# Principal Ideal using coordinates?

I thought I understood principal ideals but now im stuck... I want to find the elements of the principal ideal $\langle(1,0)\rangle$ in the ring $\mathbb Z_3\times \mathbb Z_3$ with $+_3$ and $*_3$ in each coordinate. Also find the distinct cosets of $\langle(1,0)\rangle$.

I know that a principal ideal is an ideal such that there exists an $a$ that generates the entire ring. $\langle a\rangle =\{x_1ay_1+\cdots+x_n a_n y_n\}$ where $n$ is an integer.

So in this case I would set $\langle a \rangle=\langle(1,0)\rangle$, however I am confused by the wording of this question. Does this mean that the left coordinate is $+_3$ and the right is $*_3$, or does it just mean that we can perform both of these operations on both coordinates?

If it is that each coordinate has a different operation would the set $\langle(1,0)\rangle$ be $\{(1,0),(2,0),(0,0)\}$? I think that I am just confused in general on how to set up a principal idea using coordinates.

• You perform both operations on each coordinate, The answer you have is correct. The ideal is $\mathbb Z_3\times 0$. – John Douma Apr 22 '16 at 7:08
• What do you mean by "...using coordinates"? Do you mean "... In a product ring"? If you know the operations in the product ring then everything works exactly the same way as if you were talking about a single ring ( because we are talking about a single ring). – rschwieb Apr 22 '16 at 10:05

Your ring $R=\Bbb{Z}_3\times \Bbb{Z}_3$ is commutative and has unity. The principal ideal generated by $a$ is $\langle a\rangle=Ra=\{ra\mid r\in R\}$.

$(a,b), (c,d)\in \Bbb{Z}_3\times \Bbb{Z}_3$, $(a,b)+(c,d)=(a+c,b+d), (a,b)*(c,d)=(a*c,b*d)$. This is the definition of the direct product of rings. You can't operate two elements in $R$ using different operation in separate coordinate.

• So under multiplication <(1,0)>= (1,0),(2,0),(0,0). But under addition <(1,0)>= { every element of the set Z3*Z3} or are they just defining addition here because <a> is technically a summation? – p.l Apr 22 '16 at 7:23
• @p.l Your confusion is palpable because I have no idea what you are talking about :) For a commutative ring with 1, $(a)=\{ar\mid r\in R\}$ there is no "under addition" about it. Why do you think $(a)$ is a summation? An ideal is a set, not a "summation" – rschwieb Apr 22 '16 at 10:09

We don't say that an ideal under multiplication or under addition.

An ideal $I$ of a ring $R$ is a subring of $R$ which satisfying $ri\in I$ and $ir\in I$ for all $r\in R$ and $i\in I$. In general, the principal ideal generated by $a\in R$ is defined by $$\langle a\rangle=\left\{ra+as+ma+\sum_{i=1}^{n}x_i a y_i\mid r,s,x_i,y_i\in R, n\in \Bbb{N}^+, m\in \Bbb{Z}\right\}.$$ You can verify that such ideal is the smallest ideal contains $a$.

If $R$ has multiplicative identity, then the principal ideal $\langle a\rangle$ became more easily $$\langle a\rangle=\left\{\sum_{i=1}^{n}x_i a y_i\mid x_i, y_i\in R, n\in \Bbb{N}^+\right\}=\left\{x_1 a y_1 +\cdots +x_n a y_n\mid x_i,y_i\in R, n\in \Bbb{N}^+\right\}.$$ Which is the definition you remember.

If $R$ is commutative and has multiplicative identity, then $\langle a\rangle=\{ra\mid r\in R\}$.

So, in your case, $R=\Bbb{Z}_3\times \Bbb{Z}_3$, $a=(1,0)$. \begin{eqnarray*} \langle (1,0)\rangle=\{r(1,0)\mid r\in R\}&=&\{(0,0)\cdot (1,0), ~~~~~~(0,1)\cdot (1,0), ~~~~~~(0,2)\cdot (1,0),\\ && (1,0)\cdot (1,0), ~~~~~~(1,1)\cdot (1,0), ~~~~~~(1,2)\cdot (1,0),\\ && (2,0)\cdot (1,0), ~~~~~~(2,1)\cdot (1,0), ~~~~~~(2,2)\cdot (1,0)\}\\ &=& \{(0,0), (1,0), (2,0)\}. \end{eqnarray*}