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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- Oct 22nd 2008, 09:38 PMterr13Integrality
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.

- Oct 22nd 2008, 10:01 PMNonCommAlg
- Oct 22nd 2008, 10:05 PMterr13
You're right, sorry about the confusion, it should be algebraic over L.

- Oct 22nd 2008, 11:19 PMNonCommAlg
you again repeated what you already posted! (Wondering) 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?