Is $\mathbb{Z}$ an ideal of $\mathbb{Q}$?

Question

I have to decide whether or not $\mathbb{Z}$ is an ideal of $\mathbb{Q}$.

My attempt:

$\exists q\in\mathbb{Q}$ and $z\in\mathbb{Z} $ such that $qz\notin\mathbb{Z}$, for example for $q=\frac{1}{2}, z=1$.

Thus, $\mathbb{Z}$ is a subring of $\mathbb{Q}$, but not its ideal.

Am I right?

How do you know that $\exists q\in\mathbb{Q}$ and $z\in\mathbb{Q} $ such that $qz\notin\mathbb{Z}$? — Another User, Jul 24 '22 at 18:15
Also, $\mathbb{Q}$ is a field. Does that mean anything to you? (It mightn't; this isn't a rhetorical question.) — Brian Tung, Jul 24 '22 at 18:18
@FirdousAhmadMala Do you know what "simple rings" are ? . A field is a simple ring. — Mr.Gandalf Sauron, Jul 24 '22 at 18:23

score 6 · Accepted Answer · edited Jul 24 '22 at 22:08

6

Your demonstration that $\mathbb{Z}$ is not an ideal of $\mathbb{Q}$ is sufficient, since it shows that $\mathbb{Z}$ does not "absorb" multiplication by $\mathbb{Q}$; in fact, it spills out to the entirety of $\mathbb{Q}$.

It is also enough to observe that $\mathbb{Q}$ is a field; fields only have two ideals—the trivial zero ring and the entire field. (In fact, the implication goes both ways—if a commutative ring has only two ideals, it is a field. See this question on this site, for example.)

edited Jul 24 '22 at 22:08

egreg

238,574

answered Jul 24 '22 at 18:25

Brian Tung

34,160

egreg: Thanks for the edit, totally missed that. – Brian Tung Jul 25 '22 at 03:45

Is $\mathbb{Z}$ an ideal of $\mathbb{Q}$?

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