Let $n\geq 0$ be an integer, $R\to R'$ a ring homomorphism, and $M$ an $R$-module. Then the following holds: $$\bigl(\bigwedge^n_R M\bigr)\otimes_r R' \cong \bigwedge_{R'}^n\, (M\otimes_r R').$$ I want to prove this statement in the following way:
We have an exact sequence $$0\to N\to M^{\otimes n}\to \bigwedge_R^n M\to 0,$$ where $N$ denotes the ideal generated by elements of the form $m_1\otimes\cdots\otimes m_n$ such that $m_i = m_j$ for two different indices. Taking the tensor product $-\otimes_R R'$ we obtain the exact sequence $$N\otimes_R R'\to (M\otimes_R R')^{\otimes n}\to \bigl(\bigwedge^n_R M\bigr)\otimes_R R' \to 0,$$ where we have made the identification $M^{\otimes n}\otimes_R R'\cong (M\otimes_R R')^{\otimes n}$. I am not convinced that the image of $N\otimes R'$ inside $(M\otimes_R R')^{\otimes n}$ is the ideal we need to quotient out. For example, why is every element of the form $(\ldots,m\otimes r, m'\otimes r',\ldots)$ with $m\otimes r = m'\otimes r'$ contained in this image? Can someone make this proof work? Thanks in advance.
AYK