What are the all maximal ideals of $2 \mathbb Z$ ? what are the all prime ideals of $2 \mathbb Z$ ? We know that if $R$ is a commutative ring with multiplicative identity $1$ and $M$ is a maximal ideal then $M$ is a prime ideal . But the problem here is $R$ does not have any multiplicative identity ... Please help


1 Answer 1


An ideal of $2\mathbb{Z}$ must be an additive subgroup, to begin with; so it is of the form $2k\mathbb{Z}$. Any such subset (with $k\ge0$ for a unique representation) is indeed an ideal, as it is readily checked.

Saying that $2k\mathbb{Z}\subsetneq 2l\mathbb{Z}\subsetneq 2\mathbb{Z}$ means $l>1$ and $l\mid k$ with $k\ne l$. Thus the maximal ideals are those of the form $2p\mathbb{Z}$, with prime $p$.

Let's now assume $2k\mathbb{Z}$ is a nonzero ideal ($0\mathbb{Z}$ is prime, of course). If $k=ab$ is composite ($a>1$ and $b>1$), then $2a\cdot2b\in 2k\mathbb{Z}$, but neither $2a$ nor $2b$ belong to $2k\mathbb{Z}$, so this ideal is not prime.

We're left with prime $k$. Can you go on?

  • $\begingroup$ should it not be $l|k$ ? $\endgroup$
    – user123733
    Dec 12, 2014 at 13:59
  • $\begingroup$ @user123733 Yes, of course. Thanks for noting. $\endgroup$
    – egreg
    Dec 12, 2014 at 15:29

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