Complex K-theory cohomology ring $K^*(T^2)$ I was considering the complex K-theory of $T^2$ and I found, using the split exact sequences associated with the pair $(T^2,S^1 \vee S^1)$ that
$\widetilde{K}^0(T^2)\cong\mathbb{Z}$ and $\widetilde{K}^{-1}(T^2)\cong\mathbb{Z}\oplus\mathbb{Z}$,
so that one arrives at
$K^*(T^2)\cong \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\oplus \mathbb{Z}$.
Now my question is about the ring structure here. We know that in $K(T^2)$ we have the same ring structure as in $K(S^2)\cong \mathbb{Z}[H]/(H-1)^2$, where $H$ is the standard degree one line bundle over $S^2$, and hence we understand how to multiply some factors since the product in $K^*$ reduces to that in $K$ when multiplying degree zero elements, but I can't seem to understand how it works in general. I would appreciate any pointers on how to derive the ring structure in this case.
Thanks in advance!
 A: You have to take more information from the exact sequence. Indeed it tells you what $K^*(T^2)$ is but also where its elements "come from".
For instance, the sequence tells you that $\tilde K^0(S^2)\to \tilde K^0(T^2)$ induced by the quotient map $p:T^2\to S^2$ is an isomorphism.
Similarly, by Bott periodicity, this holds in all even degrees.
In odd degrees, you know that the restriction map $K^{2i+1}(T^2)\to K^{2i+1}(S^1\vee S^1)$, induced by the inclusion $i:S^1\vee S^1\to T^2$, is an isomorphism.
But both $K^*(S^2)\to K^*(T^2)$ and $K^*(T^2)\to K^*(S^1\vee S^1)$ are also ring maps.
So for instance let's take an odd degree element $x\in K^{2i+1}(T^2)$ and an even degree one, $y\in K^{2j}(T^2)$. Then $y= p^*(z)$ for some $z\in K^{2j}(S^2)$, therefore $i^*(xy) = i^*(x) i^*(p^*(z))$. Now $p\circ i$ is the constant map, so $i^*p^*$ is exactly the map induced by $S^1\vee S^1\to *\to S^2$, so if $z \in \mathbb Z[H]/(H-1)^2$, $i^*p^*(z)$ corresponds to taking the degree $0$ monomial of $z$, so it's just $i^*(x) \epsilon(z) = i^*(\epsilon(z)x)$ because $\epsilon(z)\in\mathbb Z$
But $i^*$ is an isomorphism in odd degrees, so $xy = \epsilon(z)x$. So you're fixing the isomorphism $K^{2j}(T^2)\cong \mathbb Z^2$, you forget one of the coordinates of $y$ and you multiply $x$ by that (which is an integer)
The rest of the ring structure is determined similarly : if you multiply two elements of even degree, they'll come from elements of even degree in $S^2$ and so this is entirely controlled by $K^*(S^2)$.
The last thing is what happens when you multiply two odd degree elements?
EDIT : as Tyrone pointed out, the argument I had written here does not work to compute the product of odd degree elements.
In fact, as he suggested, there is a Künneth formula, as $K^*(S^1)$ is flat over $K^*$, that gives $K^*(T^2)\cong K^*(S^1)\otimes K^*(S^1)$, as rings and this allows for a good understanding of the ring structure on $K^*(T^2)$, better than what I tried to do here.
For a reference for that Künneth formula, one can check out the references of this nLab page, see e.g. Theorem (Boa95, 4.19) 2.12
So, to summarize :
-multiplication of even degree elements is controlled by $K^*(S^2)$
-multiplication of elements of different parity is "trivial" : if $x$ is odd and $y$ is even, then $xy = \epsilon(y)x$, where $\epsilon$ is the composite $K^*(T^2)\to K^*(\mathrm{pt})\to K^*(T^2)$
