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Thread: Integrality

  1. #1
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    Integrality

    Let R be a domain with fraction field L, and assume a is algebraic over L, show that $\displaystyle \{r \in R | ra \,\, integral\,\, over \,\, R\}$is a nonzero ideal in R.
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    Quote Originally Posted by terr13 View Post
    Let R be a domain with fraction field L, and assume a is algebraic over L, show that $\displaystyle \{r \in R | ra \,\, integral\,\, over \,\, R\}$ is a nonzero ideal in R.
    are you sure the L is not R and i guess you're choosing $\displaystyle a \in L,$ right?
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    You're right, sorry about the confusion, it should be algebraic over L.
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    Quote Originally Posted by terr13 View Post
    You're right, sorry about the confusion, it should be algebraic over L.

    the L here should be R. otherwise the problem wouldn't make sense!
    you again repeated what you already posted! you should be more careful with posting your question! the correct and complete version of your question and the solution:

    Problem: let $\displaystyle R$ be a domain and $\displaystyle L$ its field of fractions. show that if $\displaystyle a \in L$ is algebraic over $\displaystyle R$, then the set $\displaystyle I=\{r: \ \ ra \ \ \text{is integral over} \ R \}$ is a non-zero ideal of $\displaystyle R$.

    Proof: $\displaystyle I$ is obviously closed under addition because if $\displaystyle r, s \in I,$ then $\displaystyle ra, \ sa$ are integral over $\displaystyle R,$ i.e. $\displaystyle ra, \ sa$ are in $\displaystyle \overline{R},$ the integral closure of $\displaystyle R$ in $\displaystyle L.$ since $\displaystyle \overline{R}$ is a ring, we have

    $\displaystyle (r+s)a \in \overline{R}.$ thus $\displaystyle r+s \in I.$ now suppose $\displaystyle s \in I$ and $\displaystyle r \in R.$ we must show that $\displaystyle rs \in I$: so $\displaystyle sa \in \overline{R}$ and $\displaystyle r \in R \subseteq \overline{R}.$ thus $\displaystyle rsa \in \overline{R},$ i.e. $\displaystyle rs \in I.$ this proves that $\displaystyle I$ is an ideal

    of $\displaystyle R.$ the only thing left is to show that $\displaystyle I \neq 0$: since $\displaystyle a$ is algebraic over $\displaystyle R$, we will have: $\displaystyle r_0a^m + r_1a^{m-1} + \cdots + r_m=0,$ for some integer $\displaystyle m \geq 1$ and $\displaystyle r_j \in R, \ r_0 \neq 0.$ hence

    $\displaystyle (r_0a)^m + r_1r_0(r_0a)^{m-1} + \cdots + r_m r_0^m = 0.$ so $\displaystyle r_0a$ is integral over $\displaystyle R,$ i.e. $\displaystyle 0 \neq r_0 \in I. \ \ \ \Box$

    something for you to think about:

    did we use the assumption that $\displaystyle R$ is a domain and $\displaystyle L$ is its fraction field? is the claim in the problem true for any (unitary) commutative ring R and any ring extension L of R?
    Last edited by NonCommAlg; Oct 24th 2008 at 10:15 PM. Reason: simplifying the proof
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