2
$\begingroup$

How many ideals are there in the $\mathbb Z_{28}$?

$\mathbb Z_{28}=\{0, 1, 2, 3, 4, 5, ..., 27\}$ is a semigroup under multiplication modulo 28.

$\endgroup$
  • 5
    $\begingroup$ Do you understand the question? Any ideas? What have you tried? Show your work. $\endgroup$ – anon Oct 16 '13 at 5:06
2
$\begingroup$

As a ring, $\mathbb{Z}_{28}$ is isomorphic to $\mathbb{Z}_{7} \times \mathbb{Z}_{4}$. The multiplicative structure of $\mathbb{Z}_{7}$ is a cyclic group of order $6$ plus a zero. It contains two (nonempty) ideals: $0$ and $\mathbb{Z}_{7}$. The multiplicative structure of $\mathbb{Z}_{4}$ is a semigroup with a chain of three $\mathcal{J}$-classes: a cyclic group of order $2$ ($\{1, 3\}$), a nonregular $\mathcal{J}$-class ($\{2\}$) and a zero $(\{0\})$. It contains three (nonempty) ideals: $\{0\}$, $\{0, 2\}$ and $\mathbb{Z}_{4}$. Thus the multiplicative semigroup of $\mathbb{Z}_{28}$ has $2 \times 3 = 6$ (nonempty) ideals.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.