Prove that $\Bbb C[x,y]/(x^3+y^3-1)$ is not a UFD

Question

I am posting this question on MO since I haven't received any answers on MSE.

Below is my (very elementary) attempt. Feel free to post a solution using facts in algebraic geometry and facts about elliptic curves.

Start by noticing $x^3=(1-y)(1+y+y^2)$ in $\Bbb C[x,y]/(x^3+y^3-1)$. I am trying to show first that $x$ is irreducible modulo $(x^3+y^3-1)$.

Every polynomial in $\Bbb C[x,y]$ is of the form $$p_1(x)+p_2(x)y+p_3(x)y^2+r(x,y)(x^3+y^3-1)$$ for some polynomials $p_1(x),p_2(x),p_3(x),r(x,y)$ in $\Bbb C[x,y]$. So that given $$x=(p_1(x)+p_2(x)y+p_3(x)y^2)(q_1(x)+q_2(x)y+q_3(x)y^2)+r(x,y)(x^3+y^3-1)$$ one needs to show that either $q_1(x)+q_2(x)y+q_3(x)y^2$ or $p_1(x)+p_2(x)y+p_3(x)y^2$ is a unit modulo $(x^3+y^3-1)$, i.e for example $$(p_1(x)+p_2(x)y+p_3(x)y^2)f(x,y) = k+g(x,y)(x^3+y^3-1)$$ for some constant $k$ and polynomials $f,g\in\Bbb C[x,y]$ is the same as saying $p_1(x)+p_2(x)y+p_3(x)y^2$ is a unit modulo $(x^3+y^3-1)$.

This gets extremely messy afterwards. Is there a better way I could approach this?

Answer 1

Here is a proof using a bit of commutative algebra and algebraic geometry, elaborating on comments by @abx and @ChrisWuthrich. The ring $$R=\mathbb{C}[x,y]/(x^3+y^3-1)$$ is a Dedekind domain, because it is the coordinate ring of the affine elliptic curve $$E=\{(x,y)\in\mathbb{C}^2:x^3+y^3=1\}.$$ As a result, if $R$ is a UFD, then it is also a PID, hence the Picard group $\mathrm{Pic}(E)$ is trivial. We shall show that this is not the case.

Consider the projective closure of $E$: $$\tilde E=\{(x:y:z)\in\mathbf{P}(\mathbb{C}^3):x^3+y^3=z^3\}.$$ We have an embedding $f:E\to\tilde E$ given by $(x,y)\mapsto(x,y,1)$, and in fact $\tilde E$ is the disjoint union of $f(E)$ and the set of points at infinity, $$S=\{(-1:1:0), (-1:e^{2\pi i/3}:0), (-1:e^{4\pi i/3}:0)\}.$$ Restricting divisors of $\tilde E$ to $f(E)=\tilde E\setminus S$ gives rise to an exact sequence $$0\to\mathrm{Pic}(\tilde E,S)\to\mathrm{Pic}(\tilde E)\to\mathrm{Pic}(E)\to 0,$$ where $\mathrm{Pic}(\tilde E,S)$ is the group of divisors of $\tilde E$ supported on $S$ modulo principal divisors of $\tilde E$ supported on $S$. Note that $\mathrm{Pic}(\tilde E,S)$ is a finitely generated abelian group, while $\mathrm{Pic}(\tilde E)$ is isomorphic to $\mathbb{Z}\times\tilde E$ if we regard $\tilde E$ as an abelian group in the usual way. Hence $\mathrm{Pic}(E)$ has cardinality continuum, and $R$ is not a UFD.

Remark 1. The field $L=\mathbb{C}(x,y)/(x^3+y^3-1)$ is a cubic Galois extension of $K=\mathbb{C}(x)$. The ideal $(x)$ of $R$ factors into prime ideals as $$(x)=(x,y-1)(x,y-e^{2\pi i/3})(x,y-e^{4\pi i/3}),$$ and the three factors are cyclically permuted by $\mathrm{Gal}(L/K)$. In fact these factors are not principal as the answer of user516477 shows below. Equivalently, $x$ is irreducible (but not prime) in $R$.

Remark 2. Note that $x+y$, $x+e^{2\pi/3}y$, $x+e^{4\pi/3}y$ are units in $R$ as their product is $x^3+y^3=1$. It can be derived from the calculations of user516477 that these three units and $\mathbb{C}^\times$ generate $R^\times$. It follows that $\mathrm{Pic}(\tilde E,S)$ is isomorphic to $\mathbb{Z}\times(\mathbb{Z}/3)$, and $\mathrm{Pic}(E)$ is isomorphic to $\tilde E$ divided by a certain 3-element subgroup.

Answer 2

There are likely more-elementary approaches, but here is an argument using some basic function-field arithmetic. Let's work a bit more generally than in the original question, considering a base field $F$, not of characteristic $3$. Let $R$ be the ring $F[x,y] / (x^3 + y^3 - 1)$, and let $L$ be the fraction field of $R$. Let's assume furthermore that $F$ contains an element $\omega$ that is a primitive cube root of $1$. We will show that $R$ is not a UFD by showing that the ideal $(x, y - 1)$ is not principal.

The ring $R$ has three evident units, $\eta_1 = x + y$, $\eta_2 = \omega x + y$, and $\eta_3 = \omega^2 x + y$, satisfying $\eta_1 \eta_2 \eta_3 = 1$. This rank-$2$ group of units complicates the arithmetic of $R$, as discussion in comments elsewhere noted. The field $L$ is something like a totally real cubic number field. (If $F$ did not contain a primitive cube root of $1$, then we would only have a rank-$1$ group of units, and the arithmetic would be simpler. The field $L$ would be something like a cubic number field with one real and one complex infinite place.)

The function field $L$ over $F$ has three infinite places (where the valuation of $x$ is negative), all of which have degree $1$. We can describe these places explicitly using the coordinates $u = 1/x$ and $v = y/x$. The infinite places are $P = (u, v + 1)$, $Q = (u, v + \omega)$, and $R = (u, v + \omega^2)$. Using these places, we find that the divisor of $\eta_1$ is $2P - Q - R$; that the divisor of $\eta_2$ is $-P + 2Q - R$; and that the divisor of $\eta_3$ is $-P-Q+2R$. As a consequence of these divisor calculations, we have $2P - Q \sim R$ and $3P \sim 3Q$, where $\sim$ denotes linear equivalence of divisors in the function field $L$ over $F$.

Let $T$ be a degree-$1$ place of $L$, and suppose that we have $T \sim aP + bQ + cR$ for some integers $a, b, c$, where $\sim$ denotes linear equivalence of divisors in $L$ as above. A bit of algebra with the divisors of the $\eta_i$ shows that $T$ is equivalent to one of $P, Q, R$. Indeed, using $-P + 2Q \sim R$, we can eliminate $R$ to get $T \sim a'P + b'Q$. Using $3P \sim 3Q$, we then have $T \sim a'' P + b''Q$, where $b''$ is $0, 1,$ or $-1$. Since the total degree of $a''P + b''Q$ is $1$, in the first case we have $T \sim P$. In the same way, the second case gives $T \sim Q$, and, using $2P - Q \sim R$, the third case gives $T \sim R$.

Let $\mathfrak{p}$ be a prime ideal of $R$ whose residue field is $F$, such as $(x, y - 1)$. We show that $\mathfrak{p}$ is not principal. Suppose for contradiction that $\mathfrak{p}$ were principal, generated by an element $f$ of $R$. The divisor of $f$ would then be $T - (aP + bQ + cR)$, where $T$ is the degree-$1$ place of $L$ corresponding to $\mathfrak{p}$, and where $a,b,c$ are suitable integers. By the argument above, we could then conclude that one of $T\sim P$, $T \sim Q$, or $T\sim R$ holds. Since $T$ is a finite place, it is distinct from the three infinite places, and so these linear equivalences are impossible, as the genus of $L$ is $1$.

(There are various ways to see that these linear equivalences are impossible. Suppose, for example, that $g$ were an element of $L$ with divisor $T - P$. Then the $1$-form $g \frac{dx}{y^2}$ would have a single singularity, a simple pole at $P$ with non-zero residue. But this is impossible, since the sum of residues of a meromorphic differential for $L$ must be $0$.)