# Integrality

• Oct 22nd 2008, 09:38 PM
terr13
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.
• Oct 22nd 2008, 10:01 PM
NonCommAlg
Quote:

Originally Posted by terr13
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?
• Oct 22nd 2008, 10:05 PM
terr13
You're right, sorry about the confusion, it should be algebraic over L.
• Oct 22nd 2008, 11:19 PM
NonCommAlg
Quote:

Originally Posted by terr13
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! (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?