Quick question, mostly just for my knowledge, but I'm working on a problem:

Determine whether the indicated set $A$ is an ideal in the indicated ring $R$:

$$A = \{0,2,4,6,8\},~~R = \mathbb{Z}/10\mathbb{Z}$$

For short hand, can I denote $A$ as $2\mathbb{Z}/10\mathbb{Z}$?

  • $\begingroup$ Do you mean to have $2\Bbb Z$ and $10\Bbb Z$ switched? $\endgroup$ – pjs36 Nov 30 '15 at 3:48
  • 1
    $\begingroup$ No, probably a stupid question anyway. We haven't really covered quotient spaces, and intuitively, I know: $$\Bbb{Z}/10\Bbb{Z} := \{0,1,2,3,4,5,6,7,8,9\}$$ and: $$2\Bbb{Z} := \{\ldots, -4, -2, 0, 2, 4, \ldots\}$$ So I figured $$2\Bbb{Z}/10\Bbb{Z} \overset{?}{:=} \{0,2,4,6,8\}$$ $\endgroup$ – HDM Nov 30 '15 at 3:51
  • $\begingroup$ Ah, I see. It's definitely nonstandard notation (as far as I know) but not a stupid question (in my opinion). I was just expecting "the bigger number on top" I suppose. $\endgroup$ – pjs36 Nov 30 '15 at 3:57

Yes. $A=2\mathbb{Z}/10\mathbb{Z} = \left\{ 2k+10\mathbb{Z}\;|\; k \in \mathbb{Z} \right\} = \left\{ 10\mathbb{Z}, 2+10\mathbb{Z}, \dots, 8+10\mathbb{Z}\right\}$.

Also, by the third isomorphism theorem: $R/A = (\mathbb{Z}/10\mathbb{Z})/(2\mathbb{Z}/10\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$ (the field of order 2).


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