Singular

Thank you very much for the example!

So it seems that the result is true only for algebraically closed fields.

For me it is really enough to know that the result is true for K=C and not true for K=Q and K=R.

Sorry, my example had some typos and was wrong.
The following should however work:

f=x^2*(x-1)^2*(y^2+1)+y2=0 defines two points (0,0),(1,0) in R^2 which are the union of x=y=0 and x=1,y=0,
but f is irreducible in R[x,y] (even in C[x,y]).
Adding squares of new variables makes this an example in any R^n, n>=2.

Is it ment that in the example that V(i) and V(i1)\cup V(i2) are birationally equivalent

V(i1)\cup V(i2) \simeq V(i) ?


In this way I can think the varieties are the same, but I think V(i)=V(xy-1) can not be represented as

V(i)=V(i1)\cup V(i2), ( V(i1)\nsubseteq V(i2) , V(i2)\nsubseteq V(i1) )

i1\in R[x,y] & i2\in R[x,y] ?

Sorry in last comment I ment hyperboloids defined by x-z^2=0 and x+z^2=0.

Hi and thank you for the answer!

I am sorry I do not fully understand this example.

Isn't V(i1), V(i2) varieties in R^3 (space curves which are intersection of paraboloids z-x^2=0 and z+x^2=0 and hyperboloid yx-1=0) ?

Do you mean projections \pi(V(i1)) and \pi(V(i2)) into (x,y)-plane and

V(i)=\pi(V(i1))\cup \pi(V(i2)) ?

Then I think \pi(V(i1)) (y positive) and \pi(V(i2)) (y negative) represent the two branches of hyperbola V(xy-1)\subset R^2, but \pi(V(i1)) and \pi(V(i2)) are not affine varieties ?

I am sorry if this is a stupid question .

Here is an example in R^2 where the ideal is prime but the variety is the union of two disjoint affine varieties:
ring r = 0,(x,y,z),dp;
ideal i = xy-1;
ideal i1 = i,x-z2; //means just that x is not negative
ideal i2 = i,x+z2;
V(i) = V(i1) union V(i2), with i a prime ideal,
all is understood in R^2 resp. in in R[x,y,z].

First of all thank you very much for your answer !

At least now I am more convinced that the problem is not as simple as it first seemed

However this is not a big issue for me I just have to refrace some sentences and make sure I do not claim something that is not true.

I am not an expert in algebraic geometry but have applied it to some engineering problems.

It seems that in most elementary level books this problem has not been discussed, but usually it is assumed in the beginning that K is algebraically closed...

In some books it seems that this is omitted and one just says that variety is irreducible if it is not union of two proper affine varieties.

The modern approaches uses scheam theory which I am not so familiar with.

In real algebraic geometry (for example Basu Saugatas and Richard Pollacks Algorithms in Real Algebraic Geometry) it seems that semi algebraic sets are more natural objects than varieties.

I also thought of similar example.

If f(x,y)=x^2-y^2-1 and I=<f>\subset R[x,y] then V(I)\subset R^2 (hyperbola) consists (euclidean) topologically of two components.

However if <f>\subset C[x,y] then V(I)\subset C^2 probably consists of one component (euclidean topologically).

Also as a real variety V(I) can not be written as a non-trivial union of two or more affine varieties.

Thank you again for the answer!

Hello,

this seems to be an interesting question, and we have also thought about it a bit. Your counterexample over Q is quite nice.

However, it seems that the definition of "irreducible variety" is somehow not adequate over non-closed fields such as R. E.g., the variety V(J) of
J = <x^2+y^2-z^2+1>
naturally separates into two parts (one for z >= 1 and one for z <= -1). The polynomial is irreducible, hence J is a prime ideal. Yet, I suspect that one is not able to write V(J) as a non-trivial union of two or more affine varieties.
So, all I am saying is that the usual definition for "irreducible variety" may not make so much sense, geometrically, over non-closed fields. I guess in this case, some topological definition of "irreducible variety" would be much more adequate.
(I know that this does not answer your question. )

Regards,
Frank

I dare anybody to comment my problem !!!

I don't care if your comment is false or true. I just want some feedback. Any comment would be greatly appreciatied!!!

I might have found an counterexample if K=Q.

Remeber Fermat's last theorem;

f(x,y,z)=x^n+y^n-z^n=0 does not have integer solutions

so that x*y*z\neq 0 when n>2 and integer.

Divide this by z^n and get;

f(x,y,z)=(x/z)^n+(y/z)^n-1=X^n+Y^n-1=g(X,Y)=0

does not have rational solutions other than

(X,Y)=(\pm 1, 0)
(X,Y)=(0, \pm 1)

if n is even.

Consider then polynomial

f(X,Y)=X^4+Y^4-1

and ideal I=<Y^4+X^4-1>\subset Q[X,Y].

Now the ideal I is prime and

V(I)=(1,0)\cup (-1,0)\cup (0,1) \cup (0,-1) \subset Q^2

which is not irreducible variety.

Thus it seems that Conjecture 1. is false.

However a question still remains; Is the conjecture true
only for closed fields or is it true for example if K=R
or other fields ?

The case K=R intrest me the most.

I have a small problem which concerns prime ideals and varieties. I ask if the following statement (Conj.1.) is true for arbitrary field K.

I stated some definitions and theorems (that I know of) after the Conj.1. and proved it if K is algebraically closed.

I have not been able to find the result in any book and could not think of a counterexample.

If somebody knows this result, a book reference would be greatly appreciated !!!
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Conjecture 1.

Let I\subset\K[x_1,...,x_p] be a prime ideal where K is arbitrary field. Then V(I)\subset K^p is an irreducible variety.




The converse of Conj.1. can not be true since V(I^2) could be irreducible and I^2 is not necessarily a prime ideal.

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Def.1.

Variety V\subset K^p is irreducible if V=V_1\cup V_2 implies V=V_1 or V=V_2.

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Def.2.

Ideal J\subset K[x_1,...x_p] is prime if fg\in J implies f\in J or g\in J.

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Def.3.

By variety V(J) we mean the point set V(J)={a\in K^p | f(a)=0 \forall f \in J } \subset K^p.

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Theorem 1.

Variety V(J) is irreducible if and only if I(V(J))={f\in K[x_1,...,x_p] | f(a)=0 \forall a \in V(J)}\subset K[x_1,...,x_p] is prime ideal .

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Theorem 2. (Hilbert's Nulstellensatz)

Let K be algebraically closed and I\subset\K[x_1,..,x_p] an ideal. Then I(V(J))=\sqrt{J}.

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Now if K is algebraically closed and J is prime ideal then I(V(J))=J since J is prime so that \sqrt{J}=J and so I(V(J))=J is prime and V(J) is irreducible and Conj.1. follows.

If K is not algebraically closed we can not use Nulstellensatz but anyway it seems that the Conj.1. is true!