The rationals as a direct summand of the reals

Question

The rationals $\mathbb{Q}$ are an abelian group under addition and thus can be viewed as a $\mathbb{Z}$-module. In particular they are an injective $\mathbb{Z}$-module. The wiki page on injective modules says "If $Q$ is a submodule of some other left $R$-module $M$, then there exists another submodule $K$ of $M$ such that $M$ is the internal direct sum of $Q$ and $K$, i.e. $Q + K = M$ and $Q \cap K = \{0\}$."

Take the reals $\mathbb{R}$ as a $\mathbb{Z}$-module and $\mathbb{Q}$ as a submodule of $\mathbb{R}$. What can the submodule $K$ be in this case? If $\mathbb{Q}$ and $K$ are supposed to have trivial intersection then it seems like the only thing $K$ can be is a set of irrational numbers together with $0$, but I don't see how that can be a submodule of $\mathbb{R}$.

What am I missing here? Thanks.

Answer 1

Consider the short exact sequence of $\mathbb Z$-modules $$ 0 \to \mathbb Q \hookrightarrow \mathbb R \xrightarrow{\varphi} \mathbb R / \mathbb Q \to 0. $$

Since $\mathbb Q$ is injective, the sequence splits. Thus, there exists a homomorphism $\lambda: \mathbb R / \mathbb Q \to \mathbb R$ such that $\varphi \circ \lambda = \operatorname{id}_{\mathbb R / \mathbb Q}$. Let $K = \lambda(\mathbb R / \mathbb Q)$. Then $\mathbb R = \mathbb Q \oplus K$.

The axiom of choice is usually used to show that $\mathbb Q$ is injective, typically via Baer's criterion. I don't think one can find an explicit description of $K$. What we can say is that it consists of a choice of coset representatives in $\mathbb R$ of the quotient $\mathbb R / \mathbb Q$.

Answer 2

Essentially you are asking how to write down a homomorphism of abelian groups $f : \mathbb{R} \to \mathbb{Q}$ such that $f|_{\mathbb{Q}}=\mathrm{id}_{\mathbb{Q}}$. Because then $\mathbb{R} = \mathbb{Q} \oplus \ker(f)$. Notice that $f$ is automatically $\mathbb{Q}$-linear. Existence of $f$ follows directly from linear algebra when one chooses a $\mathbb{Q}$-basis of $\mathbb{R}$ containing $1$. However, such a basis cannot be written down explicitly, and in fact ZF alone cannot prove the existence of $f$. The axiom of choice AC is enough (but it is not correct that one needs AC for this). Notice that $f$ cannot be continuous, since otherwise $f(r)=r$ for all $r \in \mathbb{R}$ and $f$ would not be well-defined. In other words, $\ker(f) \subseteq \mathbb{R}$ is not closed.