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Determine whether a subring is an ideal


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.9

Solution: 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.


Linearity

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