# The ring of real trigonometric polynomials

## The ring

Throughout we consider the polynomial ring $R=\mathbb{R}[\cos{x},\sin{x}].$ This ring has a lot of non-trivial properties which give us a good chance to study commutative ring theory.

First of all, note it is immediate that $R \cong \mathbb{R}[X,Y]/(X^2+Y^2-1)$ if the map is given by $X \mapsto \cos x$ and $Y \mapsto \sin x$. Besides, in $R$ we have $\sin^2x=(1-\cos{x})(1+\cos{x})=\sin{x}\cdot\sin{x}$ which indicates us to study whether $R$ is a UFD. In fact, it is not, because the ideal class group is $\mathbb{Z}/2\mathbb{Z}$. A Dedekind domain is a UFD if and only if its ideal class group is trivial (corollary 3.22).

This blog post is inspired by an exercise on Serge Lang's Algebra. But when writing this blog post, I found some paywalls. It would be absurd of me to direct a random reader to these paywalls. So it is very likely that I will include proofs as many as possible (when there is an absurd paywall, and chances are I will rework them for readability). But I can't remove the assumption that the reader has finished Atiyah-MacDonald full book or equivalences at the very least. I will add more topics in the future but that is not an easy job.

### Normal Noetherian ring

By Hilbert's basis theorem, $\mathbb{R}[\cos{x},\sin{x}]$ is Noetherian because it is a finite $\mathbb{R}$-algebra. Now we are interested in the normality of it. Since $\mathbb{R}[X,Y]/(X^2+Y^2-1) \cong \mathbb{R}[X][Y]/(Y^2-(1-X^2))$ and $2$ is a unit, $1-X^2$ is square-free but not a unit, we can apply the following lemma to show that $R$ is a normal Noetherian ring (integrally closed in its field of fraction). For the definition and properties of a normal ring, please refer to the stack project.

(Lemma 1) Let $A$ be a factorial ring with the field of fraction $K$ in which $2$ is a unit, $a$ in $A$ a square-free element (i.e., if $p$ is a prime element in $A$, then $a \not\in p^2A)$ which is not a unit. Then $A[T]/(T^2-a)$ is normal.

Click to expand the proof. Proof. By definition, we need to show that $A[T]/(T^2-a)$ is integrally closed in its field of fraction $L$. First we investigate the structure of $L$.
Let $t$ be the image of $T$ in $A[T]/(T^2-a)$ and in $L$. Then it is clear that $A[t] \cong A[T]/(T^2-a)$ and we can write $L=K(t)$. Note an element in $K(t)$ is of degree at most $1$, which is to say every element in $L$ can be written uniquely as a sum $r+st$ where $r,s \in K$. To prove integral closeness, we need to find minimal polynomial of $r+st$.
Next we show when $A[t]$ is integrally closed. Note \begin{aligned} \left[(r+st)-r\right]^2=(st)^2 &= s^2[T^2+(T^2-a)]\\ &= s^2[a+T^2-a+(T^2-a)]\\ &= as^2 \end{aligned} Hence $f(X)=(X-r)^2-as^2$ sends $r+st$ to $0$. For polynomial of degree $1$, we can only write $g(X)=X-X$ such that $g(r+st)=0$, which is absurd. Hence $f(X)$ is the minimal polynomial of $r+st$. With these being said, $r+st$ is integral over $A[t]$ if and only if $-2r \in A[t]$ and $r^2-as^2 \in A[t]$. We need to show this implies $r+st \in A[t]$. Since we can consider $A$ to be a subring of $A[t]$, it suffices to show that $r,s \in A$, provided $-2r \in A$ and $r^2-as^2 \in A$ when $s \ne 0$.
Since $2$ is a unit in $A$, $-2r \in A$ clearly implies $r \in A$. It remains to prove that $-as^2 \in A$. For $s \in K$, we can write $s=s_1/s_2$ with $s_1,s_2 \in A$ relatively prime. We shall show that $s_2$ will always be a unit, which implies that $s \in A$. Write $as^2=h$, then we have $as_1^2=hs_2^2$. Assume $s_2$ is not a unit, then there is a prime $p$ divides $s_2$ as $A$ is a factorial ring. hence $as_1^2 = hs_2^2 \in p^2A$. Since $s_1$ and $s_2$ are relatively prime, $p$ and $p^2$ do not divide $s_1$, hence $a \in p^2A$, a contradiction (we have assumed $a$ to be square-free. Also, the assumption that $a$ is not a unit is used here to reach the contradiction). Hence $s_2$ is a unit, $s \in A$ and therefore $-as^2 \in A$. The proof is complete. $\square$

Of course, I shan't be this lazy. It is clear that in the factorial ring $A=\mathbb{R}[X]$, $2$ is a unit. By square-free, we mean, if $p \in A$ is prime, then $a \not \in p^2A$. For example, in $\mathbb{Z}$, $12$ is not square-free because $12=2^2 \times 3 \in 2^2\mathbb{Z}$ while $14$ is square-free because $14=2 \times 7$ and square does not appear. And for $1-X^2$ things is clear because we only have $1-X^2=(1-X)(1+X)$ - there is no square. We require $2$ to be a unit because if not this argument becomes much more difficult to prove. We shall return to normality after we study the irreducible elements.

To conclude we have got a satisfying result:

(Proposition 1) $R$ is a normal Noetherian ring.

### Irreducible Elements

With help of Fourier transform or elementary trigonometric relations, every polynomial in $R=\mathbb{R}[\cos{x},\sin{x}]$ can be written in the form $P(x) = a_0+\sum_{k=1}^{n}(a_k\cos{kx}+b_k\sin{kx})$ where $a_0,a_k,b_k \in \mathbb{R}$. Define the degree $\delta(P)$ to be the maximum of integers $r,s$ where $a_r,b_s \ne 0$. Then a direct computation shows that $\delta(PQ)=\delta(P)+\delta(Q)$. This is not the case when the scalar is complex. For example, $(\cos{x}+i\sin{x})(\cos{x}-i\sin{x})=1$.

If $\delta(P)=0$, then $P(x)=a_0$ is zero or a unit. If $\delta(P)=1$, then if we have $P=P_1P_2$, then $\delta(P_1)+\delta(P_2)=1$. One of them has to be unit, hence $P$ is irreducible. If $\delta(P)=2$, then $P$ is reducible because we can solve equations in the expansion of the product $(a+b\sin{x}+c\cos{x})(a'+b'\sin{x}+c'\cos{x}).$ By induction all polynomials of degree $\ge 2$ is reducible. Hence irreducible elements are of the form $a+b\sin{x}+c\cos{x} \quad (b,c) \ne (0,0).$ But since $R$ is not a UFD, we cannot work on the ideal $(a+b\sin{x}+c\cos{x})$ directly. We need to dive into abstraction for a long time.

### Dedekind Domain

We now proceed to another satisfying result.

(Proposition 2) $R$ is a Dedekind domain.

Proof. Throughout, we work on the form $R \cong \mathbb{R}[X,Y]/(X^2+Y^2-1)$. Since $\mathbb{R}[X,Y]$ is of Krull dimension $2$ (see Atiyah-MacDonald exercise 11.7, where a solution is almost given), $X^2+Y^2-1$ is irreducible, we have a prime ideal $(X^2+Y^2-1)$, and all prime ideals $P \subset \mathbb{R}[X,Y]$ strictly containing $(X^2+Y^2-1)$ are maximal. Next, let the canonical map $\pi:\mathbb{R}[X,Y] \to \mathbb{R}[X,Y]/(X^2+Y^2-1)$ be given. By proposition 1.1 of Atiyah-MacDonald, $\pi(P)$ are maximal ideals in $\mathbb{R}[X,Y]/(X^2+Y^2-1)$ provided that $P \supsetneq (X^2+Y^2-1)$ is prime. If nontrivial ideal $Q \subset \mathbb{R}[X,Y]/(X^2+Y^2-1)$ is prime, then $\pi^{-1}(Q)=Q^c$ is also prime, and it contains $(X^2+Y^2-1)$ strictly, which implies that $Q$ is maximal. Hence $R$ is of Krull dimension $1$. By proposition 1, $R$ is integrally closed, hence it is Dedekind. $\square$

### Krull Domain, Ideal Class Group, Half-Factorial Domain

Let $A$ be an integral domain and $P$ be the set of all prime ideals of height $1$, i.e. the set of all prime ideals that only contain itself as a nonzero prime ideal. Then $A$ is a Krull domain if

(KD1) $A_{\mathfrak{p}}$ is a discrete valuation ring for all $\mathfrak{p} \in P$. (KD2) $A$ is the intersection of these discrete valuation rings (all considered as subrings of the field of fraction of $A$. (KD3) Any nonzero element of $A$ is contained in only a finite number of height $1$ prime ideals.

To proceed our study of $R$, we need a lemma:

(Lemma 2) If $A$ is a Dedekind domain, then $A$ is also a Krull domain.

Click to expand the proof Proof. Since in a Dedekind domain, every prime ideal is of height $1$, (KD1) coincides with one of the equivalent definition of Dedekind domain.
Next we prove (KD3). Pick any nonzero $a \in A$. If $a$ is a unit, then it is contained in $0$ ideals. If not, consider the ring $(a)=aA$. We have a unique factorisation as a product of prime ideals: $(a)= \mathfrak{p}_1^{r_1}\cdots\mathfrak{p}_n^{r_n} \subset \bigcap_{j=1}^{n}\mathfrak{p}_j.$ Hence (KD3) is proved.
For (KD2), note first $A \subset \bigcap_{\mathfrak{p}}A_{\mathfrak{p}}$ because the natural map $A \to A_{\mathfrak{p}}$ is injective for all $\mathfrak{p}$. Hence it suffices to prove the reverse. But elements in $A_{\mathfrak{p}}$ are of the form $a/s$. Hence we expect those elements of the form $b/1$ to be in $A$. Therefore it suffices to prove that $b/1 \in (a/1)A_{\mathfrak{p}}$ for all prime $\mathfrak{p}$ implies $b \in aA$ for all $a,b \in A$, $a ,b\ne 0$. Put $(a)=\mathfrak{p}_1^{r_1}\cdots\mathfrak{p}_n^{r_n}$ we see $\mathfrak{q}_i = \mathfrak{p}_i^{r_i}$ is $\mathfrak{p}_i$-primary and we obtain a primary decomposition. Note we in particular have $b \in \bigcap_{j=1}^{n}\left(aA_{\mathfrak{p}_i} \cap A \right) = \bigcap_{j=1}^{n}\mathfrak{q}_i = aA$ because each $\mathfrak{p}_i$ has height $1$. $\square$

Which is to say that

(Proposition 3) $R$ is a Krull domain.

We know that since $R$ is Dedekind, its fractional ideals form an abelian group. This gives rise to the ideal class group. By a result of Samuel, we have a shockingly simple fact:

(Proposition 4) The ideal class group $Cl(R) \cong \mathbb{Z}/2\mathbb{Z}$.

Which can be considered as a corollary to this following statement:

(Samuel) Let $F$ be a non-degenerate quadratic form in $k[X_1,X_2,X_3]$. Let $A_F=k[X_1,X_2,X_3]/(F)$. Then $Cl(A_F)=\mathbb{Z}/2\mathbb{Z}$ if and only if there is a nontrivial solution to $F(X_1,X_2,X_3)=0$ in $k$.

One can find this result via this link, and refer to study of plane conics.

With these being said, by theorem 8 of Zaks' paper, one sees that $R$ is a HFD domain. To be precise, for polynomials $x_1,x_2,\cdots,x_n$ and $y_1,y_2,\cdots,y_m$, if $x_1x_2\cdots x_n=y_1y_2\cdots y_m$, then $m=n$. I may recover the proof here one day, but it would be much more difficult than writing everything you have seen here. This ring $R$​ also shows that HFD is not necessarily UFD.

### Maximal ideals of $R$

Since $Cl(R) \cong \mathbb{Z}/2\mathbb{Z}$, for any maximal ideal $M \subset A$, either it is principal or $M^2$ is principal. If $M$ and $M'$ are two non-principal ideal, then $MM'$ is principal. Conversely, for any irreducible $z \in R$, either $(z)$ is maximal or $(z)=MM'$ for some maximal ideal $M$ and $M'$, and $M$ and $M'$ may coincide. We have given the form of irreducible elements $z = a+b\sin{x}+c\cos{x},\quad (b,c) \ne (0,0).$ So we are now interested in these $a,b,c$​. We will do some high school trick first. If we put $\begin{cases} k = \frac{a}{\sqrt{b^2+c^2}} \\ b' = \frac{b}{\sqrt{b^2+c^2}} \\ c' = \frac{c}{\sqrt{b^2+c^2}} \end{cases}$ then $z= \sqrt{a^2+b^2}(\sin(x+\alpha)+k)$​ where $b'=\cos\alpha$​ and $c' = \sin\alpha$​. Since $\sqrt{a^2+b^2} \in \mathbb{R}$​​ it suffices to study ​elements of the form $\sin(x+\alpha)+k$​.

Define a shift morphism $h:R \to R$ by $h(\cos{x})=\cos(x+\alpha), \quad h(\sin{x}) = \sin(x+\alpha), \quad h(t) = t$ This map is clearly an isomorphism. More importantly, since $h(\sin{x}+k)=\sin(x+\alpha)+k,$ the primary decomposition of $(\sin(x+\alpha)+k)$ and $(\sin{x}+k)$​ are of the same form. We are interested in the ring $R/(\sin{x}+k)$, where it is natural to study the behaviour of $\cos{x}$​. For this reason we consider the substitution morphism \begin{aligned} g:\mathbb{R}[X] & \to R \\ X & \mapsto \cos{x}. \end{aligned} We first compute the inverse image $g^{-1}[(\sin{x}+k)]$​. It is natural to think about cancelling $\sin x$ into $\cos x$. Note $(\sin x + k)( \sin x - k) = (\sin^2x -k^2) = (1- \cos^2x-k^2)$, pick whichever $P(X) \in (1-k^2-X^2)$, we have $g(P(X))=P(\cos{x}) = (1-\cos^2x-k^2)Q(\cos{x})=(\sin{x}+k)(\sin{x}-k)Q(\cos{x})$ Hence $(1-k^2-X^2)\subset g^{-1}[(\sin{x}+k)]$​. For the converse, note that if nonzero $P \in g^{-1}[(\sin x + k)]$, we have $\deg P > 1$ because trigonometric polynomial of the form $a+b\cos x$ can never be divided by $\sin x + k$​. By Euclidean algorithm, we find $Q(X)$, $R(X)$ such that $P(X)=Q(X)(1-k^2-X^2) + R(X)$ with $\deg R \le 1.$​ But when $P \in g^{-1}[(\sin x + k)]$​, we must have $R(X)=0$​​, according to our study of the degree earlier. Hence we must have $P(X) \in (1-k^2-X^2)$​, which is to say $g^{-1}[(\sin x + k)]= (1-k^2-X^2).$ This induces an isomorphism $\mathbb{R}[X]/(1-k^2-X^2) \cong R/(k+\sin x).$ And it is much easier to study the ideal $1-k^2-X^2$. To be precise,

1. $k^2=1 \iff (1-k^2-X^2)=(X)^2 \iff (k+\sin x)=M^2$ for some maximal ideal $M$, because $(X)$ is a maximal ideal.
2. $k^2<1 \iff (1-k^2-X^2)$​ is a product of two distinct maximal ideals $\iff (k+\sin x)$ is a product of two distinct maximal ideals $M$ and $M'$.
3. $k^2>1 \iff (1-k^2-X^2)$ is maximal $\iff$ $(k+\sin x)$ is maximal.

Therefore maximal ideals of $R$​ are determined by $k$, or more precisely the relation between $c^2$ and $a^2+b^2$​. Moreover, let $M$ be a maximal ideal, we have

1. If $M$​ is principal, then there exists $\alpha$ and $k$ such that

$M = (\sin(x+\alpha) + k)$

and $R/M \cong \mathbb{C}$​.

1. If $M$​ is not principal, then there exists $\alpha \in \mathbb{R}$ such that

$M = (\sin(x+\alpha)+1,\cos(x+\alpha)), \quad M^2 = (\sin(x+\alpha)+1).$

and $R/M \cong \mathbb{R}$.

## References

1. Robert M. Fossum, The Divisor Class Group of a Krull Domain.
2. M. F. Atiyah, FRS & I. G. MacDonald, Introduction to Commutative Algebra.
3. Macro Fontana, Salah-Eddine Kabbaj, Sylvia Wiegand, Commutative Ring Theory and Applications.
4. Hideyuki Matsumura, Commutative Ring Theory.
5. P. Samuel, Lectures on Unique Factorization Domains.
6. A. Zaks, Half Factorial Domains.

Desvl

2021-07-12

2021-12-24