Cohomology ring of complex grassmanian as a quotient

Question

This is about the complex version of this question,

I am interested in understanding the integral cohomology ring of $\mathrm{Gr}_k(\mathbb{C}^n)$ as the following quotient of the integral cohomology ring of $\mathrm{Gr}_k(\mathbb{C}^{\infty})$: $$ H^*(\mathrm{Gr}_k(\mathbb{C}^n))\cong H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))/\langle \bar{c}_{n-k+1},\dots,\bar{c}_n\rangle. $$ Here the $\bar{c}_i$ are polynomials in the Chern classes $c_i$ of the tautological $k$-plane bundle over $\mathrm{Gr}_k(\mathbb{C}^{\infty})$, and are defined by the relation $$ c\bar{c}=1. $$ (Recall that $H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))\cong \mathbb{Z}[c_1,\dots,c_k].$)

The approach suggested for the real case (with $\mathbb{Z}/2$ coefficients) in the linked question is to show that

1. the inclusion $\mathrm{Gr}_k(\mathbb{C}^{n})\subset \mathrm{Gr}_k(\mathbb{C}^{\infty})$ induces a surjection $H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))\to H^*(\mathrm{Gr}_k(\mathbb{C}^n))$;
2. the polynomials $\bar{c}_{n-k+1},\dots,\bar{c}_n$ are zero in $H^*(\mathrm{Gr}_k(\mathbb{C}^n))$, hence the above map induces a surjection from the quotient $H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))/\langle \bar{c}_{n-k+1},\dots,\bar{c}_n\rangle \to H^*(\mathrm{Gr}_k(\mathbb{C}^n))$;
3. use a dimension count to deduce that the latter map is also injective, hence an isomorphism.

Question: Since we are working with integral cohomology, which is merely a ring (not a $\mathbb{Z}/2$-vector space), my question is how one could carry out or alter step (3) to obtain injectivity.

Thanks for the help. Any references are appreciated.

Answer 1

For each $i$, $H^i(\mathrm{Gr}_k(\mathbb{C}^n))$ is a finitely generated free abelian group, as is the $i$th degree of $H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))/\langle \bar{c}_{n-k+1},\dots,\bar{c}_n\rangle$. So, if you can show they have the same rank, any surjective homorphism between them is automatically injective as well (any surjection $\mathbb{Z}^n\to\mathbb{Z}^n$ is injective; more generally, this is true for any Noetherian module over a ring, by considering the kernels when you iterate the homomorphism). If you like, this means you can count their dimensions after tensoring with $\mathbb{Q}$ (or equivalently, use cohomology with coefficients in $\mathbb{Q}$) and check that they are the same.