Quotient field of the integral closure of a ring

[coquelicot]

This is probably a stupid question.
Let R be a domain, K its field of fractions, L a finite (say) extension of K, and S the integral closure of R in L.
Is the quotient field of S equal to L ?
I believe that not, but I have no counter-example.

 

[mathwonk]

the answer is yes. let a be any element of L. we want to express a as a quotient of elements of S. since L is finite over K, it is also algebraic over K, so there is a polynomial satisfied by a, with coefficients in K, and multiplying out the denominators, which lie in R, we get coefficients in R. Suppose c is the lead coefficient, and the polynomial has degree n. i.e. we have c a^n +...=0. multiplying through by c^(n-1) then gives an equation satisfied by (ca), of degree n, and with coefficients in R, and lead coefficient = 1, hence ca is integral over R, i.e. ca belongs to S. Thus a = (ca)/c is a quotient of elements of S, one of which is actually in R.

This is proposition 1, in Lang's chapter on integral ring extension, and is thus essentially the first fact about them.

 

[coquelicot]

Thx. This was not a stupid question, I am stupid.

 

[mathwonk]

well, rather this is a vote for lang's book, in putting basic things at the beginning. the fact that you asked this shows you identified correctly a basic question. that is an intelligent trait.

 

[blue_raver22]

No, the quotient field of S is not necessarily equal to L. In fact, there are counter-examples where the quotient field of S is strictly contained in L. For example, consider the ring R = Z[x], the field of fractions K = Q(x), and the finite extension L = Q(x^1/2). The integral closure of R in L is the ring S = Z[x^1/2], which has a quotient field of Q(x^1/2), which is strictly contained in L. Therefore, it is important to carefully consider the properties of the integral closure of a ring and its relationship to its quotient field.