Let $\newcommand{\ID}[1]{\langle#1\rangle}F$ be the quotient ring $\mathbb{Q}[x]/\ID{x^3}$, where $\mathbb{Q}$ is the field of rational numbers.
Then find out which are correct
(i) There are exactly three distinct proper ideals of $F$
(ii) There is only one prime ideal in $F$
(iii) $F$ is an Integral Domain
Answer:
The ideals of $F$ are $\ID{0}$, $\ID{x}$, $\ID{x^2}$ and $\ID{x^3}$
The proper Ideals are $\ID{x}$ and $\ID{x^2}$
Thus there are two distinct ideals in $F$
But the option (i) is given to be correct.
Further $\ID{x}$ is the only prime ideal of $F$.
Thus option (ii) should be correct.
But how option (i) is correct?
Why $\ID{x^3}$ is a proper ideal?
Help me out, please.
I wouldn't say that $\newcommand\ID[1]{\langle #1\rangle}\ID{x}$ is an ideal in $F$, although it's a common abuse of language.
Precisely, the ideals of $F$ are in one-to-one correspondence with the ideals of $\mathbb{Q}[x]$ that contain $\ID{x^3}$. Since this ring is a principal ideal domain, such ideals are $\ID{1}$, $\ID{x}$, $\ID{x^2}$ and $\ID{x^3}$.
If $I\supseteq\ID{x^3}$ is an ideal in $\mathbb{Q}[x]$, the corresponding ideal in $F$ is $I/\ID{x^3}$. So you have three proper ideals. A proper ideal is one that is not the whole ring and certainly $\ID{x^3}/\ID{x^3}$ (the zero ideal in $F$) is not the whole ring.
Option (ii) is also correct, because $\ID{x}$ is the only prime ideal in $\mathbb{Q}[x]$ containing $\ID{x^3}$.
Option (iii) is incorrect, because $\ID{x^3}$ is not a prime ideal.