Are there complements of $\mathbb{Q}$ in $\mathbb{R}$ that are (non-unital) subrings? Because $(\mathbb{Q}, +)$ is a divisible group, $\mathbb{R} = \mathbb{Q} \times H$, for some subgroup $H$ (if one sees $\mathbb{R}$ as a $\mathbb{Q}$-vector space, for instance one can choose a basis containing $1$ and then get rid of it).
Is there such an $H$ that is a ring?

EDIT: I do not mean it as a direct product of rings, I am only asking if there is a complement of $\mathbb{Q}$ as additive subgroup that is also closed by multiplication.

EDIT2: Eric Wofsey solved it below. What if now we replace $\mathbb{Q}$ with the algebraic numbers?
 A: Suppose you had such a complement $H$ and consider any irrational algebraic number $\alpha$.  Then for some $q\in\mathbb{Q}$, $\alpha-q\in H$; replacing $\alpha$ with $\alpha-q$ (which is still an irrational algebraic number) we may assume $\alpha$ itself is in $H$.  But now we have a problem: $\alpha$ is a root of some polynomial $p(x)$ with rational coefficients and nonzero constant term $c$.  The equation $p(\alpha)=0$ then writes $c$ as a $\mathbb{Q}$-linear combination of powers of $\alpha$, and so would imply that $c\in H$.  This is a contradiction since $c$ is a nonzero rational.
(Slightly more strongly, this actually shows that no complement of $\mathbb{Q}$ in $\mathbb{R}$ can be closed under taking natural number powers.)

Here's another argument.  Suppose such a complement $H$ existed.  Note then that $H$ would be closed under multiplication by arbitrary elements of $\mathbb{R}$, since it is closed under multiplication both by elements of $H$ and by elements of $\mathbb{Q}$.  That is, $H$ would be an ideal in $\mathbb{R}$.  But $\mathbb{R}$ has no nontrivial proper ideals, so this is impossible.
(More generally, this shows that if $K\subset L$ is a nontrivial field extension then $K$ cannot have a $K$-linear complement in $L$ that is closed under multiplication.)
A: There are not! You can easily check this if $H$ is a ring with unit: $(1, 0) \cdot (0, 1)=(0,0)$ will be zero. Therefore, $\mathbb{Q} \times H$ will have zero-divisors.
A similar construction applies even if $H$ has no zero-divisors. Let $x \in H$ be non-zero, and consider $(1, 0) \cdot (0, x)$.
A: I think the answer is no: let $H$ such that $\mathbb{R}=\mathbb{Q}\times H$, that is $\forall x\in\mathbb{R}$ $\exists q\in\mathbb{Q}$, $h\in H$ such that $x=q+h$.
Assume now $H$ is a ring, that is $$h\cdot k\in H,\quad \forall h,k\in H.$$ It follows that $h$ has to be trascendent: indeed, assume it exists a polynomial $p\in\mathbb{Z}[x]$, $p(x)=p_n x^n+\cdots+p_\ x+p_0$ such that $p(k)=0$. In particular, we can assume $p_{0}\:=p(0)\ne0$, up to take $p(x)/x^{m}$ instead of $p$, where $m=\min\{l\in\mathbb{N},\,p_l\ne0\}$.
It follows $$\mathbb{Z}\ni -p_0=p(k)-p_o=\underbrace{k^{n}+\cdots k^n}_{\text{$p_n$ times}}+\underbrace{k^{n-1}+\cdots k^{n-1}}_{\text{$p_{n-1}$ times}}+\cdots k^2+k\in H,$$
which is absurd since $\mathbb{Z}\cap H\subseteq\mathbb{Q}\cap H=\{0\}$ and we assumed $p_0\ne0$. So $H$ contains only trascendent numbers.
Now pick an algebraic number $a\in\mathbb{R}\setminus\mathbb{Q}$, by previous assumption, $a=q+h$, that is $h=a-q$, which is absurd since $h$ is trascendent and $a-q$ is algebraic.
