Dimension of the tangent space of an algebraic set

can you give some hints for the following exercise? Thank you.

Let $k$ be an algebraic closed field, $X=V(x^2+y^2+z^2-1,x-1)\subset \mathbb A^3_k$.

[(a)] Find the dimension of $X$.

[(b)] Describe the tangent space of $X$ at the point $P=(1,0,0)$. What is its dimension? The point $P$ is a singular point?

[(c)] Determine all the singular points of $X$.

I have done this

[(a)] $X$ is the intersection of the unitary sphere $S=\{(x,y,z)\in\mathbb A_k^3|x^2+y^2+z^2=1\}=V(x^2+y^2+z^2-1)$ with the plane $\Pi=\{(x,y,z)\in\mathbb A_k^3|x=1\}=V(x-1)$, ie, $X=V(x^2+y^2+z^2-1)\cap V(x-1)=\{(1,0,0)\}\cup V(y^2+z^2)$, thus dim $X=1$

[(b)] Let $X=V(F_1,F_2)$, where $F_1(x,y,z)=x^2+y^2+z^2-1$ and $F_2(x,y,z)=x-1$, then the Jacobian matrix is

$\left(\cfrac{\partial F_i}{\partial a}\right) _{i=1,2, a=x,y,z}(x,y,z)= \left( \begin{array}{ccc} 2x & 2y & 2z \\ 1 & 0 & 0 \\ \end{array} \right)\Longrightarrow \left(\cfrac{\partial F_i}{\partial a}\right) _{i=1,2, a=x,y,z}(1,0,0)= \left( \begin{array}{ccc} 2 & 0 & 0 \\ 1 & 0 & 0 \\ \end{array}\right),$ and it has rank 1.

Thereforedim $T_pX=3-1=2$.

[(c)] Taking partial derivaties, to $f=y^2+z^2$, we get $f_x=0,f_y=2y,f_z=2z$, them the only singular point is $(1,0,0)$

Is that correct?

Thanks

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Your $k$ is not $\mathbb{R}$ (which isn't even algebraically closed!). So your (a) and (c) have problems. – user27126 Dec 4 '13 at 2:55
You are right, i am going to edit it – Miguemate Dec 4 '13 at 3:07
@Sanchez I have edited it, can you see it please? – Miguemate Dec 4 '13 at 3:13
You claim that $X$ has dimension $2$: that is not true. – Georges Elencwajg Dec 4 '13 at 18:01
@GeorgesElencwajg Thanks for your help. Do you mean that dim \$T_pX=2? – Miguemate Dec 5 '13 at 4:09