This question is motivated from here

Let $S=\Bbb{C}[x,y]$. Consider the degree $0$ homogeneous elements of the localization at $x$, denoted $S_{(x)}.$ Then it is alleged that

$S_{(x)}\cong \Bbb{C}[y/x] \cong \Bbb{C}[s]$

Where I imagine the second isomorphism takes $y/x\mapsto s$.

Now I have two concerns (that I suspect are the one concern)

  • The first concern is that $y/x$ has degree $0$, and is sent to $s$ which has degree $1$.

  • The second concern is that $\Bbb{C}[s]$ has two homogeneous prime ideals at all, the irrelevant ideal $(s)$ and the trivial ideal $(0)$ (which is generic), the latter being the only homogeneous prime ideal. Whereas $\Bbb{C}[y/x]$ has numerous homogeneous prime degree $0$ ideals right, say $\frac{y+3x}{x}$?

I imagine that the homogeneous prime ideals should be in correspondence, between two isomorphic rings.

  • 3
    $\begingroup$ When you take degree zero elements, there is no more grading left. So, your worry about degree zero element going to a degree one element is moot. If I understand you correctly, a typical generator of your ring is of the form $f(x,y)/x^d$ where $f$ is homogeneous of degree $d$. Then, it is same $f(1,s)\in\mathbb{C}[s]$. No homogeneous issues come up. $\endgroup$ – Mohan Oct 4 '17 at 15:04

To expand on my previous answer, say we take a graded ring $S$. Then $Proj(S)$ is a scheme covered by the open affines $D(f)$. It is a theorem that the homogeneous primes in $D(f)$ are in bijection with the primes of degree zero in the localization $S_f$.

If $S=k[x,y]$, then the homogeneous primes of $k[x,y]_x$ that lie in degree $0$ are those of the form $(\frac{ay+bx}{x})$. These are clearly in bijection with primes of $k[y/x] \cong k[s]$. So we are not looking for homogeneous primes in $k[s]$, but rather primes generated by elements of degree zero in the localized ring $S_x$. Does this make sense?

  • $\begingroup$ As for your first concern, the isomorphism is not required to be graded, since we only care about the prime spectrum of the two rings $\endgroup$ – leibnewtz Oct 4 '17 at 15:50

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.