Techniques to prove that two field extensions are distinct

I have been trying to create a family of pairwise distinct field extensions from a one-parameter family of irreducible polynomials, but have no idea how to prove that they are distinct. One pair is $f = ( x \mapsto x^5 + 8 x^2 - 2 )$ and $g = ( x \mapsto x^5 + 64 x^2 - 2 )$ as polynomials over $\mathbb{Q}$. Given $a,b \in \mathbb{R}$ such that $f(a) = g(b) = 0$, is it true that $\mathbb{Q}(a) \ne \mathbb{Q}(b)$?

I know that $\mathbb{Q}(a),\mathbb{Q}(b)$ are either equal or intersect only in $\mathbb{Q}$, and hence I also know that the choice of root does not affect the answer to the question since there is an automorphism that cycles the roots of each polynomial. But I only have introductory level knowledge of Galois theory, and do not know any tools to tackle such a question.

• The choice of root may easily matter. For example adjoining different roots of your polynomial $f$ yields trivially intersecting fields, because the Galois group is full $S_5$. – Jyrki Lahtonen Apr 22 '15 at 10:39
• Discriminants may help. The discriminants of your two example polynomials share factors $2^4\cdot 7$.In addition to that $Disc(f)$ has (Mathematica says so) factors $131\cdot479$, and $Disc(g)$ has factors $17\cdot121811021$. The discriminant of the minimal polynomial of $\alpha$ need not be the dsicriminant of the field $\Bbb{Q}(\alpha)$, but the difference is a square factor. That gives a lot of information, but, of course, it is possible for distinct fields to share the same discriminant. For example when they are isomorphic embeddings of the same field. – Jyrki Lahtonen Apr 22 '15 at 10:48
• @JyrkiLahtonen: Why is it that the discriminant of the polynomials must differ by a square factor if they generate the same extension? Also, is there a more general method? The general polynomial in the family I am considering is $x \mapsto ( x^5 + 2^{3k} x^2 - 2 )$ for $k \in \mathbb{N}^+$, and I know next to nothing about how to compute discriminants. – user21820 Apr 22 '15 at 11:40
• There are two kinds of discriminants: of polynomials and of orders inside number fields. Assume that $m(x)$ is a monic irreducible polynomial of degree $n$ with $a$ as one of its zeros. Then the discriminant of the polynomial $m$ is equal to the discriminant of the order $$\Bbb{Z}[a]=\Bbb{Z}\oplus \Bbb{Z}a\oplus\cdots \oplus\Bbb{Z}a^{n-1}.$$ This is not an invariant of the field. The discriminant of a number field is the discriminant of its maximal order, i.e. its ring of algebraic integers $\mathcal{O}$. – Jyrki Lahtonen Apr 22 '15 at 12:54
• (cont'd) If the order $\Bbb{Z}[a]$ is an index $r$ subring of $\mathcal{O}$, then $$Disc(\mathcal{O})\cdot r^2=Disc(\Bbb{Z}[a]).$$ My point is that the discriminant of the field is an invariant, so two number fields with different discriminants cannot be isomorphic. Therefore if two irreducible polynomials have discriminants whose ratio is not a perfect square, then we can conclude that the corresponding number fields cannot be isomorphic. – Jyrki Lahtonen Apr 22 '15 at 12:57

Your question is to take $f=x^5+8x^2-2$ and $g=x^5+64x^2-2$, polynomials in $\Bbb Q[x]$, with a root $\alpha$ of $f$ and a root $\beta$ of $g$, and hope to show that $\Bbb Q(\alpha)$ is not isomorphic ot $\Bbb Q(\beta)$.

First I’ll tell you the moral of the sermon, and then justify it. I looked for a suitable prime $p$, and found that $3$ would do the trick: since $\widetilde f\in\Bbb F_3[x]$ (what you get by reducing the coefficients of $f$ modulo $3$) factors into an irreducible quadratic times an irreducible cubic, while $\widetilde g\in\Bbb F_3[x]$ factors into a linear times an irreducible quartic, the fields must be nonisomorphic.

To draw this kind of conclusion from the simple difference in pattern of the factorizations, you need a few special things to be going for you, so I have to spend a lot of time justifying my reasoning.

First, some notations: for a prime $p$, $\Bbb Z_p$ is the $p$-adic integers, but we don’t need to use them. So my first comment above was something of a red herring. But $\Bbb Z_{(p)}$ is the ring localized at the prime $p$, that is, the set of rational numbers whose denominators are prime to $p$. You see that this is a subring of $\Bbb Q$ with only the one maximal ideal $(p)$, in other words, it’s a local ring. Then its residue field $\Bbb F_p$ will be the finite field with $p$ elements, $\Bbb Z/p\Bbb Z$.

Now, if $K$ is a finite extension of $\Bbb Q$, it has its ring of integers, the subset of things in $K$ integral over $\Bbb Z$, and I’ll call this $\mathcal O_K$. When you take instead the set of elements of $K$ that are fractions of the form $u/m$ where $u\in \mathcal O_K$ and $m$ is an element of $\Bbb Z$ not divisible by $p$, this too is a good ring, I’ll denote it $\mathcal O_{K,p}$ even though I should write it $(\mathcal O_K)_{(p)}$, but I’m lazy. This is called the localization of $\mathcal O_K$ at $p$, and it’s not local, but “semilocal”, it has only finitely many maximal ideals, and the magical thing is that if $p\mathcal O_K$ decomposes as $\mathfrak p_1^{e_1}\cdots\mathfrak p_g^{e_g}$, then these $\mathfrak p$’s are, each of them, the intersection of $\mathcal O_K$ with one of those maximal ideals of $\mathcal O_{K,p}$ that I mentioned above. Complicated sounding, I know, but the idea is that “localization with respect to $p$” erases from the picture all primes of $\mathcal O_K$ not above $p$, so that you can just concentrate on the $p$-behavior. I think you might get some insight by looking at the case $K=\Bbb Q(i)$, $\mathcal O_K=\Bbb Z[i]$, with $p=5$. You’ll see that the only maximal ideals of $\mathcal O_{K,5}$ in this case are $(2+i)$ and $(2-i)$, corresponding to the factorization of the ideal $(5)\subset\Bbb Z[i]$ as the product of the ideals $(2+i)$ and $(2-i)$.

If you have some commutative algebra behind you, I’ve been describing the process $R\mapsto R\otimes_{\Bbb Z}\Bbb Z_{(p)}$. The utility of the process is that primes foreign to $p$ are no longer in the picture.

So much for notation and background. Let’s call $K=\Bbb Q(\alpha)$ and $L=\Bbb Q(\beta)$. We want to show that they are essentially different, i.e. nonisomorphic.

First I want to look more closely at your $f$ and $g$. Eisenstein tells us right away that they’re irreducible, so that $[K\colon \Bbb Q]=5$, similarly for $L$, as of course you know.

But it would be an amazing coincidence if $\mathcal O_K=\Bbb Z[\alpha]$. If that were to happen, then we could conclude that the discriminant of $K$ over $\Bbb Q$ (more properly of $\mathcal O_K$ over $\Bbb Z$) would be equal to the discriminant of the polynomial $f$, which is not so bad to calculate. Now here is the utility of localization: it is going to turn out that $\mathcal O_{K,3}=\Bbb Z_{(3)}[\alpha]$. The equation we couldn’t hope for globally will turn out to be true locally. And here’s why: We certainly have the inclusion $\mathcal O_{K,3}\supset\Bbb Z_{(3)}[\alpha]$. I’m going to look at the discriminant ideals of the two rings over $\Bbb Z_{(3)}$. For the right-hand member of the inclusion, the discriminant is generated by the discriminant of $f$ over $\Bbb Z_{(3)}$, and this is a story in $\Bbb Q$, the algebraic extension doesn’t come into the picture, and maybe you can convince yourself that this discriminant is the $3$-part of the global discriminant, but also can be decided upon by looking at $\widetilde f=x^5+2x^2+1$. As you’ve seen, this polynomial over $\Bbb F_3$ has no roots there, and the derivative $\widetilde f'=-x^4-x=-x(x+1)^3$. All factors of $\widetilde f'$ are linear, $\widetilde f$ has no linear factors, so the two are relatively prime, the discriminant is $1$ (up to constant factors), and the same is true of the discriminant of $f$ as a polynomial over $\Bbb Z_{(3)}$. This requires a little argument, I suppose, using the localness of $\Bbb Z_{(3)}$, but I’m going to skip it. And going back to our inclusion, you see that it has to be equality, because the discriminants are equal, and the reason for this is that there’s no ideal of $\Bbb Z_{(3)}$ bigger than $(1)$.

What have I shown? That $\mathcal O_{K,3}$ is unramified over $\Bbb Z_{(3)}$, which means that the ideals above $3$ are unramified, i.e. the exponents $e_i$ that I mentioned in the decomposition of $(p)$ in $\mathcal O_K$ (and here, $p=3$) are all equal to $1$. So we have the result that $3\mathcal O_{K,3}$ is just the product of primes of $O_{K,3}$, no exponents different from $1$. Each $\mathfrak p_i$ is maximal in the semilocal ring, and the residue field is finite, of cardinality $3^{f_i}$ (sorry for the duplication of notation, but $f_i$ is traditional here. And $\sum_if_i=n=5$, the degree of $K$ over $\mathbb Q$. The important thing is that these numbers will be calculated from $f$, but are, as a set, an invariant of the field $K$. Don’t depend on anything but the decomposition of your prime ($3$ in our case) in $K$. Now, I’m already running far too long, but I’m going to have to skip over the argument that these $f_i$ are just the degrees of the (known to be distinct) irreducible factors of $\widetilde f$.

Now, for completeness, I want to factor $\widetilde f$ over $\Bbb F_3$. Luckily, there are only three irreducible quadratics, they’re $x^2+1$, $x^2+x-1$ and $x^2-x-1$, and you try divisions, and see that $x^5+2x^2+1=(x^2+x-1)(x^3 - x^2 -x - 1)$. The upshot is that we have $\mathfrak p_1$, with residue field $\Bbb F_9$ and $\mathfrak p_2$, with residue field $\Bbb F_{27}$.

The story for $g$ and $L$ is similar, except that $\widetilde g=(x-1)(x^4+x^3 + x^2 -x -1)$. The only root of $\widetilde g$ is $1$, and you see that $\widetilde g'=-x^4-x=-x(x+1)^3$, again all factors linear, with roots $0$ and $-1$, so the two polynomials are relatively prime. Again $\mathcal O_{L,3}$ is unramified over $\Bbb Z_{(3)}$, but now the degrees of the primes above $3$ are $1$ and $4$ because that quartic factor is irreducible, as you see by trying division by irreducible quadratics. Anyhow, a different pattern of decomposition of $3$, again not dependent on $g$ itself, the argument just uses $g$ as a tool.

This has been TOO LONG, and I’m sure that if I had chewed over it for a while it could have been less than half as long. Maybe I could have avoided the whole technique of localization, for instance. But I’m sure that the people who are really skilled in this kind of thing can do better, and may jump in and shortcut many of my arguments. I think the essential thing is that $f$ and $g$ are unramified polynomials, i.e. over the field $\Bbb F_3$, they have simple roots. And that $f$ and $g$ have their irreducible factors of different degrees. More precisely, the partitions of $5$ that $f$ and $g$ induce by way of their factorizations are different.

Now I’m going to walk out on thin ice and say that I think that it’s an open question whether, if two fields $K$ and $L$ have the property that for every $p$, the decompositions of $p$ in the two induce the same partition of the common degree $[K\colon\Bbb Q]=[L\colon\Bbb Q]$. then it must be that $K$ and $L$ are isomorphic.

(End of sermon)

• Thanks a lot for your detailed answer. I see, however, that I know too little at the moment to understand it. Anyway would the same technique work for the general family of polynomials $x \mapsto ( x^5 + 2^{3k} x^2 - 2 )$ or any infinite subset of them? The root that would be chosen in each case is the positive real root. – user21820 Apr 26 '15 at 4:44
• Yes, I assumed that much of this would go over your head, which is why I gave the “method” first, and then its justification. I was thinking of @JyrkiLahtonen when I prepared this. But the method should always be useful: find a prime $p$ with three special properties: $\widetilde f\ne\widetilde g$; $\widetilde f$ and $\widetilde g$ are both “unramified”, by which I mean that each has no repeated roots (over $\Bbb F_p$, you recall); and the two of these are not only unequal, but the degrees of their splitting into irreducibles over $\Bbb F_p$ induce different partitions of the degree $n$. – Lubin Apr 26 '15 at 21:09
• Hmm but I don't really see how such a technique would extend to prove pairwise distinctness for infinitely many such field extensions, since it seems to be an ad-hoc technique for these two particular polynomials. Somehow @JyrkiLahtonen's method seems more plausible to attack, in the sense that there intuitively seems to be some infinite family of the form $50000 - 216 \cdot 2^{15k}$ that have non-square pairwise products. – user21820 Apr 27 '15 at 6:11
• Yes, your objection is definitely well-taken. Would you like to e-mail me, and we can continue this discussion that way? Find my e-mail on my page, and the link is on my profile. – Lubin Apr 27 '15 at 17:13
• Okay I've emailed you, but there's no hurry. Also, do you mind checking my proof at math.stackexchange.com/q/1242019/21820? Thanks a lot for your help! – user21820 Apr 30 '15 at 7:39