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lines changed Original file line number Diff line number Diff line change @@ -116,7 +116,7 @@ \subsection{Examples}
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their product $ a_n\in A$ , which becomes $ X_n^2 $ modulo $ Q$ , so all
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of the $ X_{i}$ will be algebraically independent in $ L/K$ and $ X_{i}^2 \in K$ .
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- \section {The extension $ B/A$ . }
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+ \section {The extension \texorpdfstring { $ B/A$ }{B/A} . }
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The precise set-up we'll work in is the following. We fix $ G$ a finite group acting
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on $ B$ a commutative ring, and we have another commutative ring $ A$ such
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\begin {proof } Use $ M_b$ .
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\end {proof }
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- \section {The extension $ (B/Q)/(A/P)$ . }
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+ \section {The extension \texorpdfstring { $ (B/Q)/(A/P)$ }{(B/Q)/(A/P)} . }
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Note that $ Q$ is prime, so $ B/Q$ is an integral domain and hence nontrivial.
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Furthermore, all our polynomials are monic and hence nonzero (indeed they
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$ X=\beta $ shows that $ \beta $ divides $ \alpha $ .
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\end {proof }
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- \section {The extension $ L/K$ . }
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+ \section {The extension \texorpdfstring { $ L/K$ }{L/K} . }
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\begin {theorem }
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\label {foo1 }
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