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I am looking at a problem in Hartshorne's Algebraic Geometry book. (Chapter $1$, Section $2$, $10$th problem) I am stating it as it is given in the book.

Let $Y\subset \mathbb{P}^n$ be a nonempty algebraic set, and let $\theta :\mathbb{A}^{n+1}\setminus\{(0,\ldots,0)\} \to \mathbb{P}^n$ be the map which sends the point with affine coordinates $(a_0,\ldots,a_n)$ to the point with homogeneous coordinates $(a_0,\ldots,a_n).$ We define the affine cone over $Y$ to be $C(Y)=\theta^{-1}(Y)\cup \{(0,\ldots,0)\}.$

(a) Show that $C(Y)$ is an algebraic set in $\mathbb{A}^{n+1},$ whose ideal is equal to $I(Y),$ considered as an ordinary ideal in $K[x_0,\ldots,x_n].$

(b) $C(Y)$ is irreducible iff $Y$ is.

(c) $\dim C(Y)=\dim Y+1.$

I got stuck at part (c), while part (a) and part (b) follows from the following observation

If $\mathfrak{a}$ is a homogeneous ideal in $K[x_0,\ldots,x_n]$ with $Z_{\mathbb{P}^n}(\mathfrak{a})\neq\emptyset,$ then $Z_{\mathbb{A}^{n+1}}(\mathfrak{a})=\theta^{-1}(Z_{\mathbb{P}^n}(\mathfrak{a}))\cup \{0,\ldots,0\}$.

Part (c) is clear when $Y$ is a projective variety since $\dim C(Y)=\dim K[x_0,\ldots,x_n]/I(Y)=\dim Y+1.$

My question is what happens if $Y$ is just projective algebraic set not necessarily irreducible ?

Help me. Many thanks.

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    One idea: dimension of a space is the supremum of dimensions of its irreducible components. By breaking your algebraic set into irreducibles you may be able to reduce to the case of an algebraic variety.2017-02-01
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    Since I am following Hartshornes book I want to apply his definition as the supremum of lengths of chain of irreducible closed set as dimension.2017-02-01
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    Andrew's suggestion uses the same definition, and is an extremely common tool. If $X$ is irreducible contained in $Y_1 \cup Y_2$, where each $Y_i$ is irreducible, then $X$ is contained in either $Y_1$ or $Y_2$. Therefore, the dimension of a space is the supremum of dimensions of its irreducible components.2017-02-02
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    I got it..thanks.2017-02-02

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