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I am trying to prove whether or not the ideal generated by $\langle (2,2)\rangle$ is a prime ideal of $\mathbb Z_4\times \mathbb Z_4$?

My issue is I'm not sure how to do the coordinate multiplication: would the ideal look like $\{(0,0),(2,2)\}$ or would it be $\{(0,0), (2,2), (2,0), (0,2) \}$? I think that either way it is a prime ideal, but I want to make sure that I am doing the multiplication for the ideal correctly.

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  • $\begingroup$ I hope you mean $\mathbb Z_4\times \mathbb Z_4$ and not some other ring. I edited your question but the previous formatting didn't allow me to see what symbol you had used. Please check that I haven't altered your question in any way. $\endgroup$ – R_D May 6 '16 at 2:37
  • $\begingroup$ Yes this is correct, thank you. Do you know the answer to my question? $\endgroup$ – K.M May 6 '16 at 2:51
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Given an element $a$ in a ring $R$ the ideal $\langle a\rangle=\{ra\ |\ r\in R\}$

So in your case you just have to take each element of $\mathbb Z_4 \times \mathbb Z_4$ and multiply it with $(2,2)$ which will give you the ideal that you have written second.

What you wrote first is the cyclic subgroup of $\mathbb Z_4\times \mathbb Z_4$ generated by $(2,2)$

Also note that the given ideal is not prime for the following reason - we know that an ideal $P$ is prime if whenever $fg\in P\implies f\in P$ or $g\in P$. However your given ideal contains the product $(3,0)\cdot (0,3)=(0,0)$ but neither $(0,3)$ nor $(3,0)$ is in it.

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  • $\begingroup$ This is why im confused though (0,1) is an element in Z4xZ4, and (0,1)*(2,2)=(0,2) $\endgroup$ – K.M May 6 '16 at 3:09
  • $\begingroup$ Yes it is an element of the ideal. Like I said, the second set that you have written is the correct answer. $\endgroup$ – R_D May 6 '16 at 3:27
  • $\begingroup$ okay thank you, I misread what you had written. $\endgroup$ – K.M May 6 '16 at 3:31

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