Determine whether a subring is an ideal

Decide which of the sets in this previous exercise are ideals of the ring R = ^{[0,1]}\mathbb{R}.


We have already seen which of these are subrings.

  1. Let S = \{ f \in R \ |\ f[\mathbb{Q} = 0 \}. We saw that this set is a subring; now we need to show that it absorbs R on the left and the right. Let f \in S and g \in R. Then for all x \in \mathbb{Q}, (fg)(x) = f(x)g(x) = 0 \cdot g(x) = 0, and likewise (gf)(x) = 0. Thus fg, gf \in S, and S is an ideal
  2. Note that \sin is not a polynomial while 1 is. Since \sin \cdot 1 = \sin is not a polynomial, this set is not an ideal.
  3. This set is not a subring, hence not an ideal.
  4. This set is not a subring, hence not an ideal.
  5. Consider f(x) = 1-x if 0 \leq x < 1 and f(1) = 1. “Clearly” \lim_{x \rightarrow 1^-} f(x) = 0, so that f \in S. Now consider g(x) = 1/(1-x) if 0 \leq x < 1 and g(1) = 1. We have gf = 1 \notin S, so that S is not an ideal.
  6. Note that \sin(x) \in S and x \in R. Suppose x\sin(x) \in S; that is, that x\sin(x) 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, x\sin(x) is not bounded. Thus we have a contradiction, and this subring is not an ideal.
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Comments

  • John McPerson  On July 18, 2011 at 3:53 pm

    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.

  • nbloomf  On July 20, 2011 at 12:48 pm

    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!

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