Integral Extensions, Finitely Generated = Integral

Finitely Generated = Integral

Warning, this proof uses a matrix to represent an endomorphism from a finitely generated module into itself, and then takes the determinant of that matrix. If these concepts are foreign to you, you may want to review module endomorphisms, and determinants.

Let t be an element in the ring extension S/R. If t is integral then the extension R[t] is generated by the powers of t, from t⁰ up to t^n-1, where the monic polynomial of t has degree n. In other words, the ring R[t] is a finitely generated R module inside S.

Note, if t were merely algebraic, like 2t²-3, we would have to replace t², wherever it occurs, with 3/2, and 3/2 is not part of our base ring. So we really need t to be integral over R.

Now for the converse. Let M be a finitely generated R module in S, where M contains R[t]. Note that M could be R[t], as above, or it could be something larger.

Furthermore, let multiplication by t map M into itself. This is true when M is R[t], or S, and perhaps several modules in between. Since R and t both act on M, M is an R[t] module, as well as an R module.

Since M contains R[t] contains 1, the only element from R[t] that drives all of M into 0 is 0. Using terminology from the world of modules, the annihilator of M is 0.

Give M a basis b₁ b₂ b₃ … b_n. This is the finite set of generators that spans all of M. Now each member of M is a linear combination of these basis elements. However, the representation may not be unique; B may not be a basis in the truest sense of the word. I'm abusing the terminology just a bit. But B does span, and everything in M has at least one representation relative to the basis B. Since M is an R module, multiplication by x in R multiplies the coefficients on the basis elements by x. However, multiplication by t replaces each b_j with its image, which is a prescribed linear combination of the n basis elements. In other words, multiplication by t is described by an n×n matrix, where the j^th row defines tb_j. Since we've already used the letter M, let's call this matrix Y.

To recap, we can represent an element from M as a vector v, where v holds the coefficients on the basis B. Then v*Y, using matrix multiplication, produces the coefficients of t*v. There may be other linear combinations of the generators in B that produce tv, but vY is definitely one of them.

Now subtract t from the main diagonal of Y. So Y isn't a matrix over R any more, it has become a matrix over R[t]. And vY isn't vt any more, it is vt-vt, or 0. It doesn't matter what v is, run it through Y and get 0. You don't get a zero vector, but you do get a column vector from R[t] whose elements sum to 0.

Let d be the determinant of Y, which is an element of R[t].

Build a matrix Z as follows. Start with the identity matrix, then replace the first column with Z_i,1 = b_i. In other words, the first column contains our basis.

Consider the product Y*Z, using matrix multiplication. The first row of Y is dotted with the first column of Z. Ignore the -t in the upper left of Y, and the first row of Y contains coefficients of R, that are applied to the basis elements in the first column of Z. But this is, by definition, the image of b₁ when multiplied by t. So without the -t, the upper left entry of the product is b₁t.

Now bring in -t and subtract b₁t, leaving 0. Thus the upper left entry in the product matrix is 0.

Multiply the second row of Y by the first column of Z, and obtain b₂t-b₂t, or 0. This continues all the way down the column. I don't know what the other columns look like; it doesn't matter. The first column is 0, hence the determinant of the product is 0. In other words, det(Y)det(Z) = 0. The determinant of Y is d, and the determinant of Z is b₁. Thus db₁ = 0.

Leave the first column of Z alone, and rearrange the ones in the remaining columns. This time arrange the ones so that the j^th row is missing a one. Last time the first row was missing a one. This time the determinant of Z is ±b_j. Here's an example of Z with determinant b₃.

b1 1 0 0 0
b2 0 1 0 0
b3 0 0 0 0
b4 0 0 1 0
b5 0 0 0 1

Evaluate the product YZ, and once again the first column is zero. This means db_j = 0 for every j.

Remember that d is an element in R[t], and it maps every generator b_j to 0. By linearity it sends the entire module M to 0. Yet only 0 satisfies 0*M = 0, therefore d = 0.

Remember, d is the determinant of Y, which is zero. Expand the determinant of Y into a polynomial in t, with coefficients in R. This polynomial is monic, and it is equal to zero. Therefore t is the root of a monic polynomial in R, and is integral.

You might think R needs to be an integral domain, so that the product of the determinants is the determinant of the product, but this resul holds for all commutative rings, thanks to a rather technical proof. Thus there are no restrictions on R, other than the usual restrictions for this topic, i.e. being commutative and having 1.

Happy Corollaries

1. Let S be a finitely generated extension of R. Every t in S belongs to a finitely generated R module inside S, namely S. Therefore every t is integral, and S is an integral extension of R.

2. Let u and v be integral elements in a ring extension S/R, and consider R[u,v]. Cross the powers of u with the powers of v to show this is finitely generated as an R module. Therefore R[u,v] is an integral extension. Use induction to generalize this to a finite set of integral elements.

3. If u and v are integral as above, then R[u,v] is an integral extension that contains u+v, u-v, and uv. These are all integral over R.

4. Let W be the set of integral elements in S/R. We just showed W is closed under addition and multiplication, hence W forms a ring. This is the largest integral extension of R inside S.

5. Let S be an integral extension of R, and let T be an integral extension of S. Let v be an element of T, and let p(x) be the monic polynomial that proves v is integral over S. Let a₀ a₁ a₂ … a_n be the coefficients of p, taken from S. Each of these is integral over R. Adjoin these to R and find an integral extension, a finitely generated R module, which we will call G. Now G(v) is a finitely generated G module, which we will call H. Cross the generators of H over G with the generators of G over R to show H is a finitely generated R module. Thus v is integral over R, and since v was arbitrary, T is an integral extension of R. The composition of integral extensions is integral.

6. Let R be an integral domain, let F be the fraction field of R, let E be an algebraic extension of F, and let S be the integral extension of R inside E. Let B be a basis for E over F. Multiply each basis element by something in R, so that the basis elements are all integral, hence they all lie in S. Now the fraction field of S includes F, and B, hence the fraction field of S is all of E.

7. Let E/F and S/R be as above. Find a basis for E/F in S and represent the elements of S as unique linear combinations of basis elements with coefficients in F. The tensor product S×F is generated by elements of S crossed with reciprocals of R. Use the basis B, that lies in s, rather than all of s. If q is an arbitrary member of our basis B, consider (x/y)q crossed with 1/z. This is the same as xq cross 1/yz. The tensor product is the free R module generated by B, with denominators from R. This is the linear combinations of B with coefficients in F, which is E. Therefore S×F = E.

8. Let the ring extension S be a finitely generated R module, which becomes an integral extension. Let S be an integral domain, whence R is also an integral domain. Let F be the fraction field of R and let E be the fraction field of S. Assume S contains a free R module of rank n, and anything in S, outside this free R module, is a linear combination of basis elements with coefficients in F. This reproduces the conditions found above, so S embeds in S tensor F, which lies in E.

   Let b₁ through b_n be a basis for the free R module inside S. Let x be an element of S, and build a matrix M that represents multiplication by x, relative to this basis. In other words, the first row of M contains the coefficients for b₁ through b_n that produce x*b₁, and so on. Let d be the determinant of M. If d is 0 then there is a vector y such that yM = 0. You might need to dip into the fraction field F to build y, but you can always multiply through by a common denominator, so that y lies in R. View y as an element of S, and yx = 0, which is a contradiction. Therefore d is nonzero.

   Let W be the inverse of M, using the common denominator d. The entries of W lie in F. Let each row of W represent a linear combination of our basis, using coefficients in F. This is an element of S×F, which lives in E. Let the i^th row of W represent v_i. In other words, v_i is the sum over j of W_i,jb_j. Now, multiplying W_i by M effectively multiplies v_i by x. The result is the identity matrix, hence v_i*x = b_i. Write 1 as the sum over c_ib_i, and the sum of c_iv_i is the inverse of x. Therefore the fractions of S are all covered by the linear combinations of B with coefficients in F. More concisely, E is the tensor product of S and F as R modules. This is the same result we saw above. However, in this case S need not be integrally closed inside E.