Intersection of an ideal with a subring (B)

[coquelicot]

In a previous thread, I asked a question different from that I actually intended to ask. Since this question is licit and was answered by micromass, I open this new thread.

The right question is in fact:

If R is an integral domain, and R' is INTEGRAL over R, then the function f which assigns to an ideal I' of R' the ideal I = I' ∩ R, sends surjectively the prime ideals of R' to the prime ideals of R, the maximal ideals of R' to the maximal ideals of R, and (not nessarily surjectively) a non prime ideal of R' to a non-prime ideal of R. [NOTE: THIS LAST CLAIM WAS PROVED TO BE FALSE IN THIS THREAD]

It would be nice if it could be proved that it sends SURJECTIVELY a non-prime ideal to a non-prime ideal, or equivalently, if it could be proved that f is a surjection from the set of ideals of R' to the set of ideals of R. But for the moment, I can only see that every ideal I of R is included in a maximal ideal of R'; the example of micromass in the previous thread does not fit here since Z_(2) is not integral over Z. Any ideas for a proof or a counter example ?

N.B: It is 2:00 in my country, so, I will react to possible rapid answers in several hours.

 

[micromass]

Let $K$ be a field. Let $R = K[x]$ the polynomial ring. Let $R^\prime = K[x,y]/(y^2 - y,xy-1)$. This is an extension of $R$ which is integral since $y$ is the root of the polynomial $Z^2 - Z= 0$.

Consider the ideal $I = (x)$ in $R$. Any ideal in $R^\prime$ which contains $I$ must contain the invertible element $x$. Thus this ideal must be entire $R^\prime$.

 

[Terandol]

It is not true that non-prime ideals get sent to non-prime ideals under this map even for an integral extension. The contraction of a prime ideal will be prime again but this does not imply the contraction of a non-prime ideal is non-prime. As an example just take the rings $R'=\mathbb{Z}$ and $R=\mathbb{Z}$. Then the ideal $I=(2)_{\mathbb{Z}<i>} </i>$ gets sent to the ideal $\mathfrak{p}=(2)_{\mathbb{Z}}$ when intersected with $\mathbb{Z}$ but $\mathfrak{p}$ is prime whereas $I$ is not a prime ideal.

So while the map definitely is not a surjective map from non-prime ideals to non-prime ideals, this is not equivalent to saying that we don't get a surjective map from all ideals to all ideals. In the example of the Gaussian integers I gave the map is surjective on all ideals since any ideal is principle and we always have $(a)_{\mathbb{Z}<i>}\cap \mathbb{Z}=(a)_{\mathbb{Z}} </i>$ for any $a\in \mathbb{Z}$.

[edited out the rest since the above post gives a counterexample.]

 

[coquelicot]

TERANDOL : Indeed, I made a stupid mistake in my proof. I have corrected my post in the previous thread so that further readers will not be influenced by this mistake, and also added a note in the subject of this thread.

MICROMASS : I don't understand something in your example : clearly, in R', xy=1 and $y^2-y = 0$, hence multiplying this last equation by x² leads to $x^2y^2-x^2y=0= 1 - x$; so x = y = 1. How can R' be an extension of K[x] ?

Also, in order to give the maximal extent to the counter example I am looking for, let me formulate my question in a stronger way :

Let R be an integrally closed integral domain, K its fraction field, L a finite algebraic extension of K, and R' the integral closure of R in L. To prove or to disprove : the function that sends an ideal I' of R' to the ideal I' ∩ R of R is surjective from the set of ideals of R' to the set of ideals of R.

 

[micromass]

MICROMASS : I don't understand something in your example : clearly, in R', xy=1 and $y^2-y = 0$, hence multiplying this last equation by x² leads to $x^2y^2-x^2y=0= 1 - x$; so x = y = 1. How can R' be an extension of K[x] ?

Yes, you are right. It is not a valid counterexample. I'll think about it tomorrow if nobody else has posted since then.