Confusion about genus-degree formulaMy question regards a perplexity I have on how to apply the genus-degree formula for irreducible, projective, complex plane curves. Consider first the affine complex plane curve given by the equation C : ( x − 2 ) ( x − 1 ) ( x + 1 ) ( x + 2 ) − y 2 = x 4 − 5 x 2 + 4 − y 2 = 0.The Jacobian is given by ( x ( 2 x + 10 ) ( 2 x − 10 ) , − 2 y ), so it never vanishes on C. Let us now look at the compactification C ^ , which has the same points as C plus a point at infinity with coordinates [ x : y : z ] = [ 0 : 1 : 0 ]. This point lies in the affine chart with y = 1, and the affine equation for C ^ in terms of x and z in that chart is x 4 − 5 z 2 x 2 + 4 z 4 − z 2 = 0.The differential ( 4 x 3 − 10 x z 2 , − 10 x 2 z + 16 z 3 − 2 z ) vanishes at ( x , z ) = ( 0 , 0 ), hence C ^ has one singularity: the point at infinity. If we pretend for a moment that it doesn't (i.e., that it is smooth), one can do the "usual construction" to see that it is topologically a torus: one can draw two cuts along [-2,-1] and [1,2] on two copies of the Riemann sphere and glue them together with the right orientation. So if C ^ were regular, it would be an elliptic curve, and in particular have genus 1. However, the genus-degree formula for projective plane curves predicts genus 3, since the equation of C ^ has degree 4. But the Wikipedia article on the genus-degree formula also mentions that the formula actually gives the arithmetic genus and that for every ordinary singularity of multiplicity r, the geometric genus is smaller than the arithmetic genus by 1 2 r ( r − 1 ). Now, I am not really sure about how to measure the multiplicity of a singularity, but in this case it seems that for any value of r ≥ 0, we never have that 3 − 1 2 r ( r − 1 ) = 1. So the geometric intuition and the formula seem to disagree. The only other thing that comes to my mind is that I have not checked yet that C ^ is irreducible, but this can be checked on C using Eisentein's criterion applied to the polynomial ring ( C [ x ] ) [ y ] using the prime ideal p = ( x + 1 ).Reassuming, my question is: what is the genus of C ^ ? If it is 1, why is the genus-degree formula wrong? If it is 3, why is the geometric intuition wrong? After all, also the article of Wikipedia on elliptic curves seems to confirm that C ^ should have genus 1.

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Confusion about genus-degree formulaMy question regards a perplexity I have on how to apply the genus-degree formula for irreducible, projective, complex plane curves. Consider first the affine complex plane curve given by the equation  C  :     (  x  −  2  )  (  x  −  1  )  (  x  +  1  )  (  x  +  2  )  −      y    2    =      x    4    −  5      x    2    +  4  −      y    2    =  0.The Jacobian is given by   (  x  (  2  x  +      10    )  (  2  x  −      10    )  ,  −  2  y  ), so it never vanishes on C. Let us now look at the compactification             C      ^      , which has the same points as C plus a point at infinity with coordinates   [  x  :  y  :  z  ]  =  [  0  :  1  :  0  ]. This point lies in the affine chart with   y  =  1, and the affine equation for             C      ^       in terms of x and z in that chart is      x    4    −  5      z    2        x    2    +  4      z    4    −      z    2    =  0.The differential   (  4      x    3    −  10  x      z    2    ,  −  10      x    2    z  +  16      z    3    −  2  z  ) vanishes at   (  x  ,  z  )  =  (  0  ,  0  ), hence             C      ^       has one singularity: the point at infinity. If we pretend for a moment that it doesn&#039;t (i.e., that it is smooth), one can do the &quot;usual construction&quot; to see that it is topologically a torus: one can draw two cuts along [-2,-1] and [1,2] on two copies of the Riemann sphere and glue them together with the right orientation. So if             C      ^       were regular, it would be an elliptic curve, and in particular have genus 1. However, the genus-degree formula for projective plane curves predicts genus 3, since the equation of             C      ^       has degree 4. But the Wikipedia article on the genus-degree formula also mentions that the formula actually gives the arithmetic genus and that for every ordinary singularity of multiplicity r, the geometric genus is smaller than the arithmetic genus by       1    2    r  (  r  −  1  ). Now, I am not really sure about how to measure the multiplicity of a singularity, but in this case it seems that for any value of   r  ≥  0, we never have that   3  −      1    2    r  (  r  −  1  )  =  1. So the geometric intuition and the formula seem to disagree. The only other thing that comes to my mind is that I have not checked yet that             C      ^       is irreducible, but this can be checked on C using Eisentein&#039;s criterion applied to the polynomial ring   (      C    [  x  ]  )  [  y  ] using the prime ideal       p    =  (  x  +  1  ).Reassuming, my question is: what is the genus of             C      ^      ? If it is 1, why is the genus-degree formula wrong? If it is 3, why is the geometric intuition wrong? After all, also the article of Wikipedia on elliptic curves seems to confirm that             C      ^       should have genus 1.

artirw9f · Accepted Answer

Step 1I think the reason the formula does not apply is that the singularity is not an &quot;ordinary singularity of multiplicity r&quot; (a.k.a. r distinct lines crossing at a point).From your second chart, we have a cusp at the origin, and if we blow it up (basically substitute xz for z and factor out an       x    2   - this is equivalent to enlarging the coordinate ring by adjoining z/x, which is integral over it), we&#039;re left with  4      x    2        z    4    −  5      x    2        z    2    +      x    2    −      z    2    =  0And the lowest degree part is       x    2    −      z    2    =  0  =  (  x  −  z  )  (  x  +  z  ), so locally the singularity is now ordinary of multiplicity 2. If this were the entire singularity, we would just be subtracting 1, but since we also needed the function z/x to resolve the cusp, we also subtract one more.More details to justify the last part. For any smooth projective curve C, let   f  :            C      ~        →  C be the normalization. There is a short exact sequence  0  →            O        C    →      f    ∗              O                                C          ~                      →  F  →  0,where F is supported only along the singularities of C, and is a finite-length             O        C  -module. Let its length be   ℓ. The cohomology gives  0  →      H    0    (      F    )  →      H    1    (            O        C    )  →      H    1    (      f    ∗              O                                C          ~                      )  →  0,so in particular       g    a    (  C  )  +  ℓ  =  g  (            C      ~        ), where       g    a    (  C  ) means the arithmetic genus and   g  (            C      ~        ) is the geometric genus. So   ℓ is basically &quot;how many extra regular functions&quot; the normalization has.Step 2The claim is that   ℓ  =  2. Resolving the cusp introduced z/x to the coordinate ring. I think z/x satisfies a quadratic polynomial (this is true at least locally, you can check that  (  1  −  4      z    2    )  (  z      /    x      )    2    +  (  5      z    2    −      x    2    )  =  0  ,and since the leading coefficient doesn&#039;t vanish at the origin, this is as good as a monic polynomial.)So we&#039;ve only introduced one more function in this step. Then, the second step introduces the             1      2        r  (  r  −  1  )  =  1 additional function to separate the two lines.

My question regards a perplexity I have on how to apply the genus-degree formula for irreducible, projective, complex plane curves. Consider first the affine complex plane curve given by the equation C: (x-2)(x-1)(x+1)(x+2)-y^2=x^4-5x^2+4-y^2=0.

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