# Some Galois Theory: Degree of elements in Q

## Main Question or Discussion Point

Hello everyone.
I'm trying to figure out a trick to finding the degree of
(81)^1/5+29(9)^1/5+17(3)^1/5-16 over Q.

I know that the degree of this number, call it a, is equal to the degree of the field extension
[Q{a}:Q], and also the degree of the minimal polynomial which has this as a root. Unfortunately, I can't seem to find an answer for it.

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The idea may be to show that $$\mathbb{Q}(a) = \mathbb{Q}(\sqrt[5]{3})$$.

The minimal polynomial may take too long to find.

At first, excuse me for my bad english.

There are simple algorithm to find minimal polynomial of $$a+b$$ (or $$ab$$) if minimal polynomials of $$a$$ and $$b$$.

Let $$p(a)=0$$, $$q(b)=0$$, $$deg(p)=m$$, $$deg(q)=n$$. Then, we can express any power of $$a$$ as linear combination of $$1,a,\dots,a^{m-1}$$ and simmilar for $$b$$ Using this idea and Newtone-s binomial formula, we can express any power of $$a+b$$ as linear combination of numbers of form $$a^ib^j$$ for $$i\in\{0,\dots,m-1\}$$, $$j\in\{0,\dots,n-1\}$$. There are $$mn$$ numbers of such form. We can denote them as $$c_1,\dots,c_{mn}$$.

We can form the matrix $$A$$ with $$mn$$ columns and $$mn+1$$ rows as follows: $$i$$-th row of matrix $$A$$ is $$a_{i,1},\dots,a_{i,mn}$$ where $$(a+b)^{i-1}=a_{i,1}c_1+\cdots+a_{i,mn}c_{mn}$$. The matrix $$A$$ has linear dependent rows (besause number of rows is greather than number of columns). Let we denote $$i$$-th row of matrix $$A$$ with $$r_i$$. Let $$\lambda_0,\dots,\lambda_{mn}$$ are such that $$\lambda_0r_1+\cdots+\lambda_{mn}r_{mn+1}=0$$. Then, for polynomial $$s(x)=\lambda_0+\cdots\lambda_{mn}x^{mn}$$ holds $$s(a+b)=0$$.

Using Kronecker's algorithm one can express the polynomial $$s(x)$$ as product of irreducibile polynomials. One of factors is minimal polynomial of $$a+b$$. Using numerical computations, all factors but one can be eliminated.

Yes, this is very hard for manual computations, but it can be implemented as automatic procedure.

mathwonk
Homework Helper
thank you for bringing this point out. i have long been interested in algorithms for problems in galois theory.

forgive me for playing devil's advocate, but how practical is this? kroneckers algorithm is incredibly lengthy. surely better factoring algorithms exist.

i would suspect this is not just very hard but completely unfeasible manually.

indeed even factoring large integers, which kronecker assumes, is quite challenging and slow.

moreover deciding which irreducible factor is satisfied by your element can be non trivial. e.g. try to check whether the unique real number given by cardano's formula for the solutions of X^3 + X -2 = 0, actually works.

i.e. why does (1 + (2/3)[7/3]^(1/2))^(1/3) + (1- (2/3)[7/3]^(1/2))^(1/3) = 1?

perhaps this is easy on a computer, but

how do you handle such difficulties?

have you implemented this procedure? e.g. does it give a reasonable solution to finding the minimal poly of the element posed here by the OP?

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let b = (3^(1/5))^4 + 29*(3^(1/5))^2 + 17*3^(1/5) - 16. Q(b) is contained in Q(3^(1/5)) and [Q(3^(1/5):Q] = 5, so the degree of Q(b) over Q divides 5, so it's 1 or 5, but it can't be 1 so must be 5

mathwonk
Homework Helper
of course we understood that clever and trivial solution to the degree problem, [every element of an extension of prime degree p has degree 1 or p], but now we are exploring the more difficult problem of finding the minimal polynomial.

in fact you omit the only non trivial part of your solution, checking that it cannot be 1.

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thank you for bringing this point out. i have long been interested in algorithms for problems in galois theory.
Then, I am glad to inform you that THERE IS an algorithm for computing Galois group (over Q) of ARBITRARY polynomial with rational coefficients. Do you know it?

...,but how practical is this?
I don't know for any practical application of the Galois theory. For me, it is just one elegant and beauty mathematic theory. Historic, Galois theory made very great and important revolution in mathematics. But, I don't know for it's practical applications. Look at it from aesthetic viewpoint. There are algorithm's! Great!

why does (1 + (2/3)[7/3]^(1/2))^(1/3) + (1- (2/3)[7/3]^(1/2))^(1/3) = 1?
Let we denote $$p(x)=x^3+x-2$$ $$q(x)=x^2+x+2$$ and

$$a=\sqrt[3]{\sqrt{\frac{28}{27}}+1}-\sqrt[3]{\sqrt{\frac{28}{27}}-1}$$

We can see that $$p(x)=(x-1)q(x)$$. You can compute $$a$$ numerical. You will obtain $$|a-1|<\varepsilon$$ for some $$\varepsilon>0$$ depending on your numeric computation. If $$\varepsilon$$ is enough small, one can conclude (by automatic procedure) that $$q(a)\neq 0$$. Then $$0=p(a)=(a-1)q(a)$$ implies $$a=1$$.

have you implemented this procedure? e.g. does it give a reasonable solution to finding the minimal poly of the element posed here by the OP?
Procedure is not implemented by me. I don't understand rest of the question because I don't know what is OP.

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mathwonk
Homework Helper
what does it mean to compute a galois group numerically? do you give a multiplication table? or do you identify it with a subgroup of S(n), or of a free group by generators and relations?

and OP is presumably "original poster", so refer to post 1.

mathwonk
Homework Helper
isnt an impractical algorithm sort of like a wrench that does not turn the nut? or a plane that does not fly? where is the aesthetic beauty in that?

of course we understood that clever and trivial solution to the degree problem, [every element of an extension of prime degree p has degree 1 or p], but now we are exploring the more difficult problem of finding the minimal polynomial.

in fact you omit the only non trivial part of your solution, checking that it cannot be 1.
"[every element of an extension of prime degree p has degree 1 or p]"
i wasn't thinking of it that way, but yes i agree with your statement

"in fact you omit the only non trivial part of your solution, checking that it cannot be 1"
well i guess i made the mistake and assumed it to be trivial, but surely there must be a way,
if b= (3^(1/5))^4 + 29*(3^(1/5))^2 + 17*3^(1/5) - 16 were rational, then... , don't see where to go with this atm, there must be a way though, there must be

and finding the minimal polynomial seems intimidating, i would think this way(ie showing it can't be 1) would be easier, the only issue left is whether or not b is rational, but how to justify it rigorously?

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what does it mean to compute a galois group numerically? do you give a multiplication table? or do you identify it with a subgroup of S(n), or of a free group by generators and relations?
The algorithm output can be multiplication table or subgroup of $$S_n$$.

For given polynomial $$p$$, one can by automatic procedure find finite number of disjoint compacts $$K_1,\dots,K_s$$ (in $$\mathbb C$$) such that any compact contains exactly any root of polynomial $$p$$ is contained in $$K_i$$ for some $$i$$. Let we call the closed ball with rational radius and both real and imaginary part of center as fine ball. Compacts can be fine balls. Let we call this procedure as localization.

Any algebraic number $$a$$ can be expressed as root of an polynomial $$p\neq 0$$ over $$\mathbb Q$$ contained in certain fine ball which does not contain any other root of $$p$$. Let we call this form of algebraic numbers as poly form. The algebraic number given in poly form can be computed numerical by automatic procedure.

We can calculate with algebraic numbers in poly form. Using previously explained procedure we can compute polynomial $$p\neq 0$$ over $$\mathbb Q$$ such that $$p(a+b)=0$$, where $$a,b$$ are algebraic numbers given in poly form. After localization of polynomial $$p$$ we can eliminate all compacts but one by numeric computing of $$a+b$$. Then, poly form of $$a+b$$ is computed. The cases of multiplication, division and subtraction are analogous. Case of the n-th root is easer. these are left to reader.

Comparing algebraic numbers given in poly form is reduced to case of comparing with zero. If an algebraic number $$a$$ is given as root of polynomial $$p$$ contained in compact $$K$$, then $$a=0\Leftrightarrow(p(0)=0\land 0\in K)$$.

General case of computing Galois group of given polynomial can be easy reduced to the case of polynomial with no multiple roots.

Let $$a_1,\dots,a_n$$ are all different roots of polynomial $$p$$. there is the automatic procedure for computing rationals $$r_1,\dos,r_n$$ such that $$b=r_1a_1+\cdots+r_na_n$$ be an primitive element. Then, one can express all conjugates of $$b$$ and all $$a_i$$ via $$b$$. Automorphisms are determined by image of primitive element. Image of primitive element can be any of its conjugates. Image of $$a_i$$ can be determined by image of $$b$$ and compared with $$a_j$$. Permutation of $$\{a_1,\dots,a_n\}$$ is computed.

This is a sketch. I hev no more time now.

mathwonk
Homework Helper
thank you. i still wonder if you can produce the minimal polynomial of the element originally listed in post 1.

Let $$a=\sqrt[5]{3}$$ and $$b=a^4+29\,a^2+17\,a+16$$. You ask me for minimal polynomial of $$b$$. Let $$b^k=c_{k;0}+c_{k;1}a+c_{k;2}a^2+c_{k;3}a^3+c_{k;4}a^4$$. for example $$b^0=1$$, $$b^1=16+17\,a+29\,a^2+a^4$$. For higher values of $$k$$ use the identity $$a^5=3$$. Consider the matrix

$$A= \left[ \begin{array}{ccc} c_{0;0} & \cdots & c_{0;4} \\ \vdots & \ddots & \vdots \\ c_{5;0} & \cdots & c_{5;4} \end{array} \right].$$

Because number of rows is greater than number of columns, the rows of matrix $$A$$ are linear dependent. Let we denote $$i$$-th row as $$r_i$$. Use the linear algebra methods to compute numbers $$p_0,\dots,p_5$$ such that $$p_0r_1+\cdots p_5r_6=0$$. For polynomial $$p(x)=p_0+\cdots p_5x^5$$ holds $$p(b)=0$$. Check that $$p$$ has no rational roots and proof that $$b$$ i irrational is completed. Because $$b\in\mathbb Q(\sqrt[5](3))$$ proof that $$deg(b)=5$$ iz done.

$$b$$ is irrational because if $$b\in\mathbb Q$$ then because $$a^4+29a^2+17a+(16-b)=0$$ the polynomial $$x^5-3$$ can not be minimal polynomial for $$a$$.

You have quite simple solution of original problem. $$b\in\mathbb Q(\sqrt[5]{3})$$ and $$b$$ is irrational.

mathwonk
Homework Helper
nedeljko, forgive me, but it seems you still resist actually finding an answer.

Solution of original problem is:

Let $$a=\sqrt[5]{3}$$ and $$b=a^4+29a^2+17a+16$$.

Statement 1. Polynomial $$x^5-3$$ is the minimal polynomial (over $$\mathbb Q$$) for $$a$$.
Proof: $$a^5-3=0$$ and polynomial $$x^5-3$$ is irreducibile (over $$\mathbb Q$$).

Statement 2. $$deg(a)=5$$.
Proof: It follows by statement 1and definition of element degree.

Statement 3. $$[\mathbb Q(a):\mathbb Q]=5$$.
Proof: It follows from statement 3.

Statement 4. $$b$$ is irrational.
Proof: Let we assume that $$b\in\mathbb Q$$ and let $$p(x)=x^4+29x^2+17x+16-b$$. Because $$b[\in\mathbb Q$$ polynomial $$p$$ has all coefficients in $$\mathbb Q$$. By definition of $$b$$ holds $$p(a)=0$$. Polynomial $$p$$ has degree 4 and it is contradiction with statement 1.

Statement 5. $$b\in Q(a)$$.
Proof: It follows by definition of $$b$$.

Statement 6. $$deg(b)=5$$.
Proof: By statement 3 and statement 5 holds $$deg(b)|5$$ and consequently $$deg(b)\in\{1,5\}$$. By statement 4 holds $$deg(b)\neq 1$$.

Is it now clear? For calculating minimal polynomial for $$b$$ I must use computer because large numbers. I posted procedure for computing it.

Correction: Statement 3 is consequence of statement 2.

$$b$$ is irrational because if $$b\in\mathbb Q$$ then because $$a^4+29a^2+17a+(16-b)=0$$ the polynomial $$x^5-3$$ can not be minimal polynomial for $$a$$.
very nice, thanks

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