I'd expect $$H^*(S^3 \times S^1,\mathbb{Z})=H^*(S^3)\otimes H^*(S^1)=[\mathbb{Z}<1>\oplus0\oplus0\oplus\mathbb{Z}<x>]\otimes[\mathbb{Z}<1>\oplus\mathbb{Z}<y>]=\mathbb{Z}<1>\oplus\mathbb{Z}<1\otimes y>\oplus0\oplus\mathbb{Z}<x\otimes 1>\oplus\mathbb{Z}<x\otimes y>=\mathbb{Z}[x]/_{(x^2)}\otimes\mathbb{Z}[y]/_{(y^2)}=\mathbb{Z}[x,y]/_{(x^2,y^2)},$$ however I saw in a book that it is $\mathbb{Z}[x,y]/_{(x^2,y^2,xy+yx)}.$
Here on math.stackexchange I learned that tensor product in the isomorphism from Künneth formula is supposed to be graded tensor product. So my question is - how do I calculate graded tensor product? For example, to get the right answer above. This page https://planetmath.org/gradedtensorproduct says that the grading is same as in regular tensor product, so I guess additively the answer still must look like $\mathbb{Z}<1>\oplus\mathbb{Z}<y>\oplus0\oplus\mathbb{Z}<x>\oplus\mathbb{Z}<x\otimes y>$. Where do I get $(xy+yx)$ from though?
Thanks.
Your result cannot be correct, as $xy = yx \in \mathbb{Z}[x,y]$, so that $xy + yx = 2xy$, and you cannot expect a 2-torsion element in the top cohomology of your space. The issue is that Künneth formula is an isomorphism of graded rings.
You are correct that $H^\bullet(S^3 \times S^1, \mathbb{Z}) \cong H^\bullet(S^3, \mathbb{Z}) \otimes H^\bullet(S^1,\mathbb{Z}) = \mathbb{Z}[x]/(x^2)\otimes\mathbb{Z}[x]/(y^2)$. However, the last tensor product is graded, so that $$ x \otimes y = (-1)^{\deg x\deg y}y\otimes x = (-1)^{3\cdot 1}y\otimes x = - y\otimes x $$ Therefore, if you wish to identify $\mathbb{Z}[x]/(x^2)\otimes\mathbb{Z}[x]/(y^2)$ with a quotient of $\mathbb{Z}[x,y]$ you need to impose the additional relation $xy = -yx$ or, in other words $xy + yx = 0$.