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Here, $\sin{x}$ and $\cos{x}$ are real functions $\mathbb{R} \to \mathbb{R}$.

First, let me explain what I mean by "as a quotient ring". It can be easily shown that $$ \mathbb{R}[\cos{x},\sin{x}] \cong \mathbb{R}[X,Y]/(X^2+Y^2-1). $$

Here $\mathbb{R}[\cos{x},\sin{x}]$ is a quotient ring of $\mathbb{R}[X,Y]$. So when the field is changed to $\mathbb{C}$, I immediately think about the ring $$ \mathbb{C}[X,Y]/(X^2+Y^2-1). $$

As people have pointed out, my argument quoted below is questionable, so I redo it in update 2. This paragraph is left here only for the record. I elaborated it in update 3.

However, $\mathbb{C}[\cos{x},\sin{x}]$ is not a UFD ($\sin^2{x}=(1+\cos{x})(1-\cos{x})$), but $\mathbb{C}[X,Y]/(X^2+Y^2-1) \cong \mathbb{C}[T,T^{-1}]$ is a UFD. Therefore $\mathbb{C}[\cos{x},\sin{x}]$ can only be a proper subring quotient ring of $\mathbb{C}[X,Y]/(X^2+Y^2-1)$. An extra restriction is missing here. Is there anyway to find it? I think there is something derived from the fact that $\mathbb{C}$ is algebraically closed.

Update 1: It is not very obvious that $\mathbb{C}[X,Y]/(X^2+Y^2-1)$ is a PID and therefore UFD, but one can find proofs in this question post: Ring of trigonometric functions with real coefficients

Update 2: It seems my argument on being UFD is a little messed up, so I will redo it. First of all, $R=\mathbb{R}[X,Y]/(X^2+Y^2-1)$ is not a UFD (one can show that its ideal class group is $\mathbb{Z}/2\mathbb{Z}$) and $S=\mathbb{C}[X,Y]/(X^2+Y^2-1) \cong \mathbb{C}[T,T^{-1}]$ is a UFD (see the link in update 1). Extending the scalar here is not a trivial matter, so the relation between $\mathbb{C}[\cos{x},\sin{x}]$ and $\mathbb{C}[X,Y]/(X^2+Y^2-1)$ is not likely to be as immediate as the case of real scalar.

Update 3: In the ring $\mathbb{R}[\cos{x},\sin{x}]$, irreducible elements are of the form $a\sin{x}+b\cos{x}+c$ where $a^2+b^2 \ne 0$; meanwhile, in the ring $\mathbb{C}[\cos{x},\sin{x}]$, irreducible elements are of the form $\cos{x}+i\sin{x}+a$ where $a \in \mathbb{C}^\ast$. I found them on this book (section: The Trigonometric Polynomial Rings). Hence $\sin{x}$, $1-\cos{x}$ and $1+\cos{x}$ are irreducible in $\mathbb{R}[\cos{x},\sin{x}]$ but not irreducible in $\mathbb{C}[\cos{x},\sin{x}]$. This is the source of my mistake. According to the comments this question have received, this change is not easily spotted. So I think I shall leave it here because it may serve as a counter-example on irreducibility.

  • How did you show that your ring is a Dedekind domain? – Severin Schraven Dec 24 '21 at 04:22
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    One can be explicit in giving an isomorphism with $\mathbb C[T,T^{-1}]$ using a projection. Also, one can observe that $\mathbb C[X,Y]/(X^2+Y^2-1)$ corresponds to an affine quadric plane curve, with a projective closure isomorphic to $\mathbb P^1$. – yearning4pi Dec 24 '21 at 04:25
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    Note that exhibiting the factorization $\sin^2{x}=(1+\cos{x})(1-\cos{x})$ is not sufficient to conclude that the ring is not a UFD: you must also show that the factors are non-associate, i.e., do not differ by a unit under multiplication. – Viktor Vaughn Dec 24 '21 at 04:26
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    @markvs $\mathbb{C}[X,Y]/(X^2+Y^2-1)$ is isomorphic to $\mathbb{C}[T, T^{-1}]$, which is a UFD, hence it is also a UFD. Matt E gave an answer here that is linked in the OP discussing a geometric proof of this. – Viktor Vaughn Dec 24 '21 at 04:41
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    @markvs I linked you an answer whose author is the user Matt E. This is (was) the account of Matt Emerton, a number theory professor at the University of Chicago. The isomorphism $\mathbb{C}[X,Y]/(X^2+Y^2-1) \cong \mathbb{C}[T,T^{-1}]$ shows that the first ring is a UFD: one map is $T \mapsto X + iY$ and the other is $X \mapsto \frac{T + T^{-1}}{2}$ and $Y \mapsto \frac{T - T^{-1}}{2i}$. I mistakenly thought that $X$ and $1-Y$ were associate, but as shown in Eric Wofsey's answer below, in fact these elements are actually not irreducible. – Viktor Vaughn Dec 24 '21 at 05:47
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    Using his observations, one sees that $X = \frac{1}{2}(X-iY)(X+iY+i)(X+iY-i)$. One can check that none of these factors is a unit, which shows that $X$ is reducible. This could also be deduced from the linked answer by Matt E, since he states that there is an equality $(X,Y-1) = (Y - 1 \pm iX)$ of ideals of $\mathbb{C}[X,Y]/(X^2+Y^2-1)$, so we must be able to write $X$ as a multiple of $Y - 1 \pm iX$ as done above. – Viktor Vaughn Dec 24 '21 at 05:47
  • In this answer I have proved that $f(\sin t, \cos t)=0$ implies that $f\in(X^2+Y^2-1)$ in $\mathbb R[X,Y]$. The proof is the same for $\mathbb C[X,Y]$ and it is immediate. – user26857 Dec 24 '21 at 07:19
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    @SeverinSchraven: $\mathbb{C}[X,Y]/(X^2+Y^2-1)$ is Dedekind for the following reasons. 1. It is Noetherian with Krull dimension 1 (this should be easy to demonstrate). 2. It is integrally closed. This follows from a lemma (11.1) of this book. –  Dec 24 '21 at 12:10
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    Lemma 11.1 Let $A$ be a factorial ring in which $2$ is a unit, $a$ in $A$ a square-free element which is not a unit. Then $A[T]/(T^2-a)$ is integrally closed. –  Dec 24 '21 at 12:11
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    Here we put $A=\mathbb{C}[Y]$. In $A$, $a=1-Y^2$ is square-free and not a unit. Therefore $A[X]/(X^2-a)$ is integrally closed. –  Dec 24 '21 at 12:21

1 Answers1

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First of all, $\mathbb{C}[\sin,\cos]$ certainly cannot be a proper subring of $\mathbb{C}[X,Y]/(X^2+Y^2-1)$ (with $X$ corresponding to $\cos$ and $Y$ corresponding to $\sin$). As a $\mathbb{C}$-algebra, $\mathbb{C}[X,Y]/(X^2+Y^2-1)$ is generated by $X$ and $Y$, so there is not any proper $\mathbb{C}$-subalgebra that contains both $X$ and $Y$. A priori, $\mathbb{C}[\sin,\cos]$ could be a proper quotient of $\mathbb{C}[X,Y]/(X^2+Y^2-1)$, since there could be more relations between $\sin$ and $\cos$ that are not generated by the single relation $X^2+Y^2-1$. However, this is not the case (and the proof is basically the same as the proof over $\mathbb{R}$ that you say you know), so $\mathbb{C}[\sin,\cos]$ really is isomorphic to $\mathbb{C}[X,Y]/(X^2+Y^2-1)$.

So what's going on? Well, the equation $\sin^2=(1+\cos)(1-\cos)$ simply does not imply that the ring is not a UFD. After all, $6^2=4\cdot 9$ but $\mathbb{Z}$ is still a UFD. To conclude that you have a failure of unique factorization, you would need to know something more about the factors $\sin,1-\cos,$ and $1+\cos$, such as that they are irreducible and not associate to each other.

In fact, none of these factors are irreducible. To see this, let us use the isomorphism $\mathbb{C}[X,Y]/(X^2+Y^2-1)\cong\mathbb{C}[T,T^{-1}]$ which is given by mapping $T$ to $X+iY$ (and $T^{-1}$ to $X-iY$). So in terms of $T$, $\sin$ (or $Y$) would be $\frac{T-T^{-1}}{2i}=\frac{T^2-1}{2iT}$. This is not irreducible, because $\frac{1}{2iT}$ is a unit (so can be ignored) and $T^2-1$ factors as $(T+1)(T-1)$ (and neither factor is a unit). Similarly, $1\pm\cos$ becomes $1\pm\frac{T+T^{-1}}{2}=\pm\frac{T^2\pm 2T+1}{2T}$ and the numerator factors as $(T\pm 1)^2$. So up to units, $\sin$ is $(T+1)(T-1)$ and $1+\cos$ and $1-\cos$ are $(T+1)^2$ and $(T-1)^2$, so the factorization $\sin^2=(1+\cos)(1-\cos)$ is just like the factorization $6^2=4\cdot 9$ of integers (with $T+1$ corresponding to $2$ and $T-1$ corresponding to $3$).

Eric Wofsey
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    Very nice answer. The observation that $X$ is not necessarily irreducible is very instructive (though it resembles the fact that the irreducible $2\in \mathbb Z[x]$ is not irreducible in the quotient $\mathbb Z[i]$). – yearning4pi Dec 24 '21 at 05:14
  • Is it legit to put $T=e^{ix}$ in your proof? I suppose your third paragraph can be understood as an Euler's formula ($e^{ix}=\cos{x}+i\sin{x}$) in disguise. –  Dec 24 '21 at 05:42
  • @ZoeDesvl: I'm not sure what you mean by that question. – Eric Wofsey Dec 24 '21 at 07:18
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    @EricWofsey: What I mean is this: is it OK to write isomorphism in this way: $\mathbb{C}[X,Y]/(X^2+Y^2-1) \cong \mathbb{C}[e^{ix},e^{-ix}] \cong \mathbb{C}[\cos{x},\sin{x}]$. By doing so, we actually get $\sin{x}=\frac{e^{ix}-e^{-ix}}{2i}$. –  Dec 24 '21 at 07:49
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    Yeah, you could do that, if you observe that the ring of functions $\mathbb{C}[e^{ix},e^{-ix}]$ is literally equal to (not just isomorphic to!) $\mathbb{C}[\cos x,\sin x]$. – Eric Wofsey Dec 24 '21 at 16:10
  • Before this question becomes inactive, I want to add something that qualifies as a complement to this answer. Irreducible elements in this ring are of the form (up to a unit) $\cos{x}+i\sin{x}+a$, where $a \in \mathbb{C}^\ast$. Meanwhile in $\mathbb{R}[\cos{x},\sin{x}]$, irreducible elements are of the form $a\sin{x}+b\cos{x}+c$ where $a^2+b^2 \ne 0$. –  Dec 25 '21 at 13:28
  • I found it on this book and they are in the section "Trigonometric Polynomial Rings". –  Dec 25 '21 at 13:29