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I'm trying to learn about deformation theory.

Consider $k[x,y]/\left< y^2 - x^2\right>$.

To deform $k[x,y]/\left< y^2 - x^2\right>$ to make it look like $k[x,y]/\left< y^2\right>$, one would introduce a parameter $k[x,y,t]/\left< y^2 - t x^2\right>$ and make $t\rightarrow 0$.

Here's my question: if $x$, $y$ and $t$ each has weight 1, then $y^2 - t x^2$ is no longer homogeneous. When doing deformation theory, does the polynomial of interest need to be homogeneous?

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In order to make $y^2 - t x^2$ homogeneous, would it be okay if I impose the following weights?

My first choice is to impose $wt(x)=wt(y)=1$ while $wt(t)=0$, but this isn't correct, right, since then $t$ is thought of as a constant? What goes wrong here?

My next choice is to impose $wt(y)=3$, $wt(x)=2$, and $wt(t)=2$.

Thanks for your time.

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1 Answer 1

up vote 3 down vote accepted

No, don't worry, no homogeneity whatsoever with respect to $t$ is required : the parameter space $T$ to which $t$ belongs is to be thought of as small or infinitesimal and thus $T$ is not projective.

In characteristic $\neq2$, your example describes two distinct points $[1:\pm \sqrt t]$ of $\mathbb P^1_k$ coalescing into the double point with support $[1:0]$: a healthy and pleasant deformation.

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Thank you Georges. This helped me a lot. –  math-visitor May 25 '12 at 8:55
    
Glad to hear that, dear math-visitor.Good luck in your learning deformation theory. –  Georges Elencwajg May 25 '12 at 8:57

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