Let $k$ be an algebraically closed field and $R$ be a $k$-algebra. Prove that $R$ is a geometrically integral $k$-algebra if and only if $R$ is an integral $k$-algebra.
My attempt: ($\Rightarrow$) part is trivial, so we only have to prove ($\Leftarrow$) part.
Let $k'$ be an extension field of $k$ such that $k \neq k'$ . Since $k$ is algebraically closed, the only algebraic extension of $k$ is $k$ itself. Therefore, $k'$ must be transcendental over $k$, and so $k'$ must be infinite-dimensional as a $k$-vector space.
Now let $R$ be an integral $k$-algebra, and we have to prove that $R\otimes_k k'$ is an integral domain.
Here is where I stuck, Unlike an ordinary integral domain, we cannot guarantee that $a \otimes b = 0$ iff $a=0$ or $b=0$. How should I deduce that $R\otimes_k k'$ is an integral domain? Does anyone have ideas?
Any hints or advices are welcome!