# Continuous functions subring?

• Feb 7th 2010, 03:44 PM
Continuous functions subring?
Let T be the ring of all continuous functions from R to R and let S={f belonging to T such that f(2)=0} Is S a subring of T? Give a proof.

I believe it is.

Closed under addition by letting f belong to T so that f(2)=0 and g belong to T so that g(2)=0. By prop. of cont. functions (f+g)(2)=0 and (fg)(2)=0.

I don't know what the 0 element is in this context. Is it just h(x)=f(2)=0?
I also am not sure about showing that the solution to a+x=0t is in S....

Thanks.
• Feb 7th 2010, 05:50 PM
Bruno J.
Well clearly the function which is identically zero is the zero of the ring and $-f(x)$ is the additive inverse of $f(x)$.

If $C^1$ denotes the ring of continuous functions on $\mathbb R$, then the ring $R$ which you have is the kernel of the ring homomorphism $C^1 \rightarrow \mathbb{R} : f \mapsto f(2)$. Automatically, since $R$ is the kernel of a ring homomorphism, it is also a ring.
• Feb 7th 2010, 06:15 PM
Okay thanks, but we didn't learn about kernel or homomorphism yet.
• Feb 7th 2010, 06:21 PM
tonio
Quote:

Let T be the ring of all continuous functions from R to R and let S={f belonging to T such that f(2)=0} Is S a subring of T? Give a proof.

I believe it is.

Closed under addition by letting f belong to T so that f(2)=0 and g belong to T so that g(2)=0. By prop. of cont. functions (f+g)(2)=0 and (fg)(2)=0.

I don't know what the 0 element is in this context. Is it just h(x)=f(2)=0?
I also am not sure about showing that the solution to a+x=0t is in S....

Thanks.

The answer may well depend on what author (what mathematician) do you ask: many mathematicians I believe, and among them I myself, would require a subset of a unitary ring to contain the ring's multiplicative unit in order to have chance to be considered as a subring, and this case S wouldn't qualify as such, though it surely qualifies as ideal and, in fact, as a maximal one.
Check what requirements your book/teacher demmands for subring.

Tonio
• Feb 8th 2010, 04:47 AM
Okay thanks, but our definition doesn't require multiplicative identity for a ring or a subring.

Subring:

Suppose R is a ring and S is a subset of R such that:
i) S is closed under addition
ii) S is closed under multiplication
iii) Zero element of R belongs to S
iv) If a belongs to S, then the solution to a+x=0 is in S.
Then S is a subring.

And the problem tells you that T is indeed a ring, and of course, S is a subset of that ring, and so we need only verify these four things. I was just unsure if I was going about iii and iv correctly.

Would iii) just be that h(x)=0 clearly belongs to S and iv) Just let a be f(2) and then we can definite x as -f(2), and hence, we have a solution?
• Feb 8th 2010, 04:56 AM
tonio
[quote=twittytwitter;453177]Okay thanks, but our definition doesn't require multiplicative identity for a ring or a subring.

Subring:

Suppose R is a ring and S is a subset of R such that:
i) S is closed under addition
ii) S is closed under multiplication
iii) Zero element of R belongs to S
iv) If a belongs to S, then the solution to a+x=0 is in S.
Then S is a subring.

And the problem tells you that T is indeed a ring, and of course, S is a subset of that ring, and so we need only verify these four things. I was just unsure if I was going about iii and iv correctly.

Would iii) just be that h(x)=0 clearly belongs to S and iv) Just let a be f(2) and then we can definite x as -f(2), and hence, we have a solution?[/qu

No, conditions (i) + (iii) + (iv) tell you that S is a subgroup od the (abelian) additive group of R , so if a= f(x) is in S, then -f(x) is in S, which is easily seen tobe true because f(2) = 0 <==> -f(2) = -0 = 0 .

Tonio
• Feb 8th 2010, 05:00 AM