Special Modules over PIDs (12.1.20-22)

Dummit and Foote Abstract Algebra, section 12.1, exercise 20-22:

20. Let $R$ be an integral domain with field of fractions $F$. Prove $\text{dim }F ⊗_R M = \text{rank }M$ for $F ⊗_R M$ a vector space over $F$ and $M$ any $R$-module.

21. Let $R$ be a PID. Prove any finitely generated projective $R$-module is free.

22. Let $R$ be a non-field PID. Prove no finitely generated $R$-module is injective.

Proof: (20) ($≤$) We see $\{f ⊗ m~|~f∈F,m∈M\}$ spans $F ⊗_R M$, so there is a basis $\{f_i ⊗ m_i\}_{i∈I}$. As well, we see $\{1⊗m_i\}$ is a basis. We claim $\{m_i\}$ is a set of linearly independent elements in $M$; otherwise, we have $∑r_im_i=0$ implying $∑r_i(1⊗m_i)=0$. ($≥$) Let $\{m_i\}$ be a set of linearly independent elements. We claim $\{1⊗m_i\}$ are linearly independent over $R$. We have $∑r_i(1⊗m_i)=∑1⊗r_im_i=1⊗∑r_im_i$; since the $m_i$ are linearly independent, we may assume that if some $r_i \neq 0$ then $∑r_im_i \neq 0$ and as well $∑r_im_i$ is nontorsion; now by 10.4.8 $1⊗∑r_im_i \neq 0$ and the $1⊗m_i$ are linearly independent over $R$. Consequently, collecting the denominators of any linear independence over $F$ leads to an equation of the form $\dfrac{1}{d}(∑r_i(1⊗m_i))=0$ so that $∑r_i(1⊗m_i)=0$ and now $r_i=\dfrac{r_i}{d}=0$ for all $i$ and the $1⊗m_i$ are linearly independent over $F$.

(21) $M$ is projective if and only if it it is a direct summand of a free module only if it is a submodule of a free module, which would imply it has no torsion elements. By the fundamental theorem this eliminates any elementary divisors of $M$ and forces $M$ free.

(22) $M≅R^n⊕R/(a_1)~⊕~...~⊕~R/(a_k)$ is injective if and only if each of its summands is injective by 10.5.4. By Baer's criterion, $M$ is injective over a PID if and only if $rM=M$ for all nonzero $r∈R$ so necessarily $R/(a_i)$ is not injective and we must only prove $R$ is not injective; indeed, for $p∈R$ a prime we see $1∉pR$.$~\square$