Dimension of irreducible component of reduced ring

Question

Let $X= SpecA$ denote the spectrum of a reduced ring $A$. Is there any way to tell the dimension of an irreducible component $Y$ of this variety? Each irreducible component corresponds to a minimal prime ideal, does this mean that the dimension is zero? (or one?)

Answer 1

It is not zero or $1$, consider the quotient $A=C[x,y,z]/x(x-1)$ where $C$ is the field of complex numbers, $Spec(A)$ has two irreducible components corresponding to the ideal generated respectively by the classes of $x$ and $x-1$ in $A$. The dimension of these components is 2.

Answer 2

Let $k$ be a field and consider the rings $A_0=k, A_j=k[x_1,...,x_j]$ and $A=\Pi^n_{i=0} A_i$.
The scheme $X=\operatorname {Spec}(A)$ has dimension $n$ but its irreducible components $X_i=\operatorname {Spec}(A_i)=\mathbb A^i_k$ have dimension $\dim X_i=i$.
So, no, the dimension of an affine scheme $X$ won't tell you the dimension of its irreducible components, which can be any number up to $\dim X$.