Deduce that the spectrum of a local ring is always connected

[mad mathematician]

Iv'e got this proof from some claim:

Any Pure maths doctors out there who can explain to me why since $V\ne \emptyset$ that $1\notin \mathfrak{b}$?

Thanks!

 

[fresh_42]

If $1\in \mathfrak{b}$ then $\mathfrak{b}=A$ and $V(\mathfrak{b)}=V(A)=\emptyset.$