Q:

Prove that if $\alpha \subseteq \mathbb{R}$ is *ideal* then $\alpha = \{0\}$ or $\alpha = \mathbb{R}$

This is a problem from an old homework assignment which I never got, and I've decided to go back and try again. As a side note, I'm quite new to writing proofs.
Definition: $\alpha \subseteq \mathbb{R}$ is ideal if

$\forall x \in \alpha, \forall y \in \mathbb{R}, xy \in \alpha$ and
$\forall x,y \in \alpha, x+y \in \alpha$

I've split up the claim as such: if $\alpha \subseteq \mathbb{R}$ is ideal then

$\alpha \neq \{0\} \implies \alpha = \mathbb{R}$ and
$\alpha \neq \mathbb{R} \implies \alpha = \{0\}$

Proof:
(first bullet) 
($\subseteq$) this is given.
($\supseteq$) Let $a \in \mathbb{R}$. Since $\alpha \neq \{0\}$, $\exists b \in \alpha$ such that $b \neq 0$. Combining (1) and (2), we get
$$b+ab = c$$
for some $c \in \alpha$. Rearranging this equation, we get 
$$a+1=\frac{c}{b}$$
(recall that $b \neq 0)$. Since $\alpha \subseteq \mathbb{R}$, we have that $b \in \mathbb{R}$, and we also know that $\frac{1}{b} \in \mathbb{R}$. Thus, by (1), 
$$\frac{c}{b} = c \cdot \frac{1}{b} = a+1 \in \alpha$$ 
But $a+1$ is still an arbitrary element in $\mathbb{R}$, so we have shown that $\mathbb{R} \subseteq \alpha$, as needed.
(second bullet) 
($\supseteq$) Let $a \in \alpha$. Then, by (1), $a \cdot 0 = 0 \in \alpha$, as needed.
($\subseteq$) Let $a \in \alpha$. Since $\alpha \neq \mathbb{R}$, $\exists b \in \mathbb{R}$ such that $b \notin \alpha$. By combining (1) and (2), we get
$$a + ab = a(1+b) \in \alpha$$

This is where I'm stuck. I know that I need to somehow show that $a$ must be equal to $0$, but I can't figure out where to go from here. Any hints?  Also, is my proof of the first bullet valid?
Thanks!

A:

For the first bullet, I think you could simplify by noting that since $b \in \alpha$ with $b\neq 0$, we have
$$1 = \frac{1}{b} \cdot b \in \alpha$$
as $\frac{1}{b} \in \mathbb{R}$. This implies $\alpha = \mathbb{R}$ by (1).
For the second bullet point, we assume  $\alpha \neq \mathbb{R}$ and so $\exists x \in \mathbb{R}$ with $x \not\in \alpha$. This implies $1 \not\in \alpha$, else we would have $x \cdot 1 \in \alpha$. This implies $\alpha \subset \{0\}$ as otherwise $a \in \alpha$ with $a\neq 0 \implies 1 \in \alpha$ as shown above. 

