To prove a set is a subring but not an ideal.
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Problem
Let $R$ be the ring of continuous functions from $mathbbR$ to $mathbbR$. Let $A=f in R mid f(0)text is an even integer$. Show that $A$ is a subring of $R$, but not an ideal of $R$.
Silly doubt
Under what binary operations this is a ring? Usually in the case of group it is composition. How to proceed here?
If that part is clear, then this problem can be solved.
abstract-algebra ring-theory
edited Sep 10 at 12:53
egreg
168k1281190
asked Sep 10 at 12:44
blue boy
1,136613
add a comment |Â
up vote
2
down vote
favorite
Problem
Let $R$ be the ring of continuous functions from $mathbbR$ to $mathbbR$. Let $A=f in R mid f(0)text is an even integer$. Show that $A$ is a subring of $R$, but not an ideal of $R$.
Silly doubt
Under what binary operations this is a ring? Usually in the case of group it is composition. How to proceed here?
If that part is clear, then this problem can be solved.
abstract-algebra ring-theory
edited Sep 10 at 12:53
egreg
168k1281190
asked Sep 10 at 12:44
blue boy
1,136613
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Problem
Let $R$ be the ring of continuous functions from $mathbbR$ to $mathbbR$. Let $A=f in R mid f(0)text is an even integer$. Show that $A$ is a subring of $R$, but not an ideal of $R$.
Silly doubt
Under what binary operations this is a ring? Usually in the case of group it is composition. How to proceed here?
If that part is clear, then this problem can be solved.
abstract-algebra ring-theory
edited Sep 10 at 12:53
egreg
168k1281190
asked Sep 10 at 12:44
blue boy
1,136613
Problem
Let $R$ be the ring of continuous functions from $mathbbR$ to $mathbbR$. Let $A=f in R mid f(0)text is an even integer$. Show that $A$ is a subring of $R$, but not an ideal of $R$.
Silly doubt
Under what binary operations this is a ring? Usually in the case of group it is composition. How to proceed here?
If that part is clear, then this problem can be solved.
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Sep 10 at 12:53
egreg
168k1281190
asked Sep 10 at 12:44
blue boy
1,136613
edited Sep 10 at 12:53
egreg
168k1281190
asked Sep 10 at 12:44
blue boy
1,136613
edited Sep 10 at 12:53
egreg
168k1281190
edited Sep 10 at 12:53
egreg
168k1281190
edited Sep 10 at 12:53
egreg
168k1281190
168k1281190
asked Sep 10 at 12:44
blue boy
1,136613
asked Sep 10 at 12:44
blue boy
1,136613
asked Sep 10 at 12:44
blue boy
1,136613
1,136613
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
The operations are addition and multiplication of functions, i.e.
$$
(f+g)(x)=f(x)+g(x) \
(fg)(x)=f(x)g(x).
$$
answered Sep 10 at 12:49
Janik
1,4352418
Thanks for answering. So if nothing is given we assume this ?
– blue boy
Sep 10 at 12:50
@blueboy Yes, you do.
– Janik
Sep 10 at 12:53
add a comment |Â
up vote
3
down vote
The implicit multiplication of the ring is pointwise multiplication of functions, so you only need to find a continuos function that multiplied pointwise with an element of the set takes you out of the set.
answered Sep 10 at 12:52
Fra
532211
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
The operations are addition and multiplication of functions, i.e.
$$
(f+g)(x)=f(x)+g(x) \
(fg)(x)=f(x)g(x).
$$
answered Sep 10 at 12:49
Janik
1,4352418
Thanks for answering. So if nothing is given we assume this ?
– blue boy
Sep 10 at 12:50
@blueboy Yes, you do.
– Janik
Sep 10 at 12:53
add a comment |Â
up vote
6
down vote
accepted
The operations are addition and multiplication of functions, i.e.
$$
(f+g)(x)=f(x)+g(x) \
(fg)(x)=f(x)g(x).
$$
answered Sep 10 at 12:49
Janik
1,4352418
Thanks for answering. So if nothing is given we assume this ?
– blue boy
Sep 10 at 12:50
@blueboy Yes, you do.
– Janik
Sep 10 at 12:53
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
The operations are addition and multiplication of functions, i.e.
$$
(f+g)(x)=f(x)+g(x) \
(fg)(x)=f(x)g(x).
$$
answered Sep 10 at 12:49
Janik
1,4352418
The operations are addition and multiplication of functions, i.e.
$$
(f+g)(x)=f(x)+g(x) \
(fg)(x)=f(x)g(x).
$$
answered Sep 10 at 12:49
Janik
1,4352418
answered Sep 10 at 12:49
Janik
1,4352418
answered Sep 10 at 12:49
Janik
1,4352418
answered Sep 10 at 12:49
Janik
1,4352418
1,4352418
Thanks for answering. So if nothing is given we assume this ?
– blue boy
Sep 10 at 12:50
@blueboy Yes, you do.
– Janik
Sep 10 at 12:53
add a comment |Â
Thanks for answering. So if nothing is given we assume this ?
– blue boy
Sep 10 at 12:50
@blueboy Yes, you do.
– Janik
Sep 10 at 12:53
Thanks for answering. So if nothing is given we assume this ?
– blue boy
Sep 10 at 12:50
Thanks for answering. So if nothing is given we assume this ?
– blue boy
Sep 10 at 12:50
@blueboy Yes, you do.
– Janik
Sep 10 at 12:53
@blueboy Yes, you do.
– Janik
Sep 10 at 12:53
add a comment |Â
up vote
3
down vote
The implicit multiplication of the ring is pointwise multiplication of functions, so you only need to find a continuos function that multiplied pointwise with an element of the set takes you out of the set.
answered Sep 10 at 12:52
Fra
532211
add a comment |Â
up vote
3
down vote
The implicit multiplication of the ring is pointwise multiplication of functions, so you only need to find a continuos function that multiplied pointwise with an element of the set takes you out of the set.
answered Sep 10 at 12:52
Fra
532211
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The implicit multiplication of the ring is pointwise multiplication of functions, so you only need to find a continuos function that multiplied pointwise with an element of the set takes you out of the set.
answered Sep 10 at 12:52
Fra
532211
The implicit multiplication of the ring is pointwise multiplication of functions, so you only need to find a continuos function that multiplied pointwise with an element of the set takes you out of the set.
answered Sep 10 at 12:52
Fra
532211
answered Sep 10 at 12:52
Fra
532211
answered Sep 10 at 12:52
Fra
532211
answered Sep 10 at 12:52
Fra
532211
532211
add a comment |Â
add a comment |Â
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