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Ler $R$ be a normal domain with its quotient field $K$. Let $L$ be a finite Galois extention of $K$. Let $T$ be the integral closure of $R$ in $L$. Then we can take a basis $t_1,\cdots t_m$ of vector space $L$ over $K$ so that each $t_i\in T$. It would be very appreciated if you prove or disprove the above.

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up vote 3 down vote accepted

This is indeed true. First, note that for every element $l \in L$, there is a non-zero element $r \in R$ such that $rl \in T$. (Mouse over empty area to see proof of this assertion)

Since $L/K$ is a finite, hence algebraic extension: for each $l$, there exists $f \in K[x]$ such that $f(l)=p_0+p_1l+\cdots p_kl^k=0$ . But we can assume this is a polynomial in $R[x]$ (by clearing denominators). Now multiply through by $p_k^{k-1}$, and see that this is a monic polynomial in $R[x]$ satisfied by $p^{k-1}l$, so $p^{k-1}l \in T$, as desired .

Now, pick a basis for $L$ over $K$, say $\{a_1,a_2 \cdots a_n\} \subset K$. Then, applying the above result to each $a_i$, we get $\{b_1,b_2 \cdots\} \subset T$, where each $b_i=r_ia_i$. Can you check that this is also a basis for $L$?

I've only sketched a proof, so if you need more details, please feel free to ask!

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Thank you. I understood. –  Tom Sep 3 '12 at 11:09
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