## Determine whether a subring is an ideal

Decide which of the sets in this previous exercise are ideals of the ring .

We have already seen which of these are subrings.

- Let . We saw that this set is a subring; now we need to show that it absorbs on the left and the right. Let and . Then for all , , and likewise . Thus , and is an ideal
- Note that is not a polynomial while 1 is. Since is not a polynomial, this set is not an ideal.
- This set is not a subring, hence not an ideal.
- This set is not a subring, hence not an ideal.
- Consider if and . “Clearly” , so that . Now consider if and . We have , so that is not an ideal.
- Note that and . Suppose ; that is, that is a finite rational linear combination of integer multiples of sine and cosine. Now every finite rational linear combination of sines and cosines is periodic, and moreover bounded. However, is not bounded. Thus we have a contradiction, and this subring is not an ideal.

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## Comments

How does a “function in R must be defined at 1” imply that “\lim_{x \rightarrow 1^-} g(x) exists”? Since we’re not told that the functions in R are continuous.

You’re right. I think the counterexample I came up with works, but my analytical intuition is terrible- let me know if I introduced a new error.

Thanks!