Ideals of $\mathbb Z/11\mathbb Z \times \mathbb Z/11\mathbb Z $?

Question

How would one go about finding all of the ideals of this ring? (with addition and multiplication defined component-wise)

Answer 1

Note that $\Bbb Z/11\Bbb Z$ is a field.

You get the more or less trivial $I_1 = \Bbb Z/11\Bbb Z\times \{0\}$ and $I_2 = \{0\}\times \Bbb Z/11\Bbb Z$ as proper ideals, and you have the zero ideal as in any ring. I claim that there are no more proper ideals.

Say we have an ideal $I$, and assume that it's not one of the three above. It is therefore not the zero ideal and thus contains some non-zero element $(a, b)$. If both $a$ and $b$ are non-zero, then $(a, b)$ is invertible, and $I$ is the whole ring.

Assume WLOG that $a = 0, b \neq 0$. Then $b$ is invertible in $\Bbb Z/11\Bbb Z$, so $(0,1)\in I$, and therefore $I_2 \subset I$. Since $I$ is not equal to $I_2$, $I$ must contain elements that have both first and second component non-zero. In other words, it must contain an invertible element, and $I$ is the whole ring.

In general, for two (commutative, unital) rings $R, S$, the ideals of $R\times S$ are of the form $I \times J$ with $I \subseteq R$, $J\subseteq S$ ideals.

Answer 2

Ideals of the ring of the form R×S, where R and S are rings , are of the form I×J , I and J are ideals of Rand S respectively. Here as Z/11Z is a field so as a ring its ideals are zero ideal and ring itself. Thus all the ideals of the ring Z/11Z×Z/11Z are (0)×(0), (0)×Z/11Z,Z/11Z×(0), and the ring itself.