Compute Hom(ZZ/(30), ZZ/(21)) over ZZ

Exhibit all of the \mathbb{Z}-module homomorphisms from \mathbb{Z}/(30) to \mathbb{Z}/(21).

By this previous exercise, we have \mathsf{Hom}_\mathbb{Z}(\mathbb{Z}/(30), \mathbb{Z}/(21)) \cong \mathsf{Ann}_{\mathbb{Z}/(21)}((30)). We claim that \mathsf{Ann}_{\mathbb{Z}/(21)} = (7)/(21). To see this, suppose first that \overline{k} \in \mathsf{Ann}_{\mathbb{Z}/(21)}((30)). In particular, we have 30 \cdot \overline{k} = \overline{30k} = 0 mod 21. Then 21|30k, and so 7|10k. Now 10k \equiv 0 mod 7, and hence k \equiv 0 mod 7. So \overline{k} \in (7)/(21). Conversely, note that 30 \cdot \overline{7k} = \overline{210k} = 0. Thus \overline{7k} \in \mathsf{Ann}_{\mathbb{Z}/(21)}((30)), and we have \mathsf{Ann}_{\mathbb{Z}/(21)}((30)) = (7)/(21) \cong \mathbb{Z}/(3).

So there are precisely three \mathbb{Z}-module homomorphisms \mathbb{Z}/(30) \rightarrow \mathbb{Z}/(21). As group homomorphisms, each of these is uniquely determined by the image of \overline{1}. These homomorphisms are given by \overline{1} \mapsto \overline{0}, \overline{1} \mapsto \overline{7}, and \overline{1} \mapsto \overline{14}.

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  • Yun  On November 14, 2011 at 10:43 pm

    Hello, I hope I’m not bothering you. In the second line, I think you misTeXed the annihilator. Also, from 21|30k shouldn’t you conclude 7|10k, not 7|15k?

    • nbloomf  On November 15, 2011 at 11:30 am

      You’re right. Thanks!

      Also, corrections never bother me- in fact I encourage them.

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