Suppose $R$ is a unitary ring and $S$ has no zero divisor. Prove that if $f$ is a homomorphic function from...
$begingroup$
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
edited Jan 3 at 14:35
rschwieb
107k12102251
asked Jan 3 at 13:36
Arman_jrArman_jr
235
$endgroup$
|
show 3 more comments
$begingroup$
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
edited Jan 3 at 14:35
rschwieb
107k12102251
asked Jan 3 at 13:36
Arman_jrArman_jr
235
$endgroup$
1
$begingroup$
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 $f(1)$?
$endgroup$
– Arthur
Jan 3 at 13:48
$begingroup$
f(1) isnt zero divisor @arthur
$endgroup$
– Arman_jr
Jan 3 at 14:17
$begingroup$
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?
$endgroup$
– Arthur
Jan 3 at 14:20
$begingroup$
@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
$endgroup$
– Arman_jr
Jan 3 at 14:23
$begingroup$
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 $(f(1))^2$?
$endgroup$
– Arthur
Jan 3 at 14:26
|
show 3 more comments
$begingroup$
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
edited Jan 3 at 14:35
rschwieb
107k12102251
asked Jan 3 at 13:36
Arman_jrArman_jr
235
$endgroup$
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
edited Jan 3 at 14:35
rschwieb
107k12102251
asked Jan 3 at 13:36
Arman_jrArman_jr
235
edited Jan 3 at 14:35
rschwieb
107k12102251
asked Jan 3 at 13:36
Arman_jrArman_jr
235
edited Jan 3 at 14:35
rschwieb
107k12102251
edited Jan 3 at 14:35
rschwieb
107k12102251
edited Jan 3 at 14:35
rschwieb
107k12102251
107k12102251
asked Jan 3 at 13:36
Arman_jrArman_jr
235
asked Jan 3 at 13:36
Arman_jrArman_jr
235
asked Jan 3 at 13:36
Arman_jrArman_jr
235
235
1
$begingroup$
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 $f(1)$?
$endgroup$
– Arthur
Jan 3 at 13:48
$begingroup$
f(1) isnt zero divisor @arthur
$endgroup$
– Arman_jr
Jan 3 at 14:17
$begingroup$
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?
$endgroup$
– Arthur
Jan 3 at 14:20
$begingroup$
@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
$endgroup$
– Arman_jr
Jan 3 at 14:23
$begingroup$
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 $(f(1))^2$?
$endgroup$
– Arthur
Jan 3 at 14:26
|
show 3 more comments
1
$begingroup$
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 $f(1)$?
$endgroup$
– Arthur
Jan 3 at 13:48
$begingroup$
f(1) isnt zero divisor @arthur
$endgroup$
– Arman_jr
Jan 3 at 14:17
$begingroup$
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?
$endgroup$
– Arthur
Jan 3 at 14:20
$begingroup$
@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
$endgroup$
– Arman_jr
Jan 3 at 14:23
$begingroup$
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 $(f(1))^2$?
$endgroup$
– Arthur
Jan 3 at 14:26
1
1
$begingroup$
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 $f(1)$?
$endgroup$
– Arthur
Jan 3 at 13:48
$begingroup$
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 $f(1)$?
$endgroup$
– Arthur
Jan 3 at 13:48
$begingroup$
f(1) isnt zero divisor @arthur
$endgroup$
– Arman_jr
Jan 3 at 14:17
$begingroup$
f(1) isnt zero divisor @arthur
$endgroup$
– Arman_jr
Jan 3 at 14:17
$begingroup$
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?
$endgroup$
– Arthur
Jan 3 at 14:20
$begingroup$
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?
$endgroup$
– Arthur
Jan 3 at 14:20
$begingroup$
@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
$endgroup$
– Arman_jr
Jan 3 at 14:23
$begingroup$
@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
$endgroup$
– Arman_jr
Jan 3 at 14:23
$begingroup$
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 $(f(1))^2$?
$endgroup$
– Arthur
Jan 3 at 14:26
$begingroup$
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 $(f(1))^2$?
$endgroup$
– Arthur
Jan 3 at 14:26
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Jan 3 at 14:33
rschwiebrschwieb
107k12102251
$endgroup$
add a comment |
$begingroup$
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 $f(1) = 1_S$.
answered Jan 4 at 15:42
Daniel W.Daniel W.
413
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
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$begingroup$
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.
answered Jan 3 at 14:33
rschwiebrschwieb
107k12102251
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 3 at 14:33
rschwiebrschwieb
107k12102251
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 3 at 14:33
rschwiebrschwieb
107k12102251
$endgroup$
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.
answered Jan 3 at 14:33
rschwiebrschwieb
107k12102251
answered Jan 3 at 14:33
rschwiebrschwieb
107k12102251
answered Jan 3 at 14:33
rschwiebrschwieb
107k12102251
answered Jan 3 at 14:33
rschwiebrschwieb
107k12102251
107k12102251
add a comment |
add a comment |
$begingroup$
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 $f(1) = 1_S$.
answered Jan 4 at 15:42
Daniel W.Daniel W.
413
$endgroup$
add a comment |
$begingroup$
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 $f(1) = 1_S$.
answered Jan 4 at 15:42
Daniel W.Daniel W.
413
$endgroup$
add a comment |
$begingroup$
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 $f(1) = 1_S$.
answered Jan 4 at 15:42
Daniel W.Daniel W.
413
$endgroup$
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 $f(1) = 1_S$.
answered Jan 4 at 15:42
Daniel W.Daniel W.
413
answered Jan 4 at 15:42
Daniel W.Daniel W.
413
answered Jan 4 at 15:42
Daniel W.Daniel W.
413
answered Jan 4 at 15:42
Daniel W.Daniel W.
413
413
add a comment |
add a comment |
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1
$begingroup$
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 $f(1)$?
$endgroup$
– Arthur
Jan 3 at 13:48
$begingroup$
f(1) isnt zero divisor @arthur
$endgroup$
– Arman_jr
Jan 3 at 14:17
$begingroup$
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?
$endgroup$
– Arthur
Jan 3 at 14:20
$begingroup$
@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
$endgroup$
– Arman_jr
Jan 3 at 14:23
$begingroup$
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 $(f(1))^2$?
$endgroup$
– Arthur
Jan 3 at 14:26