Question

Let $\mathit{I}$be an ideal in$\cancel{\mathbf{c}}$ such that $\left(3\right)\underline{\mathbf{\subset }\mathbf{I}\underline{\mathbf{\subset }\cancel{\mathbf{c}}}}$. Prove that either$\mathit{I}\mathbf{=}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$ or $\mathit{I}\mathbf{=}\cancel{\mathbf{c}}$.

Short answer

It is proved that$I=\left(3\right)$or $I=\cancel{c}$.

Step-by-step solution

Write the conditions for (3)≠I

Consider$\left(3\right)\ne I$ where$n\in I$ such that $3\xcancel{}n$.

Then we have,

$n=3k+1$or$3k+2$ for$k\in \cancel{c}$

Show that either I=3 or I=c

Consider the first case,

If$n=3k+1$ then

$n-3k=1\in I$

Thus,

$I\in \cancel{c}$

Consider the second case,

If$n=3k+2$ then

$3\left(k+1\right)-n=1\in I$

Thus,

$I\in \cancel{c}$

Therefore, if$\left(3\right)\ne I$ then we have $I\in \cancel{c}$.

Hence, the statement is proved.