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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?)

arla
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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.

Tsemo Aristide
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  • Are there criteria as to when the dimension might be 1? In my case, I need the transcendence degree of the field extension $\kappa(x) \backslash \kappa$ to be 1, where $\kappa(x)$ ist the function field of the irreducible component and $\kappa$ is the underlying field, such that $SpecA$ is a scheme over $\kappa$. This happens when the dimension of the irreducible variety is one (?) Any ideas on this? – arla May 08 '16 at 07:48
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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 $\dim X$.