# Cohomology ring of Grassmannians

I'm reading a paper called An Additive Basis for the Cohomology of Real Grassmannians, which begins by making the following claim (paraphrasing):

Let $w=1+w_1+ \ldots + w_m$ be the total Stiefel-Whitney class of the canonical $m$-plane bundle over $G_m(\mathbb{R}^{m+n})$ and let $\bar{w}=1+\bar{w_1}+\ldots+ \bar{w_n}$ be its dual. Then $H^\ast G_m (\mathbb{R}^{m+n})$ is the quotient of the polynomial algebra $\mathbb{Z}_2[1,w_1,\ldots,w_m]$ by the ideal generated by the relation $w\bar{w}=1$.

The reference provided is to Borel's La cohomolgie mod 2 de espaces homogenes. As the title suggests, this paper is in French, a language with which I am not familiar.

I'm familiar with the fact that the cohomology ring of the infinite Grassmannian $G_m(\mathbb{R}^\infty)$ is freely generated by $w_1,\ldots,w_m$ over $\mathbb{Z}_2$ (as proved in Hatcher's Vector Bundles), but I can't see how to prove this variant. Any help would be much appreciated. Perhaps someone can even translate the proof given in Borel's paper.

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One approach might be to note that the relations hold on the infinite level, so via inclusion, you have a surjection from the algebra mod the relation onto the cohomology of the m-Grassmannian. Now, use the cell structure and make a dimension counting argument to prove it must be an isomorphism. –  Justin Young May 11 '12 at 18:07
I think this should follow naturally, you may read Milnor's book. –  Kerry May 12 '12 at 5:30
It is an exercise in Milnor-Stasheff as well (problem 7B, to be precise). It is rather easier to understand than it is to prove. Justin's answer seems to provide a method of solving it, though. –  HSN May 6 '13 at 10:50