Is this the cohomology ring of $\mathbb RP^2 \times \mathbb RP^2$? Today I took a fancy in calculating the cohomology ring of $\mathbb RP^2 \times \mathbb  RP^2$. As a result, I obtained $\mathbb Z[a,b,c]/ (2a,2b,2c,a^2,b^2,c^2,ac,bc)$, where $a$ and $b$ live in degree $2$ and $c$ lives in degree $3$. For this calculation I used the Künneth formula. Is this calculation correct?
 A: Your answer is correct, but it does not follow directly from the Künneth formula. First you get from the Künneth formula that
$$ H^*(\mathbb{R}P^2 \times \mathbb{R}P^2;\mathbb{Z}/2) \cong \mathbb{Z}/2[x,y]/(x^3,y^3) $$ 
where $x$ and $y$ both have degree one. Then you can for instance compare the cellular complexes for the cohomology with coefficients in $\mathbb{Z}$ and $\mathbb{Z}/2$ through the quotient map $\mathbb{Z} \to \mathbb{Z}/2$. 
From the cellular complex with coefficients in $\mathbb{Z}$ you find first that the cohomology groups of $\mathbb{R}P^2 \times \mathbb{R}P^2$ with coefficients in $\mathbb{Z}$ are (starting in dimension $0$)
$$\mathbb{Z},0,\mathbb{Z}/2 \oplus \mathbb{Z}/2, \mathbb{Z}/2 \text{ and } \mathbb{Z}/2$$
So you need two generators $a$, $b$ in degree $2$ and one generator $c$ in degree $3$, perhaps another in degree $4$. And since they generate $\mathbb{Z}/2$, you have $2a=2b=2c=0$. Then with the comparison map of complexes, the element $a$ corresponds to $x^2$ and $b$ to $y^2$, hence $ab$ corresponds to $x^2y^2$, and so it must be the generator in degree $4$. From this you also see that $a^2=0=b^2$. The other relations follow from dimensionality.
If you want, I can write this in more detail, although you already had the right answer, so I guess you did something similar.
