I apologize if this question is too simple but I cannot see why the following is true:

Let $ K $ be a perfect field and $ \overline{K} $ a fixed algebraic closure. Let $ X \subset \mathbb{P}^n(\overline{K}) $ be a projective variety defined over $ K $ (The definition I'm using of 'defined over $ K $' is that the homogeneous ideal $ I(X) $ can be generated by homogeneous elements in $ K[X_0, \cdots, X_n] $). Then I understand that the absolute Galois group $ G = \text{Gal}(\overline{K}/K) $ acts on $ X $ in the obvious manner: $$ \sigma \cdot [x_0: \cdots : x_n] = [\sigma x_0 : \cdots : \sigma x_n] $$ and that the invariants $ X^G $ are precisely the $ K $-rational points $ X(K) $.

Further, $ G $ also acts on the function field $ \overline{K}(X) $ by acting on coefficients. My question is: Why is $ \overline{K}(X)^G = K(X) $?

I could only make the start: Let $ \frac{f}{g} $ be an element of $ \overline{K}(X)^G $, so that $ f,g $ are homogeneous polynomials of the same degree in $ \overline{K}[X_0, \cdots, X_n] $ and $ g $ is not in $ I(X) $. Then $ \sigma \cdot \frac{f}{g} = \frac{f}{g} $ implies that $ \sigma f \cdot g - f \cdot \sigma g \in I(X) $. Now I'm not sure how to proceed. To descend to the ground field, I know I have to involve Galois cohomology somehow but the map $ \sigma \rightarrow \sigma f \cdot g - f \cdot \sigma g $ is not a $ 1 $-cocycle.

Any help will be appreciated.

  • $\begingroup$ You can assume that the denominator is invariant. You will get a cocycle. $\endgroup$ – Roland Feb 28 at 7:44

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