Existence of epimorphism from a Ring to a Module

Question

So the problem is the following, given a conmutative ring $R$ and a free $R$-module let's say $M$, such that $\dim(M)=2$.

Prove there is no epimorphism from $R$ to $M$

I've been struggling for a while to solve this problem. I believe the solution has to with what elements in the ring are sent to the basis of $M$ but I literally have no clue of where to start.

Any help would be appreciated

Answer 1

Suppose that $f:R \to R \oplus R$ is a surjective $R$-module homomorphism.

Then, let $(a_1,a_2)$ be $f(1)$. By surjectivity, that there must exist elements $b_1$ and $b_2$ of $R$ for which $f(b_1)=(b_1a_1,b_1a_2)=(1,0)$ and $f(b_2)=(b_2a_1,b_2a_2)=(0,1)$.

Using commutativity of $R$, one then gets that $b_2a_1b_1=0b_1=0$, $b_2a_1b_1=b_2b_1a_1=b_21=b_2$, $b_1a_2b_2=0b_2=0$, and $b_1a_2b_2=b_1b_2a_2=b_11=b_1$, so $b_1=b_2=0$. But then, that would mean that $(1,0)=(0,1)=(0,0)$, so $R$ is the zero ring.

Answer 2

Let $\mathfrak{m}$ be a maximal ideal in $R$. Tensoring is right-exact, so if we take a surjection $R \twoheadrightarrow R^2$, we can tensor with $R/\mathfrak{m}$ to obtain a surjection $R \otimes_R R/\mathfrak{m} \cong R/\mathfrak{m} \to R^2 \otimes_R R/\mathfrak{m} \cong (R/\mathfrak{m})^2$. But $R/\mathfrak{m}=:k$ is a field and it's well-known from linear algebra that there is no $k$-linear surjection $k \to k^2$ for a field $k$.