Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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!

share|improve this answer
Thank you. I understood. –  Tom Sep 3 '12 at 11:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.