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I'm wondering what other tools there are aside from radicals can be used to extend fields in the context of solving polynomials. Since $S_5$ isn't solvable, constructing a field with a Galois group of $S_5$ with respect to $\mathbb{Q}$ can't be a radical extension, but is there some other function or operation that could be used? In other words, a quintic formula with radicals doesn't exist, but is there some function that isn't a purpose-built "this function yields solutions to a polynomial" function that could be used to solve quintics or higher polynomials?

tourist
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    This question is very vague, and will likely be closed for that reason. If you want an answer, maybe you can be clearer about what you are looking for. What do you mean "how could it be constructed"? What is wrong with, for instance, "the splitting field of $X^5-2X+1$". Why is that not satisfactory to you? – Captain Lama Mar 12 '22 at 09:06

2 Answers2

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I think this question may be a bit too vague as written, but here is a theorem one could appeal to in order to construct such a field extension. It is a fact that if $f \in \mathbb{Q}[x]$ is irreducible with $\deg f =p$ for some prime $p$ and has exactly 2 complex roots, then the Galois group of the splitting field extension is isomorphic to $S_{p}$.

Oiler
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  • Hi, thanks for the answer.

    To clarify, I mean how could you construct any field that has a Galois group of $S_5$? For example, you can construct a field corresponding to a Galois group of $Z_2$ as $\mathbb{Q}(\sqrt{2})$. I don't think that getting the splitting field would be that productive because to determine what to extend $\mathbb{Q}$ would be the solutions to a quintic equation that's not solvable with radicals.

    – tourist Mar 12 '22 at 06:55
  • @tourist you do not need full blown solutions for the zeros of a quintic polynomial to determine whether it has three real roots or not- you can literally graph it or do some analysis with calculus if you wanted to be a bit less hand-wavey. – Oiler Mar 13 '22 at 23:03
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    @tourist one could also construct as a quotient field, even a quotient of Q[x] (as you can find a primitive element) – Ariana Mar 15 '22 at 08:16
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Following up on the suggestion of Ariana. Let $$p(x)=x^5-3x^2+1,$$ when it is easy to show the Galois group of the splitting field of $p(x)$ over the rationals is $S_5$.

Assuming that after calculating the zeros of $p(x)=x^5-3x^2+1$ up to 200 decimal places makes it accurate enough, we can construct such a field $K$ as

$$K=\Bbb{Q}[x]/\langle f(x)\rangle,$$

where

$$f(x)=x^{120}+33750 x^{114}+1116000 x^{112}+3093750 x^{110}+1113560865 x^{108}+26921565000 x^{106}+604881000000 x^{104}+41167612108500 x^{102}+1431025760469805 x^{100}+14099233308750000 x^{98}+840755432509572240 x^{96}+29314339633598287500 x^{94}+838066825456291786500 x^{92}+17055343730004169787750 x^{90}+447099790210227436061835 x^{88}+16392596904188900682547500 x^{86}+524905496075736354421632150 x^{84}+7746834021558485383973580750 x^{82}+227400983306935610033504910210 x^{80}+5438203936818046114019096970000 x^{78}+161737233618167288338366207390005 x^{76}+3079140294664647133063741473941250 x^{74}+71589845771247618831377143880007030 x^{72}+1703874280827990134916438519652329750 x^{70}+37765960575062006944570272099577441065 x^{68}+671553629574346789742980518018455265750 x^{66}+13425663013791814923118775212295423224575 x^{64}+298316535549382310619276177470147839567500 x^{62}+6263322665754902028679045968192687021946417 x^{60}+113241137056665611046577975852478978853633750 x^{58}+2020482344097260796483156681326789886674641275 x^{56}+33607134753772181350504921906082658777563400000 x^{54}+550090827539075781453680855277159942042963975675 x^{52}+9645703546363285715097504428728612658988657633750 x^{50}+167589376219621416661551144351153197328368374710525 x^{48}+2591935957193938557000893428986050845007482057327500 x^{46}+37023543980552299462242594649218811608185860378407900 x^{44}+467204073589927694863531699904773814089803461948544000 x^{42}+5479652136885621906960743248094128630901237665341582235 x^{40}+60126566311045736532762730255787195693544124509832792500 x^{38}+622160800556257347660482211684147114157184154675733033935 x^{36}+5910280916004942423618805942747770827509420868385152115000 x^{34}+52106693211491045864419719017582977817447744056381867707300 x^{32}+432630947000379472605449339454013771610886258014994363460750 x^{30}+3410774365180782497074812580214046939106788562491870798046950 x^{28}+24817712899719418559089094578783591417923276718955280439927500 x^{26}+166698014114867874997691254686109786765856603463497453170639800 x^{24}+1049202472505126328549462956322404941429926499583868890532339750 x^{22}+6200097454408267204060687491470661741711907284917745068688392736 x^{20}+33063248055138491333939800416229050428492869440711652594093046250 x^{18}+160736778936639982917274226049744035868667050374165352429934665765 x^{16}+713664496954265807293435573649973706894373908524000311681369366250 x^{14}+2807053376943637105788469622562984789953566579091832894551656335585 x^{12}+9405202706949914238281707642961512160794793530858137713058729194750 x^{10}+29874334500240270890684107985135817331403471254102147813332994672740 x^8+70498858873024053142911958048672442662399336143969727445260372215750 x^6+152058512419218423143371634526878068936948757145171468466368924947025 x^4+244691421102204282324955998149483634966444221492246674851974602623750 x^2+191416245283609068525605381407966277298503971734124842355884163512481$$

has as its $120$ zeros the numbers $$x_\sigma:=2a_{\sigma(1)}+a_{\sigma(2)}-a_{\sigma(4)}-2a_{\sigma(5)}.$$ Here $a_1,\ldots,a_5$ are the zeros of the polynomial $p(x)$, and $\sigma$ ranges over all the permutations $\in S_5$.

The $120$ Galois conjugates of $x_{id}=2a_1+a_2-a_4-2a_5$ are all distinct. For otherwise there would be a linear dependency relation (over $\Bbb{Z}$) among the roots $a_i,i=1,2,3,4,5$, other than the Vieta relation $a_1+a_2+a_3+a_4+a_5=0$ — in violation of the fact that the splitting field has degree $120$.

Basic Galois correspondence then tells us that $\Bbb{Q}(x_{id})$ is the splitting field of $p(x)$. Hence all the numbers $x_\sigma$ are in there, and we can use any of them as a primitive element.

Jyrki Lahtonen
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    Computers are amazing. Cool answer! +1 – Alex Wertheim Mar 15 '22 at 09:01
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    I hope it's clear to the readers that this post is meant to be a joke rather than an insightful answer to the OP's question. A point to be taken is the difficulty of explicitly describing a 120-dimensional extension field of $\Bbb{Q}$ with the prescribed Galois group. The standard example of a field extension with Galois group $S_5$ involved transcendental elements. Five unknowns and their symmetric polynomials. – Jyrki Lahtonen Mar 15 '22 at 09:01
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    +1 and I find the standard example of "five unknowns and their symmetric polynomials" more satisfying than using 120 deg polynomial. – Paramanand Singh Mar 20 '22 at 03:56
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    But this polynomial has a charming Christmas tree shape, just right for the time of the year (as of the time of writing this comment, at least). And why not be very explicit sometimes. – Dietrich Burde Nov 28 '23 at 20:07