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I want to find the ideals of $\mathbb{Z}_4 \times \mathbb{Z}_4$.

I know how to do that for one them ($\mathbb{Z}_{12}$ or something), but not when it is a form such as above. Can anyone explain how this process is done?

alnoge
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  • Start by looking at ideals generated by single elements. There are 16 elements in your ring, so fix $(a,b)$ and look at $(a,b)\mathbb{Z}_4\times\mathbb{Z}_4$. This will give you most of your ideals (and many will be duplicates).

    Then, try looking at ideals generated by two elements, recalling that whenever you have an particular element in an ideal, you have all the elements that appear in the ideal generated by that element.

     – Michael Burr Mar 01 '15 at 20:02

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Note that $(1,0)$ and $(0,1)$ are in the product $\mathbb{Z}/4\times\mathbb{Z}/4$. Now, let $I$ be an ideal of $\mathbb{Z}/4\times\mathbb{Z}/4$. Let $(a,b)\in I$, then $(a,b)(1,0)=(a,0)$ and $(a,b)(0,1)=(0,b)$ are both in $I$.

The set of first elements of the ideal $I$, i.e., $I_1:=\{a:(a,b)\in I\}$ forms an ideal of $\mathbb{Z}/4$ and the set of second elements of the ideal, i.e., $I_2:=\{b:(a,b)\in I\}$ form an ideal of $\mathbb{Z}/4$.

It follows that $I_1\times I_2=I$. Since $\mathbb{Z}/4$ has 3 ideals, this means that there are 9 total ideals in $\mathbb{Z}/4\times\mathbb{Z}/4$.

Michael Burr
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  • Is $\mathbb{Z}/4$ an equivalent notation for $\mathbb{Z}_4$ ? – alnoge Mar 01 '15 at 20:07
  • Yes, $\mathbb{Z}/4$ is short-hand for $\mathbb{Z}/4\mathbb{Z}$. Some people prefer the quotient notation because $\mathbb{Z}_p$ is also used for the $p$-adics in number theory (when $p$ is a prime). – Michael Burr Mar 01 '15 at 20:09
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    Sorry, i'm still learning about ideals. How do you know that there are 3 ideals of $\mathbb{Z}_4$ ($\mathbb{Z}/4$) and how do we actually compute them? – alnoge Mar 01 '15 at 20:12
  • The three ideals are ${0}$, $\mathbb{Z}/4$, and $\langle 2\rangle$. The zero ideal and the whole ring are easy ideals because every ring has them. Note that if the ideal contains either 1 or 3, then it is the entire ideal because, in an ideal, you can add elements or multiply ring elements and both 1 and 3 generate $\mathbb{Z}/4$ as a group. This leaves the ideal generated by 2, which forms an ideal. – Michael Burr Mar 01 '15 at 20:19
  • Remember, $\Bbb Z_4$ is a cyclic group under addition (mod $4$), so any ideal will have to be a subgroup. There are just 3 of these, the trivial subgroup ${0}$ (also, clearly the ideal $(0)$), the subgroup ${0,2}$ (the ideal $(2)$), and $\Bbb Z_4$ itself (also an ideal, namely $(1)$). – David Wheeler Mar 01 '15 at 20:20
  • So, ideals of $\mathbb{Z}_4$ are $<1> = {0, 1,2,3}, ; <2> = { 0, 2}, ; <4> = {0}$ – alnoge Mar 01 '15 at 20:20
  • @DavidWheeler and Michael, would finding the maximal ideals be the maximals ideals which are in $\mathbb{Z_4}$ AND also in the other $\mathbb{Z}_4$? In this exercise both sides are the same, but what if we had $\mathbb{Z}_5 \times \mathbb{Z}_6$? – alnoge Mar 01 '15 at 20:29
  • @DavidWheeler The maximal ideals are maximal ideas times entire rings. So, they would be $\langle 2\rangle\times\mathbb{Z}/4$ and $\mathbb{Z}/4\times\langle 2\rangle$. Perhaps it would be better to ask your new question in a new posting? – Michael Burr Mar 01 '15 at 20:30
  • I'll give it some more thought then follow your advice on that. Thanks – alnoge Mar 01 '15 at 20:31
  • @MichaelBurr http://math.stackexchange.com/questions/1170860/maximal-and-prime-ideals-of-cross-product-of-rings#1170865 – alnoge Mar 01 '15 at 21:10
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Let $R$ be a commutative ring with unity. The ideals in $R\times R$ are all of the form $I\times J$, where $I,J$ are ideals in $R$, see also here. For $R=\mathbb{Z}/I$, the ideals of $R$ are in correspondence with the ideals $J$ of $\mathbb{Z}$ containing $I$ (i.e., ideals in $\mathbb{Z}/4\mathbb{Z}$ correspond to ideals in $\mathbb{Z}$ that contain $4\mathbb{Z}$).

So if you know already the ideals of $R$, then you also know the answer for $R\times R$.

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Dietrich Burde
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