1. ## 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.

2. 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.

3. Okay thanks, but we didn't learn about kernel or homomorphism yet.

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

5. 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?

6. [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

7. Ok, thanks again, I think I get it, even though we haven't covered groups at all. Our book starts with rings, and it is the only thing we have covered in the course besides simple modular arithmetic.