Return to Answer

(Edit: I slightly misread your question. In this answer "unique factorization" means "of ideals," not elements.)

I'll try to give a basic answer, although I'm still learning about this stuff myself. The kind of non-unique factorization you've identified is due to the fact that $\mathbb{R}$ isn't algebraically closed, and isn't as interesting as another kind of non-unique factorization, which I'll exemplify using $\mathbb{C}[x, y]/(y^2 - x^3)$. Such rings arise as rings of functions on algebraic curves, and in that case one can pinpoint exactly what causes unique factorization to fail, which is the existence of singularities (here, at the point $(x, y) = (0, 0)$).

More precisely, it's known that the ring of functions on an algebraic curve $f(x, y) = 0$ has unique factorization of ideals if and only if every point is nonsingular in the sense that the partial derivatives $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ never simultaneously vanish. The geometric intuition here is that locally, at a singular point a variety looks like the intersection of some number of lines, i.e. it is "locally reducible," so the maximal ideal associated to that point isn't generated by one element. For $\mathbb{C}[x, y]/(y^2 - x^3)$ the singular point at $(0, 0)$ is a cusp where two lines meet and the corresponding ideal is generated by $x$ and $y$ but satisfies a nontrivial relation, and this is precisely non-unique factorization. More generally I believe one can characterize the ideals without unique factorization precisely as the ideals vanishing on singular points.

Anyway, the upshot of all this is that as Greg indicates, it is possible for varieties to have lots of nasty singularities. On the other hand, it's relatively easy to fix this problem for the algebraic curve case: the integral closure of the ring of functions will have unique factorization.

(Again, I'm still learning about this stuff, so if I've misstated something please let me know!)

(Edit: I slightly misread your question. In this answer "unique factorization" means "of ideals," not elements.)

I'll try to give a basic answer, although I'm still learning about this stuff myself. The kind of non-unique factorization you've identified is due to the fact that $\mathbb{R}$ isn't algebraically closed, and isn't as interesting as another kind of non-unique factorization, which I'll exemplify using $\mathbb{C}[x, y]/(y^2 - x^3)$. Such rings arise as rings of functions on algebraic curves, and in that case one can pinpoint exactly what causes unique factorization to fail, which is the existence of singularities (here, at the point $(x, y) = (0, 0)$).

More precisely, it's known that the ring of functions on an algebraic curve $f(x, y) = 0$ has unique factorization of ideals if and only if every point is nonsingular in the sense that the partial derivatives $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ never simultaneously vanish. The geometric intuition here is that locally, at a singular point a variety looks like the intersection of some number of lines, i.e. it is "locally reducible," so the maximal ideal associated to that point isn't generated by one element. For $\mathbb{C}[x, y]/(y^2 - x^3)$ the singular point at $(0, 0)$ is a cusp where two lines meet and the corresponding ideal is generated by $x$ and $y$ but satisfies a nontrivial relation, and this is precisely non-unique factorization. More generally I believe one can characterize the ideals without unique factorization precisely as the ideals vanishing on singular points.

Anyway, the upshot of all this is that as Greg indicates, it is possible for varieties to have lots of nasty singularities. On the other hand, it's relatively easy to fix this problem for the algebraic curve case: the integral closure of the ring of functions will have unique factorization.

(Again, I'm still learning about this stuff, so if I've misstated something please let me know!)

Or maybe you just wanted to know something about your specific case. Make the substitution $x = \frac{1 - t^2}{1 + t^2}, y = \frac{2t}{1 + t^2}$; then for example $x^2 = (1 + y)(1 - y)$ can be written as $(1 - t^2)^2 = (1 + 2t + t^2)(1 - 2t + t^2)$. So here the failure of unique factorization is quite simple: certain polynomials in $t$ are being treated as prime which "shouldn't be." On the other hand any polynomial in $x, y$ which, when written as a rational function in $t$, avoids these anomalous primes, will have the usual prime factorization properties as a polynomial in $t$, but these prime factors will not necessarily always come from polynomials in $x, y$.

I'll try to give a basic answer, although I'm still learning about this stuff myself. The kind of non-unique factorization you've identified is due to the fact that $\mathbb{R}$ isn't algebraically closed, and isn't as interesting as another kind of non-unique factorization, which I'll exemplify using $\mathbb{C}[x, y]/(y^2 - x^3)$. Such rings arise as rings of functions on algebraic curves, and in that case one can pinpoint exactly what causes unique factorization to fail, which is the existence of singularities (here, at the point $(x, y) = (0, 0)$).

More precisely, it's known that the ring of functions on an algebraic curve $f(x, y) = 0$ has unique factorization of ideals if and only if every point is nonsingular in the sense that the partial derivatives $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ never simultaneously vanish. The geometric intuition here is that locally, at a singular point a variety looks like the intersection of some number of lines, i.e. it is "locally reducible," so the maximal ideal associated to that point isn't generated by one element. For $\mathbb{C}[x, y]/(y^2 - x^3)$ the singular point at $(0, 0)$ is a cusp where two lines meet and the corresponding ideal is generated by $x$ and $y$ but satisfies a nontrivial relation, and this is precisely non-unique factorization. More generally I believe one can characterize the ideals without unique factorization precisely as the ideals vanishing on singular points.

Anyway, the upshot of all this is that as Greg indicates, it is possible for varieties to have lots of nasty singularities. On the other hand, it's relatively easy to fix this problem for the algebraic curve case: the integral closure of the ring of functions will have unique factorization.

(Edit: I slightly misread your question. In this answer "unique factorization" means "of ideals," not elements.)

I'll try to give a basic answer, although I'm still learning about this stuff myself. The kind of non-unique factorization you've identified is due to the fact that $\mathbb{R}$ isn't algebraically closed, and isn't as interesting as another kind of non-unique factorization, which I'll exemplify using $\mathbb{C}[x, y]/(y^2 - x^3)$. Such rings arise as rings of functions on algebraic curves, and in that case one can pinpoint exactly what causes unique factorization to fail, which is the existence of singularities (here, at the point $(x, y) = (0, 0)$).

More precisely, it's known that the ring of functions on an algebraic curve $f(x, y) = 0$ has unique factorization of ideals if and only if every point is nonsingular in the sense that the partial derivatives $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ never simultaneously vanish. The geometric intuition here is that locally, at a singular point a variety looks like the intersection of some number of lines, i.e. it is "locally reducible," so the maximal ideal associated to that point isn't generated by one element. For $\mathbb{C}[x, y]/(y^2 - x^3)$ the singular point at $(0, 0)$ is a cusp where two lines meet and the corresponding ideal is generated by $x$ and $y$ but satisfies a nontrivial relation, and this is precisely non-unique factorization. More generally I believe one can characterize the ideals without unique factorization precisely as the ideals vanishing on singular points.

Anyway, the upshot of all this is that as Greg indicates, it is possible for varieties to have lots of nasty singularities. On the other hand, it's relatively easy to fix this problem for the algebraic curve case: the integral closure of the ring of functions will have unique factorization.

(Again, I'm still learning about this stuff, so if I've misstated something please let me know!)

I'll try to give a basic answer, although I'm still learning about this stuff myself. The kind of non-unique factorization you've identified is due to the fact that $\mathbb{R}$ isn't algebraically closed, and isn't as interesting as another kind of non-unique factorization, which I'll exemplify using $\mathbb{C}[x, y]/(y^2 - x^3)$. Such rings arise as rings of functions on algebraic curves, and in that case one can pinpoint exactly what causes unique factorization to fail, which is the existence of singularities (here, at the point $(x, y) = (0, 0)$).

More precisely, it's known that the ring of functions on an algebraic curve $f(x, y) = 0$ has unique factorization of ideals if and only if every point is nonsingular in the sense that the partial derivatives $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ never simultaneously vanish. The geometric intuition here is that locally, at a singular point a variety looks like the intersection of some number of lines, i.e. it is "locally reducible." So in this case one can characterize the ideals without unique factorization precisely: they are the ideals vanishing on the singular points.

I'll try to give a basic answer, although I'm still learning about this stuff myself. The kind of non-unique factorization you've identified is due to the fact that $\mathbb{R}$ isn't algebraically closed, and isn't as interesting as another kind of non-unique factorization, which I'll exemplify using $\mathbb{C}[x, y]/(y^2 - x^3)$. Such rings arise as rings of functions on algebraic curves, and in that case one can pinpoint exactly what causes unique factorization to fail, which is the existence of singularities (here, at the point $(x, y) = (0, 0)$).

More precisely, it's known that the ring of functions on an algebraic curve $f(x, y) = 0$ has unique factorization of ideals if and only if every point is nonsingular in the sense that the partial derivatives $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ never simultaneously vanish. The geometric intuition here is that locally, at a singular point a variety looks like the intersection of some number of lines, i.e. it is "locally reducible," so the maximal ideal associated to that point isn't generated by one element. For $\mathbb{C}[x, y]/(y^2 - x^3)$ the singular point at $(0, 0)$ is a cusp where two lines meet and the corresponding ideal is generated by $x$ and $y$ but satisfies a nontrivial relation, and this is precisely non-unique factorization. More generally I believe one can characterize the ideals without unique factorization precisely as the ideals vanishing on singular points.

Anyway, the upshot of all this is that as Greg indicates, it is possible for varieties to have lots of nasty singularities. On the other hand, it's relatively easy to fix this problem for the algebraic curve case: the integral closure of the ring of functions will have unique factorization.