$f:R \times R \rightarrow R \times R \times R$ is a R-module homomorphism. Prove f is not surjective.

Question

Let R be an integral domain and F be its field of fractions.

What I have shown: $f(R \times R)$ is finitely generated

i.e. for every $y \in f(R \times R),$ $y=f(r_1,r_2)=f(r_1(1,0)+r_2(0,1))=r_1f(1,0)+r_2f(0,1)$

Hence, $f(R \times R)=Rf(1,0)+Rf(0,1)$

What I need to show: for any $a,b \in R \times R \times R$

$R \times R \times R \neq Ra+Rb $.

I have been stuck here and I do not know how to proceed.

The clue given was: "Assume $a,b \in F \times F \times F$ and $R \times R \times R \subset F \times F \times F$ and use some result in linear algebra."

If I can prove the above statement I can show f is not surjective. Any hints or solutions?

Answer 1

For $a,b\in R\times R\times R$, consider$$ aF+bF\subset F\times F\times F$$ The space on the left is (at most) two-dimensional, that on the right is three-dimensional. Hence there exists $(u,v,w)\in F\times F\times F$ that is not in $aF+bF$. Write $u=\frac{u_1}{u_2}$ with $u_i\in R$ etc. Then $u_2v_2w_2\cdot(u,v,w)=(u_1v_2w_2, u_2v_1w_2,u_2v_2w_1)\in R\times R\times R$ is also not in $aF+bF$ (because $u_2v_2w_2\ne0$), so it is even less in $aR+bR$.

Answer 2

Let $\phi\colon R^2 \to R^3$ a morphism of $R$ modules. Denote by $f_1=\phi((1,0))$, $f_2= \phi((0,1))$, then $\phi(a,b)=a f_1 + b f_2$. Consider $v_1$, $v_2$, $v_2$ three column vectors in the image of $\phi$. So $v_i = a_i f_1 + b_i f_2$ for some $(a_1, b_1)$, $(a_2, b_2)$, $(a_3, b_3)$. Then the determinant with columns $v_1$, $v_2$, $v_3$ will be zero. This follows from the multilinearity of the determinant in columns, plus the fact that a determinant with two equal columns is $0$. Since the determinant of the basic vectors $e_1$, $e_2$, $e_3$ is $1$, so not $0$, we conclude that we cannot have all of the $e_i$ in the image of $\phi$.