[proofplan] We prove the two implications separately. If $X$ is projective, then a projective embedding pulls back the Fubini--Study form to a Kähler form on $X$, and the projective algebraic structure gives enough meromorphic functions to make $X$ Moishezon. Conversely, if $X$ is compact Kähler and Moishezon, we use as an external input the Grauert--Moishezon projectivity theorem, which says that these two hypotheses make $X$ biholomorphic to a smooth projective algebraic variety. [/proofplan] [step:Pull back the Fubini--Study form from a projective embedding] Assume first that $X$ is projective in the clarified sense of the statement. Thus there exist an integer $N \in \mathbb{N}$, a smooth projective algebraic subvariety $Y \subset \mathbb{P}^N$, and a biholomorphism \begin{align*} \Phi: X &\to Y. \end{align*} Let $\omega_{\mathrm{FS}}$ denote the Fubini--Study Kähler form on $\mathbb{P}^N$. Let $i: Y \hookrightarrow \mathbb{P}^N$ denote the inclusion map. Define the real $(1,1)$-form \begin{align*} \omega_X := \Phi^*(i^*\omega_{\mathrm{FS}}) \end{align*} on $X$. Since pullback by a holomorphic map preserves type and [exterior derivative](/theorems/1525), and since $d\omega_{\mathrm{FS}} = 0$, we have $d\omega_X = 0$. Since $\Phi$ is a biholomorphism and $i$ is an immersion, positivity of $\omega_{\mathrm{FS}}$ on complex tangent vectors implies positivity of $\omega_X$ on every nonzero vector in $T_xX$ for every $x \in X$. Hence $\omega_X$ is a Kähler form, so $X$ is Kähler. [guided] Assume $X$ is projective. Under the clarified statement this means that $X$ is not merely mapped into projective space abstractly: there are an integer $N \in \mathbb{N}$, a smooth projective algebraic subvariety $Y \subset \mathbb{P}^N$, and a biholomorphism \begin{align*} \Phi: X &\to Y. \end{align*} The standard Kähler form on projective space is the Fubini--Study form, denoted $\omega_{\mathrm{FS}}$. Let \begin{align*} i: Y &\hookrightarrow \mathbb{P}^N \end{align*} be the inclusion map. We define a real $(1,1)$-form on $X$ by \begin{align*} \omega_X := \Phi^*(i^*\omega_{\mathrm{FS}}). \end{align*} This is the natural candidate because a Kähler structure is preserved under restriction to complex submanifolds and under biholomorphic pullback. We verify the Kähler conditions. First, $\omega_{\mathrm{FS}}$ has type $(1,1)$, and holomorphic pullback preserves bidegree, so $\omega_X$ has type $(1,1)$. Second, $\omega_{\mathrm{FS}}$ is closed, so $d\omega_{\mathrm{FS}} = 0$. Since exterior differentiation commutes with pullback, \begin{align*} d\omega_X = d\Phi^*(i^*\omega_{\mathrm{FS}}) = \Phi^*(i^*d\omega_{\mathrm{FS}}) = 0. \end{align*} Third, if $x \in X$ and $v \in T_xX$ is nonzero, then $d\Phi_x(v) \in T_{\Phi(x)}Y$ is nonzero because $\Phi$ is a biholomorphism, and $di_{\Phi(x)}(d\Phi_x(v))$ is nonzero because $i$ is an immersion. Positivity of $\omega_{\mathrm{FS}}$ on nonzero complex tangent vectors gives positivity of $\omega_X$ on $v$. Therefore $\omega_X$ is a Kähler form on $X$, and $X$ is Kähler. [/guided] [/step] [step:Use the projective algebraic structure to obtain Moishezon meromorphic functions] For any compact complex manifold $Z$, write $\mathcal M(Z)$ for its field of meromorphic functions. Because $Y$ is a smooth connected projective algebraic variety of complex dimension $n := \dim_{\mathbb{C}}X$, its field of rational functions $\mathbb C(Y)$ has transcendence degree $n$ over $\mathbb{C}$. Rational functions on $Y$ define meromorphic functions on the compact complex manifold $Y$, so $\operatorname{trdeg}_{\mathbb C}\mathcal M(Y)\ge n$. On the other hand, Siegel's algebraic-dimension bound says that every compact connected complex manifold $Z$ satisfies $\operatorname{trdeg}_{\mathbb C}\mathcal M(Z)\le\dim_{\mathbb C}Z$; applied to $Z=Y$, it gives \begin{align*} \operatorname{trdeg}_{\mathbb C}\mathcal M(Y)\le \dim_{\mathbb C}Y=n. \end{align*} Equivalently, by the standard projective GAGA comparison for smooth projective complex varieties, meromorphic functions on the analytification of $Y$ are rational functions. Hence $\operatorname{trdeg}_{\mathbb C}\mathcal M(Y)=n$. Pullback by the biholomorphism $\Phi$ identifies meromorphic functions on $Y$ with meromorphic functions on $X$. Therefore the field of meromorphic functions on $X$ has transcendence degree $n$, which is the definition of $X$ being Moishezon. [guided] Let $n := \dim_{\mathbb{C}}X$. Since $\Phi: X \to Y$ is a biholomorphism, $Y$ is also connected and has complex dimension $n$. For a compact complex manifold $Z$, write $\mathcal M(Z)$ for the field of meromorphic functions on $Z$. A smooth connected projective algebraic variety is irreducible, so its rational function field is well-defined. The basic algebraic-geometric fact used here is that an irreducible projective variety of dimension $n$ has rational function field of transcendence degree $n$ over $\mathbb{C}$. Every rational function on $Y$ determines a [meromorphic function](/page/Meromorphic%20Function) on the associated compact complex manifold $Y$: it is holomorphic on the complement of its pole divisor and has meromorphic singularities along that divisor. This gives transcendence degree at least $n$ for $\mathcal M(Y)$. The reverse inequality is Siegel's algebraic-dimension bound for compact connected complex manifolds: \begin{align*} \operatorname{trdeg}_{\mathbb C}\mathcal M(Y)\le \dim_{\mathbb C}Y. \end{align*} Equivalently, in the smooth projective complex case, projective GAGA identifies meromorphic functions with rational functions. Pulling back by the biholomorphism $\Phi$ gives a field isomorphism between meromorphic functions on $Y$ and meromorphic functions on $X$. Therefore the meromorphic function field of $X$ has transcendence degree \begin{align*} \operatorname{trdeg}_{\mathbb{C}} \mathcal{M}(X) = n. \end{align*} By the definition of a Moishezon compact complex manifold, this proves that $X$ is Moishezon. [/guided] [/step] [step:Apply the Grauert--Moishezon projectivity theorem to the Kähler Moishezon case] Conversely, assume that $X$ is Moishezon and Kähler. We now invoke the Grauert--Moishezon projectivity theorem as an external prerequisite (citing a result not yet in the wiki: Grauert--Moishezon projectivity theorem): every compact complex manifold that is both Moishezon and Kähler is biholomorphic to a smooth projective algebraic variety. Its hypotheses are exactly satisfied here: compactness and complex manifold structure are part of the theorem statement, while the additional assumptions give the Moishezon and Kähler conditions. Hence $X$ is projective in the clarified sense. [guided] Now assume $X$ is Moishezon and Kähler. The required implication is supplied by the Grauert--Moishezon projectivity theorem, used here as a cited external input rather than reproved. In the form needed here, it says: if a compact complex manifold is Moishezon and admits a Kähler form, then it is biholomorphic to a smooth projective algebraic variety. We check the hypotheses one by one. The theorem statement gives that $X$ is a compact complex manifold. The present direction assumes that $X$ is Moishezon, meaning its meromorphic function field has transcendence degree $\dim_{\mathbb{C}}X$ over $\mathbb{C}$. It also assumes that $X$ is Kähler, meaning $X$ admits a positive closed real $(1,1)$-form. Therefore every hypothesis of the Grauert--Moishezon projectivity theorem is satisfied. The conclusion of that theorem is exactly that $X$ is biholomorphic to a smooth projective algebraic variety. This is precisely the clarified meaning of projective in the statement. Thus the converse implication follows. [/guided] [/step] [step:Combine the two implications] The first two steps prove that projectivity implies both the Kähler and Moishezon properties. The previous step proves that the Kähler and Moishezon properties together imply projectivity. Hence $X$ is projective if and only if $X$ is both Moishezon and Kähler. [/step]