Let $K$ be a field and let $V$ be a vector space on $K$. We consider the projective space $P(V)$ on $V$ (the set of the vector subespaces $A$ in $V$ such that $\dim A = 1$). Let $S , T$ be two vector subespaces in $V$. By definition $P(S)$ and $P(T)$ are two projective spaces in $P(V)$. I want to prove that $P(S) + P(T)$ is a new projective space in $P(V)$ (the sum of two projective spaces is a new projective space). Obviously, $S + T$ is a vector space in $V$, so $P(S + T)$ is a projective space in $P(V)$. Must I use this argument to prove that $P(S) + P(T)$ is a projective space in $P(V)$? Is the equality $P(S) + P(T) = P(S + T)$ a definition? Thank you very much.
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Can you define $P(S) + P(T)$ ? There is no addition defined in $P(V)$. – Mar 29 '17 at 17:38
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Forget that. How can I prove that sum of two projective spaces is a new projective space? – joseabp91 Mar 29 '17 at 17:40
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The expression $P(T) + P(S)$ does not make sense since there is no addition in $P(V)$. More precisely, the operation "$+$" does not pass to quotient.
So the only thing which wake sense is to look at $P(T + S)$ which is in fact the smallest projective space which contains both $P(T)$ and $P(S)$. But I won't call it $P(T) + P(S)$ since really the notion of addition does not exist when you do projectivize.
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Notice that there is nothing to prove with this definition : by definition a projective subspace $E$ is a subset which can be written $P(E')$ for some subspace $E' \subset V$. But here by definition $T+S$ is a subspace of $V$ so $P(T+S)$ is a projective subspace. – Mar 29 '17 at 17:43