Let $R$ be a P.I.D, and let $q$ be irreducible in $R$. If we have the extension $R'$ given by:

$R'= \Big\{ a+\eta \cdot b: \; a,b\in R\}$

such that $\eta\in R'$ satisfies $\eta^2=q$. I'm trying to find a nonzero prime ideal $P\lhd R$ and ideal $P'\lhd R$ such that:

  1. $P'$ is a unique ideal lying over $P$, and it satisfies $R/P\cong R'/P'$.
  2. $P'$ is a unique ideal lying over $P$, and it satisfies $R/P\ncong R'/P'$.
  3. There are multiple prime ideals in $R'$ lying over $P$.

Since $R$ is a P.I.D, I know that $P=(p)$ for some irreducible $p\in R$. I'm having problems thinking of possible ideas which will ensure any of the above properties. My original problem is for the cases $R=\mathbb{Z},R'=\mathbb{Z}[x]/(x^2-3)$ and $R=\mathbb{R}[y],R'=\mathbb{R}[x,y]/(x^2-y)$, but I think in both cases $R'$ is of the form described above.

I suspect that there is a similar method for the general case, but am unable to do so. I would appreciate hints, intuition or explanations.

  • 2
    $\begingroup$ Well clearly the ideal $(q)$ will play a special role in this situation. It is also likely that the ideals $(p)$ for $p$ a nonassociated irreducible will be "far from the trouble" : you should see what happens for these. $\endgroup$ – Max Jul 21 '19 at 16:35
  • $\begingroup$ What exactly is your goal/question here? To prove that all of 1-3 always happen? That is false in general. Consider rings of $p$-adic integers. They are PIDs. But, as they have a single prime ideal $P$, and so do their quadratic extensions. Therefore case 3 is ruled out completely, and exactly one of 1/2 will happen, but the other will not. $\endgroup$ – Jyrki Lahtonen Jul 31 '19 at 5:29
  • $\begingroup$ I'm trying to find an example for all three cases, not one satisfying all of them. Particularly for the examples given, and I hoped that looking at the general setting would give instinct on how to approach this. $\endgroup$ – Keen-ameteur Jul 31 '19 at 8:07
  • $\begingroup$ Something like: 1) If $R=\Bbb{Z}_2$, $q=2$ then $R$ and $R'$ both have a unique non-zero prime ideal, respectively generated by $2$ and $\sqrt2$. Here $R/P\cong R'/P'$. 2) $R=\Bbb{Z}_2$, $q=-3$. Both $R$ and $R'$ have a unique non-zero prime ideal generated by $2$ in both cases, $R'/P'$ is a degree two extension of $R/P$. 3)$R=\Bbb{F}_3[x]_x$ localized at the prime ideal generated by $x$, $q=x+1$. This time $x$ generates the only non-zero prime ideal of $R$, but $R'$ has two prime ideals above it corresponding to $\eta\pm1$. Or what? $\endgroup$ – Jyrki Lahtonen Jul 31 '19 at 18:00

With $R'=R[\sqrt{q}]$, you have that $R'$ is a subring of the integral closure of $R$ in $Quot(R')$, the fraction field of $R'$.

We are interested in finding prime ideals in $R'$ lying over primes in $R$ with certain properties. Since $R'$ may not be the integral closure of $R$ in $Quot(R')$, we may have some "bad" primes, for which the localisation of $R'$ is not a discrete valuation ring.

Lets restrict attention to "good" primes $P'/P$, and consider the local picture, of $R'_{P'}/R_{P}$. Your property $1$ is to say that this extension has residual degree, $1$, which is often written $f_{P'/P}$. This is the degree of the field extension $R'/P'$ over $R/P$.

Your property $2$ is to say the residual degree is greater than $1$.

Both of these can occur, and for "big" PID's they both often will, for any degree $2$ extension (not just adjoining the square root of a prime element). By big PID's we are thinking of rings with many primes, such as $\mathbb{Z}$, or $k[t]$.

Your third property is usually described as the prime $P$ "splits" in the extension. This too often happens in extensions of "big" PID's. Actually identifying which primes split, and which primes have residual degree $1$ or more requires deep knowledge of the PID in question, for $\mathbb{Z}$ this is basically quadratic reciprocity.

Not all PIDs are big however, and by localising at a prime, we can obtain a PID with a single prime ideal, and this prime is the only one we've got, so it might not satisfy one of your conditions.

All of this basically ignores your choice of irreducible element $q$, but the behaviour of $(q)$ in $R'$ gives us a canonical prime to work with. We can manually check that it has a single prime ideal lying over it, $(\sqrt{q})$, and this ideal squares to $(q)$. So we can check that for $(\sqrt{q})/(q)$, we have residual degree $1$, that is, $R'/P'\cong R/P$, and this is the only prime ideal lying over $(q)$.

So in the $\mathbb{Z}[\sqrt{3}]$ case, to find an example of case $1$, we want a prime ideal in $\mathbb{Z}$ such that it has a prime ideal lying over it in $\mathbb{Z}[\sqrt{3}]$ with an isomorphism of residue fields. In this case, we can use the general theory to tell us that $(\sqrt{3})$ will be an ideal lying over $(3)$ of the correct form. For the other cases however, you won't be able to use the ideal $(3)$, and you'll have to hope there are other primes which will have the required properties. Eg, you look around at other ideals, and see that $x^2-3$ is irreducible mod $5$, so $(5)$ remains a prime ideal in $\mathbb{Z}[\sqrt{3}]$, (Case 2) while it splits into distinct linear factors mod $7$, so $(7)$ is an example of Case 3. This stuff is gone over in most algebraic number theory books, I used "Number Rings" by Stephenhagen, or "Introduction to arithmetic geometry" for the more general dedekind domains case. Keywords, residual degree, ramification index, splitting of primes.

  • $\begingroup$ Can you explain for the examples I've given how you would apply this? I am not really seeing right now how you would actually use what you said. $\endgroup$ – Keen-ameteur Jul 28 '19 at 12:54

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