I have a few questions on rings. Some I can answer and some I am not really sure. I would appreciate your thoughts on the following questions:
In which rings is the null-ideal (0) a prime ideal?
First of all $R\neq \{0\}$, else $(0)=\{0\}$ and it is not a prime ideal by definition. Now (0) is a prime ideal iff xy=0 and x=0 or y=0. Therefore (0) is a prime ideal in every (non-trivial) integral domain.
What are the consequences of $(a)=(b)$ of two principal ideals in a integral domain?
Since $(a)\subseteq (b)$ we have $b|a$ and $(a)\supseteq (b)$ we have $a|b$. Therefore $a=b$. Is there more?
What are the prime ideals of $\mathbb{Z}$. Which of them are maximal and which are prime?
The prime ideals of $\mathbb{Z}$ are of the form $p\mathbb{Z}$ where $p$ is prime. Every prime ideal of $\mathbb{Z}$ is also a maximal ideal, since $\mathbb{Z}/p\mathbb{Z}$ is a field.
Is there a ring with just one maximal ideal, which is not a field?
Suppose $R$ is a field and $I\subsetneq R$ the only maximal ideal. Then is $R/I$ a field too. Let $[0]\neq [r]\in R/I$, so $r\in R$. Since $R$ is a field, there is a $x\in R$ with xr=1, therefore $[xr]=[1]$ and $[xr-1]=[0]$. So $xr-1\in I$. But $x,r\in R-I$. Hence $1\in I$, but then $I=R$. Contradiction to the fact that $I$ is a maximal ideal.
There is no field with one maximal ideal. There should be no fields with maximal ideals! Am I correct ?
Thanks in advance.
Yes. Also "prime ideal" is not a single word.
Your conclusion is wrong. A counterexample is $a=2$ and $b=-2$ in $\mathbb Z$. You'll have to reexamine your argument. If $a|b$ and $b|a$, you can say something very specific about the connection between the two.
Yes. $\mathbb R[x]/(x^2)$ is an example.
No. In fields, the zero ideal is the one and only maximal ideal. Rings with identity always have maximal ideals.