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Let $Y\subset\mathbb{P}_{k}^{n}$ be a projective variety. We say that Y is a linear subvariety if $I(Y)$ can be generated by linear polynomials. Now how I should show that $Y\subset\mathbb{P}_{k}^{n}$ is a linear subvariety if and only if $Y$ can be written as an intersection of hyperplanes.

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Let $p_1, \ldots, p_i$ linear polynomials which generate $I(Y)$. Then $I(Y) = \bigcap_{\mu=1}^i I(p_\mu)$. Now what is $I(p)$ for a linear $p$? –  martini Dec 13 '12 at 21:32
    
Perhaps $I(P)$ will be hyperplane? –  Iman Bahmani Dec 13 '12 at 21:35
    
Exactly. ${}{}{}$ –  martini Dec 13 '12 at 21:36
    
What I should do for it's inverse? –  Iman Bahmani Dec 13 '12 at 21:40
    
A hyperplane is of the form $I(p_\mu)$ for a linear $p_\mu$. If $Y = \bigcap_{\mu} V(p_\mu)$, then $I(Y) = \bigcup_{\mu} I(V(p_\mu)) = (p_1, \ldots, p_i)$. –  martini Dec 13 '12 at 22:12
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We assume $k$ is algebraically closed. Let $L_0,\dots,L_m$ be linear homogeneous polynomials in $k[x_0,\dots,x_n]$. Let $Y = V(L_0,\dots,L_m)$. We claim that $P = (L_0,\dots,L_m)$ is a prime ideal. We may assume that $L_0,\dots,L_m$ are linearly independent over $k$. There exist linear homogeneous polynomials $L_{m+1},\dots,L_n$ such that $L_0,\dots,L_n$ are linearly independent over $k$. Let $y_i = L_i$ for $i = 0,\dots,n$. Then $k[x_0,\dots,x_n] = k[y_0,\dots,y_n]$. Clearly $y_0,\dots,y_n$ are algebraically independent over $k$. Since $k[x_0,\dots,x_n]/P = k[y_0,\dots,y_n]/(y_0,\dots,y_m)$ is isomorphic to $k[y_{m+1},\dots,y_n]$, $P$ is a prime ideal. By the Hilbert Nullstellensatz for homogeneous ideals, $I(Y) = (L_0,\dots,L_m)$.

The converse is trivial.

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