I'm just learning blowups and resolutions of singularities and I have been unable to find a clear and concise resource on how to resolve singularities in general. I understand that the concept of "resolutions of singularities" involves adding in a relation with respect to some projective coordinates.
In this example, suppose that $\vec{x} = (x,y) \in \mathbb{R}^2$, then we introduce the projective coordinates $[X:Y] \in \mathbb{P}^1$, such that $xY = yX$, then WLOG assuming $X \neq 0$, we divide out by $X$, this leaves us with the relation $y = xY$, thus making this substitution, we find that:
$x^2 + y^2 = 0 \iff x^2 + x^2Y^2 = 0 \iff x^2(1 + Y^2) = 0$.
This does not seem to resolve my singularity, since I've still got a problem at $x = 0$, is anyone able to clarify for me what I am missing here? Most of the contructions that I've found just stop there, and it's not clear to me what I am doing, or what I am supposed to do.
Thanks.
Suppose you have two closed immersions of varieties $Z\subset Y\subset X$. If you want to compute $Bl_ZY$, the blowup of $Y$ with respect to $Z$, you can blow up $Z$ in $X$ and then take the strict transform of $Y$ in $Bl_ZX$. That is, consider the preimage of $Y\setminus Z$ in $Bl_ZX$ and then take the closure. (This is in contrast to the total transform of $Y$, which is the preimage of $Y\subset X$ under $Bl_ZX\to X$.)
In your situation, $Z=V(x,y)$, $Y=V(x^2+y^2)$, and $X=\Bbb A^2$. After blowing up $Z$, the total transform of $Y$ is locally $V(x^2(1+Y^2))$ as you've computed. To get the strict transform, divide by $x^2$, which gives you $V(1+Y^2)$, the right answer. (For a detailed explanation of why this is the right thing to do, see here.)