Let k = F_2. Let A = ∏_n k, i.e., the product of copies of F_2 indexed by the integers. Today, I am going to make a local ring R with a nonzerodivisor t in the maximal ideal such that R[1/t] is isomorphic to k((t)) ⊗_k A.

Namely, choose a basis B of A as a k-vector space such that some element b_0 ∈ B corresponds to 1 in A. Thus every element of A can be uniquely written as a finite sum of elements of B without repetitions. In particular, given b, b’ ∈ B we can write bb’ = ∑_{b” ∈ C(b, b’)} b” for a finite subset C(b, b’) of B. Then A has the presentation A = k[b]/(b_0 – 1, bb’ – ∑_{b” ∈ C(b, b’)} b”). Consider the ring

R = k[[t]][x_b]/(x_{b_0} – t, x_bx_{b’} – t ∑_{b” ∈ C(b, b’)} x_{b”})

If we invert t then we can replace x_b by x_b/t and we get a presentation of k((t)) ⊗_k A. I claim that t is a nonzero divisor in R. To show this you show that {x_b} is a basis of the quotient ring over k[[t]] (this takes a bit of work). Observe that the quotient of R by t is the ring k[x_b]/(x_{b_0}, x_bx_{b’}) whose spectrum is a singleton. A bit more work shows every prime ideal of R is contained in the ideal (t, x_b) which implies that R is local.

Let I_{fin} ⊂ A be the ideal of elements of A = ∏_n k consisting of sequences (a_n) such that all but a finite number of a_n are zero. Note that A/I_{fin} is a flat A-module as I_{fin} is generated by idempotents (every element of A is an idempotent). Let I be the unique radical ideal of R such that I[1/t] = k((t)) ⊗_k I_{fin} via the isomorphism above. Then we see that M = R/I is an R-module which is flat over the principal open U defined by t.

I constructed M to illustrate Raynaud-Gruson 5.2.2: Namely, with X = S = Spec(R) and U the open given above there is no finite type blow-up of S such that the strict transform of M becomes flat. The theorem only applies when M restricted to U is of finite presentation; an assumption which our M fails.

My reasoning is as follows. Note that the zero set of I_{fin} is nowhere dense in Spec(A). Hence also V(I) ∩ U is nowhere dense in U because U is homeomorphic to Spec(A). But if the strict transform of M becomes flat on some finite type blow up X’ of Spec(R), then M gets rank 1 over a connected component of the exceptional fibre of X’ —> X. I think this implies that M has rank 1 over a nonempty open of U as well. I haven’t check all the details so I could be wrong… let me know if so! Also, an easier example would be appreciated as well.