The ideal $(0)$ is maximal in $\mathbb{Q}$ but not $\mathbb{Z}$? I'm watching Richard Borcherds talk on Commutative algebra: https://www.youtube.com/watch?v=GTz9laU7c30&list=PL8yHsr3EFj53rSexSz7vsYt-3rpHPR3HB&index=11
And 8 minutes he says that The ideal $(0)$ is maximal in $\mathbb{Q}$ but not $\mathbb{Z}$.
So if $(0)$ is a maximal ideal then $\mathbb{Q} \backslash (0) \cong \mathbb{Q}^*$ is a field, right? How does this field relate to the field $\mathbb{Q}$.
Also, can somebody explain this claim to me without resorting to showing that the quotient rings are field (or not a field), and just show me directly $(0)$ is not a maximal ideal in $\mathbb{Z}$?
Thank you
 A: 
So if $(0)$ is a maximal ideal then $\mathbb{Q} \backslash (0) \cong \mathbb{Q}^*$ is a field, right? How does this field relate to the field $\mathbb{Q}$.

No, you would write $\mathbb{Q}/ (0) \cong \mathbb{Q}$ is a field. It looks like you might be confusing the quotient ring with the set difference. The quotient ring is the forward slash, and the backwards slash is typically used for set difference. In the context of group theory, I think perhaps sometimes the two slashes are used to indicate left/right cosets, but this isn't the case in ring theory.
Perhaps you should read this previous post with regards to gaining an understanding of the quotient argument.

just show me directly (0) is not a maximal ideal in ℤ?

Ok, pick an ideal of $\mathbb Z$ that isn't $\mathbb Z$ or $(0)$. Got it? You're done. That shows $(0)$ isn't maximal.
A: $\mathbf Q^*$ is not a field, not even a ring, since it has not neutral element for addition. $\{0\}$ is a maximal ideal in $\mathbf Q$ just because it is the only proper ideal in $\mathbf Q$ (the same is true for all fields).
In $\mathbf Z$, which is  an integral domain, not a field, $\{0\}$ is a prime ideal, as in all integral domains, which is contained in all ideals, in particular in the maximal ideals of $\mathbf Z$, which are the ideals generated by prime elements.
A: $\mathbb{Q}\setminus\{0\}$ is not a field, it doesn't even have a zero element. Don't get confused, $I$ being a maximal ideal in $R$ means that the quotient ring $R/I$ is a field, not the set $R\setminus I$.
For example, if $I=\{0\}$ then there is a natural isomorphism $\pi:R\to R/(0)$ defined by $\pi(r)=r+(0)$. So this is a not very interesting quotient, this is just the ring $R$ itself. But in particular it means that $(0)$ is a maximal ideal if and only if $R$ is a field. So $(0)$ is indeed maximal in $\mathbb{Q}$.
On the other hand, it is not maximal in $\mathbb{Z}$. For example, $I=2\mathbb{Z}$ is a proper ideal which properly contains $(0)$.
A: 
can somebody explain this claim without showing that the quotient rings are field

Hint: it's the special case ${\cal P} = 0\,$ of the following generalization of Bezout.
Lemma $\,{\cal P}\ \color{#c00}{\rm maximal}$ and $\,a\not\in {\cal P}\Rightarrow\, a\,$ is invertible $\!\bmod \cal P$
Proof $\,\ a\not\in{\cal P}\Rightarrow P\subsetneq {\cal P}+aR\color{#c00}{\Rightarrow} P+aR = (1)\Rightarrow \underbrace{p + ab = 1}_{\!\!\!\!\!\!\!\!\!\!\!\!\!\textstyle \bmod {\cal P}\!:\ ab\equiv 1}\,$ for $\,p\in {\cal P},\ b\in R.$
Remark $\ $ Note $\,r\equiv s\pmod{\! 0}\iff 0\mid r-s\iff r=s.\,$ Said structurally $\,R/0\cong R$. Therefore $\,ab\equiv 1\pmod{0}\iff ab=1\iff a\,$ is a unit (invertible).
As explained here, one reason we allow $0$ as a prime ideal is that we can often factor out prime ideals so reducing to the case ${\cal P}=0\,$ i.e. reducing to the nicer case of an integral domain.
