# The ring of real trigonometric polynomials

## The ring

Throughout we consider the polynomial ring

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

if the map is given by $X \mapsto \cos x$ and $Y \mapsto \sin x$. Besides, in $R$ we have

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

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

By induction all polynomials of degree $\ge 2$ is reducible. Hence irreducible elements are of the form

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

So we are now interested in these $a,b,c$. We will do some high school trick first. If we put

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

This map is clearly an isomorphism. More importantly, since

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

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

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

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

This induces an isomorphism

And it is much easier to study the ideal $1-k^2-X^2$. To be precise,

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

- If $M$ is principal, then there exists $\alpha$ and $k$ such that

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

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

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

## References

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

The ring of real trigonometric polynomials

https://desvl.xyz/2021/12/24/The-ring-of-real-trigonometric-polynomials/