, so $f(1) = 1_S$.






                  share|cite|improve this answer












                  Since $f(1)$ is the unit element of $im(f)$ and we assume $f$ to be non-zero, it follows that $f(1) neq 0$.
                  Let $a in S$. Then $f(1)^2 a = f(1) a$ and therefore $f(1) a = a$, since $f(1)$ is not a zero divisor. In the same way it follows that $a f(1) = a$, so $f(1) = 1_S$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 13 hours ago









                  Daniel W.

                  162




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