1
$\begingroup$

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).

  • 0
    It actually is closed under multiplication. – 2017-02-05
  • 0
    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? – 2017-02-05
  • 0
    @MattSamuel Only if you allows for rational coefficients. – 2017-02-05
  • 0
    Oh, integer coefficients. I missed that. It's closed with rational coefficients. – 2017-02-05

2 Answers 2

1

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 $$\int_{0}^{2 \pi} f(t) \overline{g(t)} dt$$

  • 0
    Can I just use this as a counterfactual? – 2017-02-05
  • 0
    I see why that is not sufficient now. OK I am working on it. – 2017-02-05
0

$ ( 1 + \cos t) . \sin t = \sin t + \sin (2t) /2 \not \in S $

  • 0
    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. – 2017-02-05
  • 0
    Yes!! Absolutely – 2017-02-05
  • 0
    Only looking at the form of elements is not an argument to conclude that you got something outside of $S$. – 2017-02-05