# Does isomorphic $\mathbb Q$-cohomology implies isomorphic $\mathbb Z$-cohomology when torsion-free?

The integral cohomology of smooth toric varieties is torsion-free, so in this case there is no loss of information when going to rational coefficients.

The above sentence may be interpreted as the following implication: if

• $$X$$ and $$Y$$ are manifolds
• $$H^*(X,\mathbb Z)$$ and $$H^*(Y,\mathbb Z)$$ have no torsion
• $$H^*(X,\mathbb Q)$$ is isomorphic to $$H^*(Y,\mathbb Q)$$ as graded rings

then

• $$H^*(X,\mathbb Z)$$ is isomorphic to $$H^*(Y,\mathbb Z)$$ as graded rings?

Question: Is this implication true?

• I'm not sure that's quite true if you want to state it for graded rings. What will probably be true however is if you start from a map $X\to Y$, then this map induces an isomorphism rationally if and only if it does so integrally (given that $X,Y$ have no torsion in their cohomology) Aug 25, 2020 at 14:55
• @MaximeRamzi I've edited the question: what matters is the discrimination power of cohomology as a topological invariant. Aug 25, 2020 at 15:04
• As stated, I don't think it's true. I don't have an example, but you could imagine an example where the multiplication of two generators $x_0,x_1$ yields $2y$ for some generator $y$, as opposed to one where the multiplication of two generators is just a third generator. Then they would be isomorphic over $\mathbb Q$, but not over $\mathbb Z$. But I don't have examples in mind Aug 25, 2020 at 18:30

Let X be a Hirzebruch surface $$H_n$$. It can be viewed as a (smooth) toric variety generated by the fans

z0=(0,1)
z1=(0,-1)
z2=(1,0)
z3=(-1,-n)
f0=[z2,z0]
f1=[z2,z1]
f2=[z3,z0]
f3=[z3,z1]


, shown in this picture for $$n=2$$. According to p.106 of Fulton's book "Introduction to toric varieties", the $$\mathbb Z$$-cohomology ring of $$X$$ is

$$\mathbb Z(z_0,z_1,z_2,z_3)/(z_0z_1,z_2z_3,z_0-z_1-nz_3,z_2-z_3)$$, and the $$\mathbb Q$$-cohomology

$$\mathbb Q(z_0,z_1,z_2,z_3)/(z_0z_1,z_2z_3,z_0-z_1-nz_3,z_2-z_3)$$. $$^\text{[remark]}$$

Therefore all the Hirzebruch surfaces $$H_n$$ have the same $$\mathbb Q$$-cohomology when $$n\geq 1$$, but their $$\mathbb Z$$-cohomology are different for even and odd $$n$$.

Remark: In fact, the Fulton book only provides $$\mathbb Z$$-cohomology directly, but the proof also works for $$\mathbb Q$$-cohomology. A more direct method of obtaining $$\mathbb Q$$-cohomology is by a SageMath computation:

n = 4  #try different values of n
cone1 = Cone([(1,0), (0,1)])
cone2 = Cone([(1,0), (0,-1)])
cone3 = Cone([(-1,-n), (0,-1)])
cone4 = Cone([(-1,-n), (0,1)])
H = Fan([cone1, cone2, cone3, cone4])
T = ToricVariety(H)
print(T.cohomology_ring().defining_ideal())