Find an ideal of $R = \mathbf{Z}/12\mathbf{Z} = \{0,1,2,\dots,11\}$ with two elements

Question

Let $R = \mathbf{Z}/12\mathbf{Z} = \{0,1,2,\dots,11\}$.

Find an ideal $I$ of $R$ which consists of two elements.

How many elements does $R/I$ have?

I thought the ideals would be $\{0\}$, $\{0,6,2\}$, and $\{0,4,3\}$.

Do the ideals not need the element $0$?

Answer 1

First you should know that $I$ has the property such that if you take an element $x \in R$ and $y \in I$ then $xr$ and $rx$ both belong to $I$

That's why you need $[0] \in I$ because $0 \times x = 0$ for any $x \in R$ Think about it , if you take $I = \{0,1 \}$ then you get that $2 \times 1 = 2 \not \in I$ and so $I \neq \{0,1\}$

Try $I = \{0,2 \}$ then you will get that $3 \times 2 = 6 \not \in I$ and so it can't be possible

Try $I = \{0,3 \}$ then you will get that $ 2 \times 3 = 6 \not \in I$ and so it can't be possible

Try $I = \{0,4 \}$ then you will get that $ 2 \times 4 = 8 \not \in I$ and so it can't be possible

Try $I = \{0,5 \}$ then you will get that $ 2 \times 5 = 10 \not \in I$ and so it can't be possible

Try $I = \{0,7 \}$ then you will get that $ 2 \times 7 = 14 = 2 \not \in I$ and so it can't be possible

Try $I = \{0,8 \}$ then you will get that $ 2 \times 8 = 16 = 4 \not \in I$ and so it can't be possible

Try $I = \{0,9 \}$ then you will get that $ 2 \times 9 = 18 = 6 \not \in I$ and so it can't be possible

Try $I = \{0,10 \}$ then you will get that $ 2 \times 10 = 20= 8 \not \in I$ and so it can't be possible

Try $I = \{0,11 \}$ then you will get that $ 2 \times 11 = 22= 10 \not \in I$ and so it can't be possible

Now you are only left with one possibility

Try $I = \{0,6 \}$ then you will get that $ 2 \times 6 = 12= 0 \in I$ and $3 \times 6 = 18 = 6 \in I$ and $4 \times 6 = 2 \times 2 \times 6 = 2 \times 0 = 0 \in I$ and so on, you will see that this is the only possible ideal as you said

You can see that $6$ worked because $6$ is a zero divisor