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Let $F$ be an algebraically closed field, and let $(V,f)$ be a quadratic space over $F$. How can one show that if $\dim V \geq 2$ then it is an isotropic space?


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up vote 4 down vote accepted

Choose an arbitrary basis of $V$ denoted by $\{v_i\}$. Consider the equation

$$ f(v_1 + k v_2) = 0 $$

which is a quadratic polynomial in $k$. (Define $q_1 = f(v_1)$, $q_2 = f(v_2)$, and $q_{12} = f(v_1 + v_2)$. You can show that $f(v_1 + k v_2) = q_1 + k^2 q_2 + k( q_{12} - q_1 - q_2 )$.)

Since $F$ is algebraically closed, you can solve for $k$ a root. Then the vector $v = v_1 + k v_2$ (which is non-zero since $v_1,v_2$ are linearly independent) is an isotropic vector.

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