# 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

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

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.

• 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

https://desvl.xyz/2021/12/29/the-ring-of-trigonometric-polynomial-with-complex-scalar/

Desvl

2021-12-29

2021-12-29