Let $R$ be the ring of functions that are polynomials in $\cos t$ and $\sin t$ with real coefficients.

  1. Prove that $R$ is isomorphic to $\mathbb R[x,y]/(x^2+y^2-1)$.

  2. Prove that $R$ is not a unique factorization domain.

  3. Prove that $S=\mathbb C[x,y]/(x^2+y^2-1)$ is a principal ideal domain and hence a unique factorization domain.

  4. Determine the units of the rings $S$ and $R$. (Hint: Show that $S$ is isomorphic to the Laurent polynomial ring $\mathbb C[u,u^{-1}]$.)

  • 1
    $\begingroup$ if you find an answer to your question helpful, you should consider accept it. See meta.math.stackexchange.com/questions/3286/… $\endgroup$ – user18119 Dec 13 '12 at 22:40
  • 2
    $\begingroup$ @QiL'8 It's verbatim from Michael Artin's Algebra without explicating any effort or explanation, therefore downvote. $\endgroup$ – Yai0Phah Jul 28 '13 at 13:55

(1) Send $X$ to $\cos$ and $Y$ to $\sin$. The kernel of this homomorphism consists from the polynomials $f\in\mathbb{R}[X,Y]$ with the property $f(\cos t,\sin t)=0$ for any $t\in\mathbb{R}$. Now prove that this implies $f$ divisible by $X^2+Y^2-1$: consider $f$ as a polynomial in $X$ with coefficients in $\mathbb{R}[Y]$ and write $f=(X^2+Y^2-1)g+r$, where $\deg_Xr\le 1$. Then $r=a+bX$, where $a,b\in\mathbb{R}[Y]$. From $f(\cos t,\sin t)=0$ for any $t\in\mathbb{R}$ we get that $r(\cos t,\sin t)=0$ for any $t\in\mathbb{R}$, that is, $a(\sin t)+b(\sin t)\cos t=0$ for any $t\in\mathbb{R}$. This gives us that $a^2(\sin t)=b^2(\sin t)(1-\sin^2t)$ for any $t\in\mathbb{R}$. But $a,b$ are polynomials and the last relation implies that $a^2(Y)=b^2(Y)(1-Y^2)$ and this is enough to deduce $a=b=0$ (why?).

(2) We show that $x$, the residue class of $X$ in $\mathbb{R}[X,Y]/(X^2+Y^2-1)$ is irreducible and does not divide $1+y$ and $1-y$, the residue classes of $1+Y$ and $1-Y$. (Note that in $\mathbb{R}[x,y]$ we have $x^2=(1+y)(1-y)$.)

Edit. If $x=z_1z_2$, then, by using (4) we get $N(x)=N(z_1)N(z_2)\Leftrightarrow Y^2-1=N(z_1)N(z_2)$. Now we have the following cases: (i) $\deg N(z_1)=0\Leftrightarrow z_1\in\mathbb R^*$, (ii) $\deg N(z_1)=2$ $\Leftrightarrow$ $\deg N(z_2)=0$ $\Leftrightarrow$ $z_2\in\mathbb R^*$, (iii) $\deg N(z_1)=\deg N(z_2)=1$. If $N(z_1)=Y-1$, then $a_1^2(Y)+b_1^2(Y)(Y^2-1)=Y-1\Rightarrow Y-1\mid a_1(Y)\Rightarrow\exists a_2\in\mathbb R[Y]$ such that $a_1=(Y-1)a_2$ and pluggin this in the foregoing equation we get $a_2^2(Y)(Y-1)+b_1^2(Y)(Y+1)=1$. Looking now at the dominant coefficients of $a_2$ and $b_1$ we find that one of these is zero (false!) or the sum of their square is zero (false!).

Assume that $x\mid 1-y$. Then $N(x)\mid N(1-y)\Leftrightarrow Y^2-1\mid (Y-1)^2$, contradiction.

(3) I've proved here all you need for this part.

(4) As one can see from $(1)$ the elements of $\mathbb{R}[x,y]$ can be uniquely written as $a(y)+b(y)x$. Define a "norm" $N:\mathbb{R}[x,y]\to\mathbb{R}[Y]$ by $N(a(y)+b(y)x)=a^2(Y)+b^2(Y)(Y^2-1)$. $N$ is multiplicative and using this we get that the units of $\mathbb{R}[x,y]$ are the non-zero elements of $\mathbb{R}$.

For $\mathbb{C}[x,y]$ we can see, via the isomorphism from part $(3)$ that the invertible elements are the non-zero elements of $\mathbb{C}$, the powers of $x+iy$ and $x-iy$, and products of the non-zero elements of $\mathbb{C}$ with the powers of $x+iy$ and $x-iy$.

  • $\begingroup$ what does the norm in (4) mean? $N((a+bx)(c+dx))=N((1-y^2)bd+ac+(ad+bc)x)=(ad+bc)^2+(ac+(1-y^2)bd)^2(y^2-1))$, which I don't think simplifies to $(ac)^2+(bd)^2(y^2-1)$, so how is $N$ multiplicative? $\endgroup$ – Miles Johnson Mar 12 '19 at 8:23

There is another way to think about this problem. Since $R:= \mathbb R[x,y]/(x^2 +y^2 -1 )$ is a smooth affine curve, it is a normal ring (i.e. integrally closed in its fraction field), and so it is factorial if and only if it has trivial class group.

Here and below I will use ideas discussed in Hartshorne, Ch.II.6, in the subsection on Weil divisors.

We may consider $U :=$ Spec $R$ as an affine open curve, and then consider its projective closure $X$. The curve $X$ is a plane conic, and so its class group (equivalently, its Picard group) is isomorphic to $\mathbb Z$, generated by the class of any rational point (e.g. the class of the point $(1,0)$).

Now $Z := X \setminus U$ is irreducible (it is a single point of $X$, which geometrically becomes two points, namely the two points at infinity $[1:\pm i: 0]$ --- note that neither of these points is individually defined over $\mathbb R$, but their union is, and so it corresponds to a single point on $X$ with residue field equal to $\mathbb C$); this is where we use that our curve is defined over $\mathbb R$ rather than $\mathbb C$. (In the latter case $Z$ is not irreducible, but is the union of the preceding two points, which are now both defined over $\mathbb C$.)

We now use the exact sequence of Hartshorne II.6, Prop. 6.5, namely

$$\mathbb Z \to \mathrm{Cl}(X) \to \mathrm{Cl}(U) \mapsto 0,$$

where the first arrow is defined by $n \mapsto $ the class of $nZ$.

Recalling that Cl$(X) = \mathbb Z$, and that $Z$ corresponds to a pair of points over $\mathbb C$, this exact sequence can be written more explicitly as $$\mathbb Z \to \mathbb Z \to \mathrm{Cl}(U) \to 0,$$ where the first map is multiplication by $2$.

Thus Cl$(R) = $ Cl$(U) = \mathbb Z/2$, and we see that $R$ is not a UFD.

Explicitly, we see that a maximal ideal in $R$ will be principal precisely if its residue field is equal to $\mathbb C$ (rather than $\mathbb R$). Thus e.g. the maximal ideal $(x,y-1)$, which cuts out the point $(0,1)$ and has residue field $\mathbb R$, is not principal.

One can think about this more geometrically:

If the maximal ideal cutting out a point $P$ over $\mathbb R$ is principal, then it is generated by some real polynomial $f(x,y)$. But then the ideal $(f)$ in $R$ is a product of maximal ideals corresponding to the intersection of the curve $f = 0$ with the curve $U$. By assumption this is just the single point $P$, with multiplicity one, and so (now passing from the affine picture to the projective picture) all the other intersections must be with the two points in $Z$. By Bezout, the total number of intersections of $f = 0$ with $X$ is even, and we are assuming the intersection of $f = 0$ with $U$ consists of the single point $P$, so in fact the number of intersections with $Z$ must be odd. But this set of intersections (counted with multiplicity) is symmetric under complex conjugation (since $f$ has real coefficients) and so it must be even (because the two points of $Z$ are interchanged by complex conjugation). This contradiction shows that the maximal ideal of $P$ is not principal. (This is more or less a rewriting of the proof of Hartshorne's Prop. 6.5 in this particular case.)

It is also easy to see what happens when we extend scalars from $\mathbb R$ to $\mathbb C$, i.e. pass from $R$ to $S$. The set $Z$ now becomes the union of two points, and so for any point $P$ of $U$ (now over $\mathbb C$) we can find a generator of the maximal ideal by choosing $f$ to be the equation of a line passing through $P$ and one of the two points in $Z$. E.g. for $P = (0,1)$, we can take a generator of the ideal $(x,y-1)$ to be $(y - 1 \pm ix)$. (Either choice of sign will do; their ratio is a unit in $S$.)

In terms of the exact sequence of class groups, $Z$ is no longer irreducible, but the union of two points each of degree one, and so the exact sequence becomes $$\mathbb Z \oplus \mathbb Z \to \mathrm{Cl}(X_{/\mathbb C}) \to \mathrm{Cl}(U_{/\mathbb C}) \to 0,$$ which more explicitly is $$\mathbb Z \oplus \mathbb Z \to \mathbb Z \to \mathrm{Cl}(U_{/\mathbb C}) \to 0,$$ with the first map being given just by $(m,n) \mapsto m+n$. Evidently this map is surjective, and so Cl$(S) =$ Cl$(U_{/\mathbb C}) = 0.$


Let me complete YACP's excellent answer by describing the units of $S=\mathbb{C}[X,Y]/(X^2+Y^2-1)=\mathbb C[x,y]$.
We will use the isomorphism $\mathbb C[U,U^{-1}]\cong \mathbb C[x,y]:U\mapsto x+iy$
First note that the units of $\mathbb C[U,U^{-1}]$ constitute the set $(\mathbb C[U,U^{-1}])^{\times}=\bigcup _{n\in \mathbb Z} \mathbb C^*U^n$ of non zero monomials.
Translating back to $S=\mathbb C[x,y]$ we obtain $$S^{\times}=\mathbb C[x,y]^{\times}=\bigcup _{n\in \mathbb Z} \mathbb C^*(x+iy)^n$$
[Don't forget that $ ((x+iy)^n)^{-1}=(x-iy)^n$ since $(x+iy)(x-iy)=x^2+y^2=1$]

Edit: graded rings
It might be of some interest to put the problem in a more general perspective.
In a $\mathbb Z$-graded domain $R=\oplus_{n\in \mathbb Z} R_n$ it is immediate that units are homogeneous and from there it is obvious that the units of $\mathbb C[U,U^{-1}]=\oplus_{n\in \mathbb Z} \mathbb CU^n$ are $(\mathbb C[U,U^{-1}])^{\times}=\bigcup _{n\in \mathbb Z} \mathbb C^*U^n$, as stated above.
Similarly one would compute that the units of $\mathbb Z[U,U^{-1}]=\oplus_{n\in \mathbb Z} \mathbb ZU^n$ are given by $(\mathbb Z[U,U^{-1}])^{\times}=\bigcup _{n\in \mathbb Z} \mathbb \lbrace U^n,-U^n\rbrace $.
However in a graded ring with zero divisors, units needn't be homogeneous : for example in the ring of dual numbers $D=\mathbb C[X]/(X^2)=\mathbb C\oplus \mathbb C[\epsilon]$, the element $1+\epsilon$ is invertible (with inverse $1-\epsilon$) although it is not homogeneous.

  • $\begingroup$ At the time I wrote this post YACP hadn't yet answered the question on the invertibles of $S$, hence my phrase "Let me complete YACP's excellent answer". $\endgroup$ – Georges Elencwajg Nov 25 '12 at 22:11
  • $\begingroup$ I'm a little confused: if $\Bbb C[X,Y]/(X^2+Y^2-1)=\Bbb C[x,y]$, wouldn't that mean it's not a PID? I'm guessing something in that line was just written incorrectly. $\endgroup$ – rschwieb Aug 1 '14 at 13:13
  • $\begingroup$ @rschwieb: the line is written quite correctly. The elements $x,y$ are the classes of $X,Y$ modulo the ideal $(X^2+Y^2-1)$, so that the ring $\mathbb C[x,y]$ is not the ring of polynomials in two independent indeterminates: $x$ and $y$ are related by the algebraic equation $x^2+y^2=1$ . $\endgroup$ – Georges Elencwajg Aug 1 '14 at 17:51
  • $\begingroup$ Dear Georges : I think using $\Bbb C[x,y]$ that way is really misleading, but it could just be me. Maybe if instead of $x,y$ there were $\alpha,\beta$ instead, or something less like an indeterminate, I would have caught on. Thanks for clarifying what you meant. Regards $\endgroup$ – rschwieb Aug 1 '14 at 17:57
  • $\begingroup$ Dear @rschwieb: I'm sorry I misled you but it is a quite common convention: the answer by user 26857 above uses it too. I find it elegant and less clumsy than putting overlines on letters or putting letters between brackets in quotients (but I too like Greek letters). However, de gustibus non est disputandum ... $\endgroup$ – Georges Elencwajg Aug 1 '14 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.