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Decide whether or not S is a subring of R, when S is the set of functions which are linear combinations with integer coefficients of the functions $\{1, \cos{nt}, \sin{nt}\}, n\in\mathbb{Z}$ and R is the set of all real valued functions of t.

It is easy to show that S is closed under addition and that it has the multiplicative identity but I cannot prove that it is not closed under multiplication (which is what I suspect).

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  • $\begingroup$ It actually is closed under multiplication. $\endgroup$ – Matt Samuel Feb 5 '17 at 18:28
  • $\begingroup$ even with integer coefficients? Let f(t)=sint and g(t)=cost then (fg)(t)=½sin(2t) can this be an integer linear combination? and can you explain? $\endgroup$ – Clade Feb 5 '17 at 18:30
  • $\begingroup$ @MattSamuel Only if you allows for rational coefficients. $\endgroup$ – N. S. Feb 5 '17 at 18:30
  • $\begingroup$ Oh, integer coefficients. I missed that. It's closed with rational coefficients. $\endgroup$ – Matt Samuel Feb 5 '17 at 18:30
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Hint Prove first that $\{1, \cos{nt}, \sin{nt}\}, n\in\mathbb{Z}$ are linearly independent over $\mathbb Q$ (or $\mathbb R$).

Then use $$\sin(t)\cos(t)=\frac{1}{2}\sin(2t)$$ to deduce that this is the only way of writing $\sin(t)\cos(t)$ as a linear combination of $S$ with rational, hence integer coefficients.

P.S. The linear independence is typically obtained for free from the fact that the set is Orthogonal with respect to the inner product $$\int_{0}^{2 \pi} f(t) \overline{g(t)} dt$$

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  • $\begingroup$ Can I just use this as a counterfactual? $\endgroup$ – Clade Feb 5 '17 at 18:37
  • $\begingroup$ I see why that is not sufficient now. OK I am working on it. $\endgroup$ – Clade Feb 5 '17 at 18:49
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$ ( 1 + \cos t) . \sin t = \sin t + \sin (2t) /2 \not \in S $

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  • $\begingroup$ Thank you this makes a lot of sense. So what you are saying is that $(1+\cos{t})$ and $\sin{t}$ are both elements of $S$ but their product is not. So therefore $S$ is not closed under multiplication. $\endgroup$ – Clade Feb 5 '17 at 18:43
  • $\begingroup$ Yes!! Absolutely $\endgroup$ – Chirantan Chowdhury Feb 5 '17 at 18:50
  • $\begingroup$ Only looking at the form of elements is not an argument to conclude that you got something outside of $S$. $\endgroup$ – user26857 Feb 5 '17 at 22:07

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