# The ring of trigonometric polynomials with complex scalars

In this post we have investigated some basic facts of \(\mathbb{R}[\cos{x},\sin{x}]\) (henceforth notations used there will be here as well). It is a half factorial domain with ideal class group \(\mathbb{Z}/2\mathbb{Z}\), and hence not a UFD. Next, we jump to the complex scalar field, and there are many nontrivial results. We will be discussing the ring

\[ R'=\mathbb{C}[\cos{x},\sin{x}]. \]

in a different style.

# Now we have a UFD (and an Euler's formula in disguise)

Again, if we consider the map

\[ \begin{aligned} \Phi:\mathbb{C}[X,Y] &\to \mathbb{C}[\cos{x},\sin{x}] \\ f(X,Y) &\mapsto f(\cos{x},\sin{x}) \end{aligned} \]

we will see that \(\ker\Phi=(X^2+Y^2-1)\) and therefore

\[ \mathbb{C}[\cos{x},\sin{x}] \cong \mathbb{C}[X,Y]/(X^2+Y^2-1). \]

Following the same step as in the previous post, we can show that \(R'=\mathbb{C}[\cos{x},\sin{x}]\) is Dedekind. However, the map

\[ \begin{aligned} \Psi:\mathbb{C}[U,V] &\to \mathbb{C}[X,Y]/(X^2+Y^2-1) \\ g(U,V) &\mapsto \overline{g(X+iY,X-iY)} \end{aligned} \]

shows that

(Proposition 1)\[ \mathbb{C}[X,Y]/(X^2+Y^2-1) \cong \mathbb{C}[U,V]/(UV-1) \cong \mathbb{C}[T,T^{-1}] \cong \mathbb{C}[T]_T. \]

The localisation of a UFD is a UFD, hence we see \(\mathbb{C}[\sin{x},\cos{x}]\) is a UFD. There are other ways to do it. For example, we can directly put \(\mathbb{C}[\sin{x},\cos{x}]=\mathbb{C}[e^{ix},e^{-ix}]\). And this is even quicker. As another way, since \(\cos{x}=\frac{e^{ix}+e^{-ix}}{2}\) and \(\sin{x} = \frac{e^{ix}-e^{-ix}}{2i}\), all trigonometric polynomials can be decomposed into the following form

\[ f(\cos{x},\sin{x}) = e^{-inx}P(e^{ix}) \]

where \(P(X) \in \mathbb{C}[X]\). Conversely, All elements of the form \(e^{-inx}P(e^{ix})\) is in \(\mathbb{C}[\cos{x},\sin{x}]\) and therefore we have an isomorphism

\[ \begin{aligned} \Lambda: \mathbb{C}[T]_{T} &\to \mathbb{C}[\cos{x},\sin{x}], \\ T &\mapsto \cos{x}+i\sin{x}. \end{aligned} \]

Note it follows that \(T^{-1}\) maps to \(\cos{x}-i\sin{x}\).

## Irreducible elements in this ring

Now we return to the identity

\[ \sin^2{x}=(1-\cos{x})(1+\cos{x}). \]

In \(\mathbb{R}[\cos{x},\sin{x}]\), since \(\sin{x}\), \(1-\cos{x}\), \(1+\cos{x}\) are all irreducible, or more precisely, elements of the form \(a+b\sin{x}+c\cos{x}\) are irreducible where \((b,c) \ne (0,0)\), we see \(\mathbb{R}[\cos{x},\sin{x}]\) is a UFD. In fact, we can deduce the fact that \(R\) is not a UFD by the fact that \(Cl(R) \cong \mathbb{Z}/2\mathbb{Z}\), i.e., the ideal class group is nontrivial (corollary 3.22).

However, since \(R'\) is a UFD, \(\sin^2{x}=(1-\cos{x})(1+\cos{x})\) tells us *nothing*. We need to figure out why and what is going on. To work with it we consider the form \(R'=\mathbb{C}[T,T^{-1}]\). What are irreducible elements in this ring? We will make use of the fact that \(\mathbb{C}\) is algebraically closed (why not!). Since \(T\) and \(T^{-1}\) are units in this ring, we can use them to modify the degree of an element. More precisely, as an application of the fundamental theorem of classical algebra,

\(P(T)=\sum_{j=m}^{n}a_jT^{j}\) (you should be reminded of Laurent series!) is irreducible where \(m,n \in \mathbb{Z}\) if and only if \(Q(T)=T^{-m}P(T)\) is irreducible. However, \(Q(T) \in \mathbb{C}[T]\) is irreducible if and only if \(Q\) is of degree \(1\)), which is equivalent to say that \(n-m=1\) in \(P(T)\).

Therefore irreducible elements in the form \(aT^m+bT^{m+1}\) where \(a,b \ne 0\) . Dropping \(bT^m\) because it is a unit, we obtain a finer result:

(Proposition 2)Irreducible elements of \(R'\) is of the form\[ \cos{x}+i\sin{x}+a, a \in \mathbb{C}^\ast. \]

With this being said, \(\sin{x}\), \(1-\cos{x}\) and \(1+\cos{x}\) are all *not* irreducible. For example, for \(\sin{x}\) we actually have

\[ \begin{aligned} \sin{x}&=\frac{1}{2i}(e^{ix}-e^{-ix})\\ & = \frac{1}{2ie^{ix}}(e^{2ix}-1) \\ & = \frac{1}{2ie^{ix}}(e^{ix}+1)(e^{ix}-1) \\ & = \frac{1}{2ie^{ix}}(\cos{x}+i\sin{x}+1)(\cos{x}+i\sin{x}-1) \end{aligned} \]

# Quotient fields

We can find some obvious facts about these two rings. For example, \(R\) is a free \(\mathbb{R}[\cos{x}]\)-algebra with basis \(\{1,\sin{x}\}\) (note all \(\sin^nx\) of even degree can be transformed into \(\cos{x}\) by the relation \(\sin^2{x}=1-\cos^2{x}\)). Likewise \(R'\) is a free \(\mathbb{C}[\cos{x}]\)-algebra with basis \(\{1,\sin{x}\}\). We can also write \(R'\) as \(R \oplus iR\) or \(R[i]\). That is, \(R'\) is a free \(R\)-algebra with basis \(\{1,i\}\). These are quite elementary and don't touch the structure of polynomial pretty much. Now we touch it by studying the quotient field of \(R\) and \(R'\) respectively.

Treating \(R\) as a free \(\mathbb{R}[\cos{x}]\)-algebra, we can write any polynomial \(f(\cos{x},\sin{x})\) as

\[ f(\cos{x},\sin{x})=P(\cos{x})+Q(\cos{x})\sin{x} \]

where \(P,Q \in \mathbb{R}[X]\). For simplicity we write \(f=P+Q\sin{x}\). Suppose we now have \(f=P_1+Q_1\sin{x}\) and \(g=P_2+Q_2\sin{x}\) with \(g \ne 0\), then

\[ \begin{aligned} \frac{f}{g} &= \frac{P_1+Q_1\sin{x}}{P_2+Q_2\sin{x}} \\ &= \frac{(P_1+Q_1\sin{x})(P_2-Q_2\sin{x})}{(P_2+Q_2\sin{x})(P_2-Q_2\sin{x})} \\ &=\frac{P_1P_2-Q_1Q_2(1-\cos^2{x})+(P_2Q_1-P_1Q_2)\sin{x}}{P_2^2+Q_2^2(1-\cos^{2}{x})} \end{aligned} \]

Therefore every element of \(K(R)\) can be written in the form \(U(\cos{x})+V(\cos{x})\sin{x}\) where \(U,V \in \mathbb{R}(\cos{x})\), the rational field of \(\cos{x}\) over \(\mathbb{R}\). Since \(\sin^2{x} \in \mathbb{R}(\cos{x})\), we obtain:

(Proposition 3)The quotient field of \(R\) is\[ K(R)=\mathbb{R}(\cos{x})[\sin{x}]. \]

Likewise,

\[ K(R')=\mathbb{C}(\cos{x})[\sin{x}] \]

can be proved in exactly the same way.

Since \(R\) is Dedekind, it is integrally closed in \(K(R)\). But what about its relation with \(K(R')\)? For this we have an elegant result:

(Proposition 4)\(R'\) is the integral closure of \(R\) in \(K(R')\).

*Proof.* Let \(C\) be the closure of \(R\) in \(K(R')\). Note \(K(R')=K(R)[i]\). For any \(f+ig \in C\), we see \(f \in R\) and \(g \in R\) and hence \(f+ig \in R'\) because \(f,g \in K(R)\) and \(R\) is integrally closed. Therefore \(C \subset R'\). Conversely, any \(f+ig \in R'\) is in \(C\) because \(f,g \in R \subset C\) and \(i \in C\). Therefore \(R' \subset C\). \(\square\)

# Study the ring in the setting of elementary algebraic geometry

*We are using the notation that Hartshorne used in his book Algebraic Geometry.*

Put \(f(X,Y)=X^2+Y^2-1\), then \(Y=Z(f)\) is an irreducible affine curve in the affine space \(A^2_{\mathbb{C}}\). This curve is non-singular everywhere because the matrix

\[ \begin{pmatrix} \partial f/\partial X \\ \partial f/\partial Y \end{pmatrix} = \begin{pmatrix} 2X \\ 2Y \end{pmatrix} \]

has rank \(1\). The coordinate ring \(A(Y)\) is exactly \(R'\).

Let \(P\) be a point on \(Y\), which, by Hilbert's Nullstelensatz, corresponds to a unique maximal ideal \(\mathfrak{m}_P \subset A(Y)\cong R'\). Since \(R'\) is a PID, and by proposition 2, \(\mathfrak{m}_P=(\cos{x}+i\sin{x}+a)\) where \(a \ne 0\). Hence \(P\) corresponds to a nonzero complex number \(a\).

(Proposition 5)Every point \(P\) on the curve \(Z(X^2+Y^2-1)\) corresponds to a unique nonzero complex number \(a \in C^\ast\).

Since \(Y\) is nonsingular, it also follows that \(\dim_{\mathbb{C}}\mathfrak{m}/\mathfrak{m}^2=\dim R'=1\) for all maximal ideal of \(R'\). This is to say, the tangent space is always of dimension \(1\) as a \(\mathbb{C}\)-vector space, or \(2\) as a \(\mathbb{R}\)-space. Besides, if we localise it at \(\mathfrak{m}_P\), we see \(\mathcal{O}_{P,Y} \cong R'_{\mathfrak{m}_P}\) is always a regular local ring.

# References / Further Readings

*Introduction to Commutative Algebra*, M. F. Atiyah & I. G. MacDonald.*Algebraic Geometry*, Robin Hartshorne.*Commutative Ring Theory and Applications*, edited by Marco Fontana, Salah-Eddine Kabbaj and Sylvia Wiegand.

The ring of trigonometric polynomials with complex scalars

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