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Let $k$ be an algebraically closed field. Define a variety as an integral scheme over $k$ whose structure morphism is separated and finite type.

Let $R = \{f \in k[x,y] : f(0,0) = f(1, 1) \}$. Is $\operatorname{Spec} (R)$ a variety?

Integral: yes, it's a subring of $k[x,y]$.

Separated: All affine schemes are separated.

Finite type: This is where I'm stuck. It contains polynomials of the form $xy (x-1)(y-1) g(x) +k $, but I don't know if that's all of it. I think we can use long division of $f \in R$ by $xy (x-1)(y-1)$ to get something with lower degree, and find our generating set that way.

KReiser
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David Lui
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  • $R=k+(x,y)(x-1,y-1)k[x,y]$. – Kenta S Jan 13 '24 at 08:41
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    You can get the proper font and spacing for operators like $\operatorname{Spec}$ using \operatorname{Spec}. – joriki Jan 13 '24 at 08:58
  • @KentaS I still can't figure out how to get a generating set from that, though. For example, if I take ${1, (x, x-1), (x, y-1), (y, x-1), (y, y-1) }$, then I can't figure out how to write $x(y-1)(x-2)$ as a sum/product of those elements. – David Lui Jan 13 '24 at 12:41
  • Dear David, have you had a chance to look at the answer below? – KReiser Jan 29 '24 at 18:27
  • @KReiser Sorry about that, I forgot about this question , but now I read it over and it makes sense, so I accepted it. – David Lui Jan 29 '24 at 18:48

1 Answers1

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Yes, it is a variety. From Kenta S's identification in the comments, $R=k+(x,y)(x-1,y-1)k[x,y]\subset k[x,y]$. I claim that $$\{x(x-1),x(y-1),y(x-1),y(y-1),x^2(x-1),x^2(y-1),y^2(x-1),y^2(y-1)\}$$ forms a set of generators for $R$ as a $k$-algebra. To show this, it suffices to prove that every element of $(x,y)(x-1,y-1)$ can be written as a polynomial in these generators.

To do this we'll work backwards. Suppose $f\in (x,y)(x-1,y-1)$ and $\deg f = d > 1$: I claim that we can write $f=f'+p$ for $p$ a polynomial in the generators, $f'\in (x,y)(x-1,y-1)$, and $\deg f' < \deg f$. To accomplish this, if the coefficient of $x^iy^j$ with $i+j=d$ is nonzero in $f$, we can remove it as follows:

  • If $i=2a,j=2b$: subtract a constant multiple of $(x(x-1))^a(y(y-1))^b$.
  • If $i=2a+1,j=2b$: note that at least one of $a,b$ is positive, so subtract either a constant multiple of $(x^2(x-1))(x(x-1))^{a-1}(y(y-1))^b$ or $y^2(x-1)(x(x-1))^a(y(y-1))^{b-1}$.
  • If $i=2a,j=2b+1$: the same logic shows we can subtract either a constant multiple of either $(x^2(y-1))(x(x-1))^{a-1}(y(y-1))^b$ or $y^2(y-1)(x(x-1))^a(y(y-1))^{b-1}$.
  • If $i=2a+1,j=2b+1$: subtract a constant multiple of $(x(x-1))^a(y(y-1))^b(x(y-1))$.

So it suffices to show that if $f=cx+dy+e$ for $c,d,e\in k$ is in $(x,y)(x-1,y-1)$, we can write it as a polynomial in the above generators. But we must have $e=0$ and $c+d=0$ for this to be true, so $f=c(x-y)=c(xy-y-xy+x)=c(y(x-1)-x(y-1))$ and we're done.

KReiser
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