# To prove a set is a subring but not an ideal.

Problem

Let $R$ be the ring of continuous functions from $\mathbb{R}$ to $\mathbb{R}$. 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.

## 2 Answers

The operations are addition and multiplication of functions, i.e. $$(f+g)(x)=f(x)+g(x) \\ (fg)(x)=f(x)g(x).$$

• Thanks for answering. So if nothing is given we assume this ? – blue boy Sep 10 '18 at 12:50
• @blueboy Yes, you do. – Janik Sep 10 '18 at 12:53

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.