Dimension of the Zariski cotangent space for the non-smooth points I'm trying to understand the following problem:

Let $k$ be an algebraically closed field of characteristic not $2$. Consider the $k$-algebra $A = k[x_1,\ldots,x_n]/(x_1^2+\cdots+x_n^2)$. For each $n \geq 1$ determine which $k$-points of $\mbox{Spec}(A)$ are smooth and for the non-smooth $k$-points determine the dimension of the Zariski cotangent space.

First of all, can anyone give me a clean definition of smooth and non-smooth points of $\mbox{Spec}(A)$? I have read on the Wikipedia that if there is an ideal given by some system of equations $g_1=0,\ldots,g_r=0$, where each $g_i$ is in the polynomial ring $k[x_1,\ldots,x_n]$, then $X$ is smooth if it has dimension at least $m$ in a neighborhood of each point and the matrix of partial derivatives $(\frac{\partial g_i}{\partial x_j})$ has rank at least $n-m$ everywhere on $X$. But what to do if on my example the ideal is given not by the system, but only by single equation $x_1^2+\cdots+x_n^2 = 0$?
It seems to me that I understand the case $n=1$ but aren't able to generalize it to greater dimensions.
So, for $n=1$ we have $A = k[x]/x^2$. The solution of $x^2=0$ over $k$ is only zero, but $\frac{d(x^2)}{dx}|_0=2x|_0=0$ and we conclude that this point is non-smooth. By the definition of Zariski cotangent space at a point $P$ it is $m_P/m_P^2$, where $m_P$ is a maximal ideal obtained by localization at $P$. In a polynomial ring the ideal is given by the polynomials without free coefficient, then $m_P/m_P^2 = x/x^2$ which may have dimension $1$ over $k$ (I can't explain why it just seems to me that it is).
Maybe it will be helpful to admit that from general theory $\dim(m/m^2) \geq \dim k = 0$ and $\dim(k[x_1,\ldots,x_n]) = \dim k + n = n$ since $k$ is a field and thus of Krull dimension $0$.
Can anyone check is my explanation for $n=1$ true and generalize it to bigger dimensions? Feel free to provide your solutions and explanations.
 A: The algebraic set $X$ defined by $g = x_1^2 + \cdots + x_n^2 = 0$ is a subset of affine $n$-space.  Since it is cut out by one equation, $X$ has dimension $n-1$.  (This can be proved by Krull's Hauptidealsatz.)  At a smooth point, the cotangent space has the same dimension as $X$ (cf., Theorem 5.1 of Hartshorne).  At a singular point, the dimension of the cotangent space is "too big," which in this case means it must have dimension $n$.  (It can't be larger than the dimension of tangent space of the ambient space $\mathbb{A}^n$.)
As you said, we can find the singular point using the Jacobian matrix.  Since $X$ is defined by just the one equation above, then the Jacobian matrix is the $1 \times n$ matrix
$$
\begin{pmatrix}
2 x_1 & 2 x_2 & \cdots & 2 x_n
\end{pmatrix} \, .
$$
At the point $(x_1, \ldots, x_n) = (0, \ldots, 0)$, we get the zero matrix, whose kernel has dimension $n$.  Thus $(0, \ldots, 0)$ is a singular point.  At any other point the Jacobian has rank $1$, hence the kernel has dimension $n-1$, which is the dimension of $X$.  Thus all other points are smooth.
Your example is correct, but it's a bit hard to visualize.  Since $A$ is a quotient of $k[x]$, $X$ is a subset of the line $\mathbb{A}^1$.  The equation $x^2 = 0$ has only the single point $x = 0$ as its solution.  But this is a "fat" point as you observed: $X$ is just a point, hence has dimension $0$, but the cotangent space $\mathfrak{m}_p/\mathfrak{m}_p^2$ has dimension $1$, since $x$ is a $k$-basis for $\mathfrak{m}_p/\mathfrak{m}_p^2$.
I think the case $n=2$ is a bit easier to visualize.  Then
$$
A = \frac{k[x_1,x_2]}{(x_1^2 + x_2^2)}
$$
and $X$ is defined by the equation $0 = x_1^2 + x_2^2 = (x_1 + ix_2)(x_1 - i x_2)$ where $i^2 = -1$.  From this equation we can see that $X$ is the union of the two lines $x_1 + ix_2 = 0$ and $x_1 - ix_2 = 0$.  These lines each have dimension $1 = n - 1$ and intersect at the point $(0,0)$.
