# 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

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

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!

• The two claims that you've written are logically equivalent to each other. If you prove any one, you've automatically proved the other. A statement of the form $P \implies Q$ is equivalent to the statement $\neg Q \implies \neg P$ (where $\neg$ implies negation, or "not"). The latter is statement is called the "contrapositive" (of the former). So, if you prove that $\alpha \ne \{0\} \implies \alpha = \mathbb R$, then of course, if $\alpha \ne \mathbb R$, then it must be that $\alpha = \{0\}$. – M. Vinay Mar 21 at 1:49
• Assume that $\alpha\neq \mathbb{R}$. If $a\in\alpha$ and $a\neq0$ then there is $a^{-1}\in\mathbb{R}$ such that $a^{-1}a=1$. Since $\alpha$ is ideal, then for all $r\in\mathbb{R}$ you hav ethat $r=r1=(ra^{-1})a\in\alpha$. Contradiction. – user647486 Mar 21 at 1:50
• Your proof of the first bullet is correct, well done! Let me suggest a simplification though: You don't need to use both 1 and 2 in that step. From 1 itself, you know that $ab = c \in \alpha$, and since $b \ne 0$, you get $a = \frac c b = \frac 1 b \times c \in \alpha$. – M. Vinay Mar 21 at 1:55
• Thanks @M.Vinay! Very helpful points. – Archie Gertsman Mar 21 at 2:04

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.
I'd attack the problem a little differently. Assume $$0 \neq x \in \alpha$$ and choose an arbitrary $$z \in \Bbb R$$. Then by property $$1, x\frac{z}{x} = z \in \alpha$$. ($$\frac{z}{x}$$ exists because $$x \neq 0$$.) But $$z$$ was an arbitrary element of $$\Bbb R$$, so $$\exists x \in \alpha, x \neq 0 \Rightarrow \alpha = \Bbb R$$.