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If you have a polynomial (say $p(x)$) which is irreducible in $\mathbb R[x]$, then is there a practical use of constructing an extension field $\mathbb R[x]/\langle p(x) \rangle$ with a zero rather than just finding the root in $\mathbb C$?

Yes, it's smaller but just because $\mathbb C$ is bigger doesn't mean you have to care about everything there when finding your root.

Is there any practical application for it? Or is it just pure mathematics without any application as yet?

user93353
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  • You should check out Evan Chen's Napkin. He has a very good motivation for splitting fields – Clemens Bartholdy Apr 18 '22 at 09:32
  • @Buraian - I check the motivation paragraph at the beginning of the chapter which covers splitting field - it didn't contain any practical motivation or use – user93353 Apr 18 '22 at 10:48
  • The idea of a root of a polynomial is best handled by constructing such an extension. Viewing it as a real or complex number destroys the whole algebraical viewpoint. A root is not some special kind of number but rather it is defined by means of the polynomial it satisfies. – Paramanand Singh Apr 18 '22 at 16:12
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    It takes some time getting used to the fact that any algebraical property of $\sqrt{2}$ is obtained by treating it as a root of $x^2-2$ and further algebraically one can't distinguish between the two roots $\pm\sqrt{2}$. Usually one learns roots as real or complex number and the algebraic aspect is presented later in courses of abstract algebra. Also are you sure you want to deal with polynomials in $\mathbb {R} [x] $ and not in $\mathbb {Q} [x] $? – Paramanand Singh Apr 18 '22 at 16:17
  • @ParamanandSingh - now this goes into why we find the root of a polynomial? A practical reason for finding a root is to sketch the curve because finding the root tells us where it intersects the axis. In such a practical endeavor, does this method of finding a root by constructing an extension really help us. That's what I mean by my question - what's the practical value of finding a root by constructing an extension? Is there a practical use for doing this? I am fine with an answer that we do this purely for math & no practical use - but I want to confirm it – user93353 Apr 18 '22 at 21:20
  • @ParamanandSingh Also are you sure you want to deal with polynomials in R[x] and not in Q[x]? - my point is why should I care at all where the root exists as long as I can find the root – user93353 Apr 18 '22 at 21:21
  • @ParamanandSingh further algebraically one can't distinguish between the two roots $\pm\sqrt{2}$ - in all practical uses of finding the roots of a polynomial like plotting the curve, you do need to distinguish between the two roots. Which goes back to my question as to the practical use of finding the root of the polynomial this way. I am just trying to step back a little & think as to why we are into finding roots at all in the first place? – user93353 Apr 18 '22 at 21:43
  • "Yes, it's smaller but just because C is bigger doesn't mean you have to care about everything there when finding your root." Mathematicians don't just care about solving a problem, they care about truly understanding the problem. That's way more important than just solving it. To truly understand a solution to a problem, you have to identify which aspects of the solution are absolutely necessary, and which aspects are just bloat. That's the reason why mathematicians try to use small extensions instead of going straight to $\mathbb C$: Most of $\mathbb C$ is bloat. – Vercassivelaunos Apr 20 '22 at 09:44
  • @Vercassivelaunos - that's exactly my question - whether this actually has some practical purpose or if it's just pure mathematics for the joy it. I am not knocking down studying stuff which doesn't have a practical purpose as yet - just wanted to know if it has or no. – user93353 Apr 20 '22 at 09:47

2 Answers2

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You are actually using the theorem that an integral domain quotiented by a maximal ideal is a field. See my answer here .

The main problem is that you have not defined what an algebraic closure is using just the axioms of field theory. Okay sure, for $\Bbb{R}$ you do have $\Bbb{C}$.

But what about arbitrary fields. What about a finite field. How do you get a field where an irreducible polynomial has a root. Say you have an irreducible polynomial like $X^{p}-X+1\in \Bbb{Z}_{p}[X]$. How do you even find a root and attach it to get a "larger" field where this polynomial has a root?

Or even worse, what about function fields? . Say you have $F(X^{n})$ as a field( the field of fractions of the ring $F[X^{n}]$ and you are given an irreducible polynomial like $X^{n}-t^{n}\in F(X^{n})[t]$ . how would you construct a "larger" field where this polynomial has a root?.

That is where this definition comes in handy. You can see my answer in the link . There I have provided a short outline to the process of "attaching a root of a polynomial to a field to get a larger field". All of this is contained in any algebra texts ( I would recomment Dummit and Foote for a reference ).

Edit:- As the conversation in the comment of this answer indicates, I think the main problem is that the op is questioning the need for arbitrary fields in general. So as far as Number theory goes or the study of primes(I don;t think I need to justify it's practical usage in the times of cryptocurrency and stuff like that) , the group $\Bbb{Z}_{p}$ is very very imporant and it is a Field( Called a finite field) . Now as it turns out, every field of characteristic $p$ contains $\Bbb{Z}_{p}$ inside it( to be more precise, an isomorphic copy of it). So this enough should tell you why we should study finite fields . You might learn this as you go forward that all finite fields are just simple extensions of $\Bbb{Z}_{p}$. That is , they occur as $\frac{\Bbb{Z_p}[x]}{(f(x))}$ for some irreducible polynomial. So this should perhaps satisfy your query as to importance of proceeding by quotienting rather than directly moving to $\Bbb{C}$ which as I pointed out is a field of characteristic $0$ and hence does not contain $\Bbb{Z}_{p}$ . See this and this for proofs. This is because we actually do not know explicitly how the root actually looks like. Sure in $\Bbb{R}$ you can say $\sqrt{2}$. But what about the root of the polynomial $x^{2}-2$ over $\Bbb{Z}_{p}[x]$? $\sqrt{2}$ has no meaning here as any extension which will contain this root does not lie in $\Bbb{R}$ as I said that $char\,0 $ fields don't contain $\Bbb{Q}$ and hence no $\Bbb{R}$. The proper mathematical justifications of $\sqrt{2}$ and irrationals are done by Dedekind cuts and the Supremum property but those things are part of Real Analysis .

  • What about a finite field. How do you get a field where an irreducible polynomial has a root. - What is the need to have a field in a finite field? What's wrong in having the root in $\mathbb C$? Is there a practical use of having the root in a finite extension rather than in $\mathbb C$? – user93353 Apr 18 '22 at 10:36
  • You are actually using the theorem that an integral domain quotiented by a maximal ideal is a field. - Yes, I understand the construction. But my question is why are we even doing this? I am ok with the answer that we are doing as pure mathematics which no purpose. If not, is there a practical use for it? – user93353 Apr 18 '22 at 10:38
  • Finite extensions and finite field are different. As far as number theory is concerned , $\Bbb{Z}{p}$ is very very important as are it's extensions. The complex numbers or in fact any field of characteristic $0$ does not have $\Bbb{Z}{p}$ inside it. – Mr. Gandalf Sauron Apr 18 '22 at 10:41
  • Secondly, Do you know about Galois theory? . If you want to prove something like insolvability of quintics then you have to deal with finite extensions. Proceeding directly to an algebraic closure is of no use. – Mr. Gandalf Sauron Apr 18 '22 at 10:43
  • @user93353 I have edited in the crux of the conversation in the comments and added a few more things. – Mr. Gandalf Sauron Apr 23 '22 at 13:57
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If you ask about fields $k$ other than $\mathbb{R}$, then I think I can give several good answers to this question.

(1) As other answers have pointed out, you may not have proved that $k$ has an algebraic closure yet.

(2) Even if you know that $k$ has an algebraic closure (perhaps $k = \mathbb{Q}$), computing in $k[x]/f(x)$ may be more convenient than computing in $k^{alg}$. For example, if I had to do a lot of computations in $\mathbb{Q}[\zeta_5]$, where $\zeta_5$ is a primitive $5$-th root of unity, I'd rather store my numbers in the form $a_1 \zeta+a_2 \zeta_2+a_3 \zeta^3+a_4 \zeta^4$ for $a_1$, $a_2$, $a_3$, $a_4 \in \mathbb{Q}$ than storing them as $x+iy$ with $x$ and $y$ floating point. (There are tradeoffs, though: As the degree of $f$ gets larger, floating point gets more useful.)

Point (2) is much more true if $k$ is not a subfield of $\mathbb{C}$. In that case, basically* the only way to compute in $k^{alg}$ is to work in finitely generated subfields of it.

(3) If you are heading for Galois theory, you want to show that, if $f(x)$ is an irreducible polynomial, and $K$ is a splitting field of $f$, then $\text{Aut}(K/k)$ acts transitively on the roots of $f$. To be concrete, you want to show that there is an automorphism of $\mathbb{Q}(\sqrt[3]{2}, \omega)$ taking $\sqrt[3]{2}$ to $\omega \sqrt[3]{2}$. This is pretty straightforward if you write an element of $\mathbb{Q}(\sqrt[3]{2}, \omega)$ as $a + b \omega + c \sqrt[3]{2} + d \omega \sqrt[3]{2} + e \sqrt[3]{4} + f \omega \sqrt[3]{4}$, for $a$, $b$, $c$, $d$, $e$, $f \in \mathbb{Q}$. It is hard (in my opinion) if you think of $\mathbb{Q}(\sqrt[3]{2}, \omega)$ as the subfield of $\mathbb{C}$ generated by $\sqrt[3]{2}$ and $\omega$. In particular, you need the Axiom of Choice if you want to prove this automorphism extends to $\mathbb{C}$.


If you insist on thinking about $\mathbb{R}$, the advantages are much less, because you'll be using floating point numbers to store the elements of $\mathbb{R}$ any way, and because $\mathbb{C}$ is so much more concrete than a general $k^{alg}$.

The one reason I can think of is that perhaps you haven't proved the FTA yet. There are proofs of the FTA where you want to first know that there is some field where $f(x)$ has roots, before proving that $\mathbb{C}$ is such a field. Several mathematicians (Euler, Lagrange, Laplace) gave flawed proofs of the FTA in the 18th century which assumed such a field existed. From a modern perspective, there is a slick Galois theory proof of the FTA; if you want to take this route, you need to be able to build splitting fields without the FTA.


* For experts, the "basically" is my way of indicating that yes, I am aware of Kedlaya and other's work on explicitly working in an algebraic closure of $\mathbb{F}_p[t]$ by generalized Puiseux series.

  • I think the main problem is that the op is questioning the need for arbitrary fields in general . The root of the question lies in the fact that the op is satisfied with just $\Bbb{Q,R,C}$. I sort of provided a short explanation as to why atleast finite fields are important. Your answer is great though(+1). But I think the op is going to ask the same thing as to what is the need for general algebraic closures(whose proof uses Zorn's lemma anyway) and why can't we just stick to $\Bbb{C}$. – Mr. Gandalf Sauron Apr 23 '22 at 13:49
  • If the OP cares about $\mathbb{Q}$, I think (2) and (3) are good answers, although maybe the OP isn't ready for them. If the OP only cares about $\mathbb{R}$ and $\mathbb{C}$, I think they may be right to think that the $k[x]/p(x)$ construction isn't important to them yet. – David E Speyer Apr 23 '22 at 14:00
  • My question why find the smallest algebraic closure at all? Why not use $\mathbb C$. Is there a practical use to finding the smallest closure? I am fine with the answer being that it's being done for doing mathematics & not any practical purpose - just want to confirm that. – user93353 Apr 26 '22 at 10:33
  • My question isn't about $\mathbb Q$, $\mathbb C$, $\mathbb Z$ or $\mathbb R$? I am saying keep your $Field(x)$ or $Ring[x]$ with field or ring as $\mathbb Q$, $\mathbb C$, $\mathbb Z$ or $\mathbb R$ & take the algebraic closure of all as $\mathbb C$. What is the practical use of a smaller closure? – user93353 Apr 26 '22 at 10:36