Find the singular point

Question

Let $f(x, y)=(x-y)^2$. We want to find the singular points. We do the following:

Let $P=(a, b)$ be the singular point.

$$f(a,b)=0 \Rightarrow (a-b)^2=0 \Rightarrow a=b \\ \frac{\partial{f}}{\partial{x}}(a,b)=0 \Rightarrow 2(a-b)=0 \Rightarrow a=b \\ \frac{\partial{f}}{\partial{y}}(a, b)=0 \Rightarrow -2(a-b)=0 \Rightarrow a=b$$

So is the singular point $P=(a,a)$ ?

score 1 · Answer 1 · answered Jan 21 '15 at 07:12

1

Right, this curve is singular everywhere, assuming you're working over $\mathbb{C}$ or at least over a reduced ring of characteristic not $2$.

answered Jan 21 '15 at 07:12

Kevin Carlson

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So all the points of the curve are Singular? Why? Are all the points of the curve of the form $P=(a,a)$ ? – Jan 21 '15 at 09:59
Yes, you already showed that in your first displayed equation. – Kevin Carlson Jan 21 '15 at 17:05

Find the singular point

1 Answers1