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I am slightly confused by some statements in Hatcher's Vector Bundle book (page 60). To start with (and I am happy with) the natural ring homomorphism $$K(S^2) \simeq \mathbb{Z}[H]/(H-1)^2$$

I am trying to understand this statement

The external product $K(S^{2k}) \otimes K(X) \to K(S^{2k} \times X)$ is an isomorphism. This follows from (2) by the same reasoning which showed the equivalence of the reduced and unreduced forms of Bott periodicity. Since external product is a ring homomorphism, the isomorphism $K(S^{2k} \smash X) \simeq K(S^{2k}) \otimes K(X)$ is a ring isomorphism.

(I am okay up to here)

For example, since $K(S^{2k})$ can be described as the quotient ring $\mathbb{Z}[\alpha]/(\alpha^2 )$, we ...

(where he defines $\alpha$ as the pullback of the generator of of $\tilde{K}(S^{2k})$ under the projection $S^{2k} \times S^{2 \ell} \to S^{2k}$).

I don't see why it is not $\mathbb{Z}[\alpha]/((\alpha-1)^2 )$?

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Certainly $\mathbb{Z}[x] = \mathbb{Z}[x-1]$. –  Dylan Wilson Nov 29 '12 at 1:59

1 Answer 1

up vote 2 down vote accepted

It's the same ring: $\mathbb Z[a]/a^2\cong \mathbb Z[H]/(H-1)^2$ ($a\mapsto H-1$).

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oh you are correct. For some reason I was thinking this would be $\mathbb{Z}[a+1]/(a^2)$ –  Juan S Jun 13 '11 at 9:24
    
@Juan: that's also the same ring. –  Qiaochu Yuan Dec 2 '13 at 8:58

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