[proofplan] We translate irreducibility of the projective [closed set](/page/Closed%20Set) into ideal-theoretic primality by using homogeneous equations only. First we record that $I_+(X)$ is homogeneous, proper, and cannot contain the irrelevant ideal because $X$ has a point. Then irreducibility rules out a factorization $fg \in I_+(X)$ with $f,g \notin I_+(X)$ by decomposing $X$ into the two closed loci where the relevant homogeneous factors vanish. Conversely, a reducible decomposition produces homogeneous forms separating the two proper closed subsets, and their product vanishes on all of $X$ while neither factor does. [/proofplan] [step:Verify that the homogeneous vanishing ideal is proper and avoids the irrelevant ideal] By definition of $I_+(X)$, each homogeneous component of an element of $I_+(X)$ again lies in $I_+(X)$, so $I_+(X)$ is a homogeneous ideal of $R$. Since $X$ is nonempty, choose a point $p=[a_0:\cdots:a_n]\in X$. At least one coordinate $a_i$ is nonzero. The constant polynomial $1\in R$ does not vanish at $p$, so $1\notin I_+(X)$; hence $I_+(X)$ is proper. Also $x_i(a_0,\ldots,a_n)=a_i\neq 0$, so $x_i\notin I_+(X)$. Therefore the irrelevant ideal $(x_0,\ldots,x_n)$ is not contained in $I_+(X)$. [guided] The first point is purely structural. The ideal $I_+(X)$ is defined degree by degree: its degree-$d$ piece consists exactly of the homogeneous degree-$d$ forms that vanish at every point of $X$. Therefore, if a polynomial lies in $I_+(X)$, each of its homogeneous pieces also lies in $I_+(X)$. This is precisely the compatibility with the standard grading of $R=k[x_0,\ldots,x_n]$, so $I_+(X)$ is a homogeneous ideal. Now use the hypothesis that $X$ is nonempty. Choose $p=[a_0:\cdots:a_n]\in X$. Since this is a point of [projective space](/page/Projective%20Space), the tuple $(a_0,\ldots,a_n)$ is not the zero tuple, so there is an index $i\in\{0,\ldots,n\}$ with $a_i\neq 0$. The constant polynomial $1$ evaluates to $1$ at every representative of $p$, so $1$ does not vanish on $X$. Thus $1\notin I_+(X)$, and $I_+(X)$ is a proper ideal. For the irrelevant ideal, the same point gives the obstruction. Since $a_i\neq 0$, the homogeneous coordinate function $x_i$ satisfies \begin{align*} x_i(a_0,\ldots,a_n)=a_i\neq 0. \end{align*} Thus $x_i\notin I_+(X)$. Because $x_i\in (x_0,\ldots,x_n)$, the containment $(x_0,\ldots,x_n)\subset I_+(X)$ is impossible. [/guided] [/step] [step:Show that irreducibility forces primality] Assume that $X$ is irreducible. It is enough to prove the prime condition for homogeneous elements. Indeed, let $f,g\in R$ satisfy $fg\in I_+(X)$, and suppose $f\notin I_+(X)$ and $g\notin I_+(X)$. Write $f=\sum_d f_d$ and $g=\sum_e g_e$ for their homogeneous decompositions, where $f_d,g_e\in R$ are homogeneous of degrees $d,e$. Choose $d_0$ minimal with $f_{d_0}\notin I_+(X)$ and $e_0$ minimal with $g_{e_0}\notin I_+(X)$. Since $I_+(X)$ is homogeneous and $fg\in I_+(X)$, the degree-$d_0+e_0$ homogeneous component of $fg$ lies in $I_+(X)$. All terms in that component except $f_{d_0}g_{e_0}$ contain either some $f_d\in I_+(X)$ with $d<d_0$ or some $g_e\in I_+(X)$ with $e<e_0$, hence lie in $I_+(X)$. Therefore $f_{d_0}g_{e_0}\in I_+(X)$. The homogeneous test proved below then forces $f_{d_0}\in I_+(X)$ or $g_{e_0}\in I_+(X)$, contradicting the choices of $d_0$ and $e_0$. Let $f,g\in R$ be homogeneous and suppose that $fg\in I_+(X)$. Define the projective closed subsets \begin{align*} X_f := X\cap V_+(f), \qquad X_g := X\cap V_+(g). \end{align*} For every $p=[a_0:\cdots:a_n]\in X$, the equality $(fg)(a_0,\ldots,a_n)=0$ holds. Since $k$ is a field, this implies $f(a_0,\ldots,a_n)=0$ or $g(a_0,\ldots,a_n)=0$. Hence \begin{align*} X = X_f \cup X_g. \end{align*} Both $X_f$ and $X_g$ are closed subsets of $X$. Since $X$ is irreducible, either $X=X_f$ or $X=X_g$. In the first case $f$ vanishes on all of $X$, so $f\in I_+(X)$; in the second case $g\in I_+(X)$. Therefore $I_+(X)$ is prime. [guided] We want to prove that $I_+(X)$ is prime. Because $I_+(X)$ is homogeneous, it suffices to test products of homogeneous elements. Indeed, the quotient $R/I_+(X)$ is a graded ring, and a homogeneous ideal is prime exactly when the product of two homogeneous classes being zero forces one of the two classes to be zero. Thus we take homogeneous polynomials $f,g\in R$ and assume \begin{align*} fg\in I_+(X). \end{align*} Now define the two closed pieces of $X$ cut out by the two factors: \begin{align*} X_f := X\cap V_+(f), \qquad X_g := X\cap V_+(g). \end{align*} These are closed in $X$ because $V_+(f)$ and $V_+(g)$ are projective closed subsets of $\mathbb{P}^n_k$. The assumption $fg\in I_+(X)$ means that for every projective point $p=[a_0:\cdots:a_n]\in X$, \begin{align*} f(a_0,\ldots,a_n)g(a_0,\ldots,a_n)=0. \end{align*} Since $k$ is a field, a product in $k$ is zero only when at least one factor is zero. Therefore every point of $X$ lies in $X_f$ or in $X_g$, and so \begin{align*} X = X_f\cup X_g. \end{align*} Irreducibility now applies directly: an irreducible [topological space](/page/Topological%20Space) cannot be written as the union of two proper closed subsets. Hence either $X=X_f$ or $X=X_g$. If $X=X_f$, then $f$ vanishes at every point of $X$, so $f\in I_+(X)$. If $X=X_g$, then $g\in I_+(X)$. This proves the prime condition. [/guided] [/step] [step:Show that reducibility produces a nonprime product] Conversely, suppose that $X$ is reducible. Then there exist proper closed subsets $Y,Z\subsetneq X$ in the Zariski [subspace topology](/page/Subspace%20Topology) such that \begin{align*} X=Y\cup Z. \end{align*} By the definition of the projective Zariski topology in the theorem statement, the closed subsets of $\mathbb{P}^n_k$ are exactly the sets $V_+(A)$ cut out by homogeneous equations $A\subset R$. Since $Y$ is closed in $X$, there is a set $A\subset R$ of homogeneous polynomials such that $Y=X\cap V_+(A)$. Because $Z\not\subset Y$, choose $q\in Z\setminus Y$. Then some homogeneous $f\in A$ satisfies $f(q)\neq 0$. Since $Y\subset V_+(f)$, we have $f\in I_+(Y)$, while $f\notin I_+(X)$ because $q\in X$ and $f(q)\neq 0$. Similarly, because $Y\not\subset Z$, choose $r\in Y\setminus Z$ and choose a [homogeneous polynomial](/page/Homogeneous%20Polynomial) $g\in R$ such that $Z\subset V_+(g)$ but $g(r)\neq 0$. Then $g\notin I_+(X)$. For every $p\in X=Y\cup Z$, either $p\in Y$, in which case $f(p)=0$, or $p\in Z$, in which case $g(p)=0$. Therefore $(fg)(p)=0$ for every $p\in X$, so \begin{align*} fg\in I_+(X). \end{align*} Thus $I_+(X)$ contains the product $fg$ but contains neither factor. Hence $I_+(X)$ is not prime. [/step] [step:Conclude the equivalence and the final ideal-theoretic properties] The previous two steps show that $X$ is irreducible if and only if $I_+(X)$ is prime. The first step already established, using only nonemptiness of $X$, that $I_+(X)$ is proper and that \begin{align*} (x_0,\ldots,x_n)\not\subset I_+(X). \end{align*} Therefore, whenever the equivalent conditions hold, $I_+(X)$ is a proper prime homogeneous ideal not containing the irrelevant ideal. [/step]