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I wonder if either the coordinate ring $A(X):=\mathbb{C}[x,y]/(x^2+y^2-1)$ or the polynomial ring $\mathbb{C}[t,t^{-1}]$ are unique factorization domains?

I know that those are isomorphic, so the answere for one of those would do it. In my trys of solving this issue, I saw that: $y^2 = (1-x)(1+x)$ ist a none-unique factorization in irreducible elements, because $y, (1-x)$ and $(1+x)$ are irreducible in $\mathbb{C}[x,y]/(x^2+y^2-1)$. But I can't even prove, that $y$ is irreducible. I know, that there are no none-units nor units, which I could multiply and get $y$ again. As far as I know, the units in $A(X)$ are the constant polynomials.

I hope that someone can understand my issue (so in advance in hope you can read my spelling w/o getting hurt :) ..) and make this topic clear to me.

Thanks

user26857
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Shalec
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  • Are you sure that $\mathbb C[x,y]/(x^2+y^2-1)$ is isomorphic to $\mathbb C[t,t^{-1}]$? I'm pretty sure the first one is not local. – Jakob Hansen Jun 09 '16 at 19:54
  • The second one is also not local. It's just a localisation. – Hmm. Jun 09 '16 at 19:57
  • Sorry, I take it back — the usual projection to a line should work. But this was not obvious at first glance. – Hoot Jun 09 '16 at 20:01
  • @Hoot, yes I do agree on that part. – Hmm. Jun 09 '16 at 20:02
  • I just saw that on that post: http://math.stackexchange.com/questions/112033/coordinate-ring-of-zero-locus-of-unit-circle and still on a proof. – Shalec Jun 09 '16 at 20:26
  • @JakobHansen Geometrically, these two rings are isomorphic because over $\mathbb{C}$, a circle and hyperbola are isomorphic as curves over $\mathbb{C}$. Note that $\mathbb{C}[t,t^{-1}] \cong \mathbb{C}[t,u]/(tu - 1)$. Letting $t = x+iy$ and $u = x - iy$ we have $x^2 + y^2 - 1 = (x+iy)(x-iy) - 1 = tu - 1$, hence this map transforms the circle into a hyperbola. – Viktor Vaughn Jun 10 '16 at 05:45
  • The localization of a UFD is a UFD : see https://math.stackexchange.com/questions/140584. – Watson Nov 01 '16 at 20:18

1 Answers1

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Observe that $\mathbb{C}[t,t^{-1}]=\mathbb{C}[t]_t$. Now, as localization of a UFD is a UFD, we are done.

In fact, we actually have a stronger result. Localization of a PID is again a PID . Since $\mathbb{C}[t]$ is a PID, $\mathbb{C}[t,t^{-1}]$ is a PID as well.

Hmm.
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  • Well, I didn't know, that $\mathbb C[t,t^{-1}]=\mathbb{C}[t]_t$ is the localizsation fo $\mathbb C[t]$ at the Element t. The last statement is known by the way.

    I wonder, because of http://www.fen.bilkent.edu.tr/~franz/ag06/ag03.pdf example 3.

     – Shalec Jun 09 '16 at 19:40
  • It's wrong. $Y$ is not irreducible. As for why $K[C] \not\simeq K[X]$, observe that the only invertible elements of $K[X]$ are the constant polynomials, hence the image of $t$ has to be a constant polynomial. But then that gives rise to a contradiction. – Hmm. Jun 09 '16 at 19:53
  • So what's the factorization of X or Y, if it's not irreducible? I see, that $y^2=1-x^2=(1-x)(1+x)$, so I know that $y$ must devide $(1-x)$ or $(1+x)$. So there is an Element $a\in A(X)$ with $y=a(1+x)$ or $y=a(1-x)$. But a should not be a unit. Units are the constants, so $a$ shouldn't be a unit. Right? – Shalec Jun 09 '16 at 20:06
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    $y$ may not divide $1-x$ or $1+x$, since $y$ is not prime in this ring. As far as irreducibility of $y$ is concerned, note that $(x+iy-1)(x-iy+1)=2iy$. – Hmm. Jun 09 '16 at 20:10
  • Well nice, that's it. Thank you. – Shalec Jun 09 '16 at 20:13
  • @Shalec, glad to help :) – Hmm. Jun 09 '16 at 20:14
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    Perhaps the author meant to work over $\mathbb R$. There's probably something interesting about factoriality to say in that case. – Hoot Jun 09 '16 at 20:16
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    @Hoot: you're right $\mathbf R[x,y]/(x^2+y^2-1)$ is not factorial. In fact its class group is isomorphic to $\mathbf Z/2$, a generator being the ideal of a real point like $(x-1,y)$. This ideal cannot be cut out by only one real polynomial. – Johannes Huisman Jun 09 '16 at 20:33
  • The author noted in the beginning, that the field are algebraically closed. Well I still got my probs with showing that:$\phi:\ A(X)\to \mathbb C[t,t^{-1}], t\mapsto (x+iy), t^{-1}\mapsto (x-iy)$ is isomorphic. The proof for injectivity is left. – Shalec Jun 10 '16 at 06:13
  • The proof was done by defining the inverse homomorphism, which maps $x \mapsto \frac{1}{2}(t-t^{-1})$ and $y \mapsto \frac{i}{2}(t+t^{-1})$. Well I'm actually not sure about the sign. If someone needs to show the bijectivity, just define this map for inverting x+iy and x-iy, so that $\psi(x+iy)=t$ and $\psi(x-iy)=t^{-1}$. – Shalec Jun 10 '16 at 11:42