Graduate 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 $V(\mathfrak{b)}=V(A)=\emptyset.$

 

[mathwonk]

This fact about local rings is obvious if you visualize the spec of a local ring as a bunch of sub varieties of different dimensions passing through a common point, i.e.:

general principles:
1) a one point set is connected.
2) the closure of a connected set is connected;
3) the union of connected sets all having a common point is connected;
4) any space is the union of the closures of its individual points.

special fact about local rings:
5) the closure of each point in the spec of a local ring contains the unique closed point, i.e. the unique maximal ideal;
hence: ????

the more general connectivity principle in post #1 is not as obvious to me though.

OK, guided by the statement, since the intersection of the sets associated to two ideals is the set associated to the sum of the ideals, it follows that if that intersection is empty, then the sum of the ideals is the whole ring, which gives the result.

But more efficiently, and to me more intuitively, although more sophisticated, is the fact that the association of a ring R to the space spec(R) is a direction reversing equivalence of categories. Hence direct sums (or "coproducts", as they are now called) of spaces, i.e. disjoint unions, correspond to direct products of rings. I.e. the correspondence between rings and specs of rings, is arrow reversing, hence products correspond to sums, (and a spec is a sum, i.e. disjoint union, of two specs, iff it is disconnected).

[For the same reason, the spec of the tensor product of two rings (coproduct in category of commutative rings) is the direct product of the two specs.]

for those who may not know "category" terminology, a product X = AxB, of two objects A,B is roughly something such that any two maps Y-->A, Y-->B define a map Y-->X.

and a coproduct X of A and B is roughly an object X = A+B, so that any two maps A-->Y and B-->Y define a unique map X-->Y.

thus any maps out of A and B define a map out of their coproduct, and any maps into A and B define a map into their product.

in particular note that any maps out of spec(R) and spec(S) define a map out of their disjoint sum.