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How can I construct a ring of order $a$, where $a$ is an odd number which is not a prime power? Thanks in advance.

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    $\mathbb{Z}/a\mathbb{Z}$ works, right? – 2017-01-06
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    What is the mean of $\mathbb{Z}/a\mathbb{Z}$? is this the set of residuals of dividing integer numbers to $a$? – 2017-01-06
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    You can think of $\mathbb{Z}/a\mathbb{Z}$ as the set of integers modulo $a$. – 2017-01-06
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    @LeonSot Thank you very much – 2017-01-06

2 Answers 2

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You can simply use for any $a$:

$$\mathbb Z/a\mathbb Z=\{\bar n, \ n\in \mathbb Z\}$$

where $\bar . $ is such that

$$\bar n=\bar m\iff n\equiv m\pmod a.$$

$\mathbb Z/a\mathbb Z$ is a ring of order $a$.

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We know that $M_n(\mathbb F_p)$ has $p^{n^2} $ elements. Now, assume that $a={p_1}^{n_1}\ldots{p_k}^{n_k} $, where each $n_i$ is a square . When, $p_i$'s are distinct primes. So, $R=M_{\sqrt n_1}(\mathbb F_{p_1})\times\ldots \times M_{\sqrt n_k}(\mathbb F_{p_k}) $ is a ring with $a$ elements. By, this method we try to construct a noncommutatie ring. Now, in arbitrary case, $R\times \mathbb Z_n$ is our Required ring