# Thread: Ideal of the Ring of Continuous Functions

1. ## Ideal of the Ring of Continuous Functions

Let $T$ be the set of continous functions on the reals ${f : R \rightarrow R}$

It's not hard to show that it forms a ring under "pointwise" addition and multiplication:

$(f+g)(x) = f(x) + g(x)$
$(fg)(x) = f(x)g(x)$

Now, Let $I = \{ f \in T : f(2) = 0\}$

I am trying to
i) Show that $I$ is an ideal in $T$.
ii) Describe T/I using the 1st isomorphism theorem.

Well, in i), I think that if we let $f \in I, g \in T$ then $(fg)(2) = f(2)g(2) = 0g(2) = 0$ so $fg \in I$ so it's an ideal? Also, it contains the 0 element and is closed under "subtraction".

and in ii), I need to think of a ring homomorphism from $T$ to some ring which has $I$ as the kernel, but I can't seem to think what that would be!!

Any help appreciated.... and if anyone could verify my reasoning in (i) that would be great!
Thanks!!

2. Originally Posted by matt.qmar
Let $T$ be the set of continous functions on the reals ${f : R \rightarrow R}$

It's not hard to show that it forms a ring under "pointwise" addition and multiplication:

$(f+g)(x) = f(x) + g(x)$
$(fg)(x) = f(x)g(x)$

Now, Let $I = \{ f \in T : f(2) = 0\}$

I am trying to
i) Show that $I$ is an ideal in $T$.
ii) Describe T/I using the 1st isomorphism theorem.

Well, in i), I think that if we let $f \in I, g \in T$ then $(fg)(2) = f(2)g(2) = 0g(2) = 0$ so $fg \in I$ so it's an ideal? Also, it contains the 0 element and is closed under "subtraction".

and in ii), I need to think of a ring homomorphism from $T$ to some ring which has $I$ as the kernel, but I can't seem to think what that would be!!

Any help appreciated.... and if anyone could verify my reasoning in (i) that would be great!
Thanks!!

Look at the ring homomorphism $\phi:T\rightarrow \mathbb{R}\,,\,\,\phi(f):=f(2)$

Tonio