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So basically some formula for irreducible number of equations would be a better 'term'.

But I also know that $y^2 = xz$ and $xy = z$ can simply be written down as $(y^2 - xz) (xy - z) = 0$.

Something like $n_{p} = no. of parameters$ and $n_e = no. of independent variables$, then number of equations is equal to $n_{p} - n_{e}$, or something like that.

Or may be not exactly a rigorous proof, but a result for me would work.

How do I prove that solution is $y^2 = xz$ and $xy = z$, -- (eq. 1) but not just $y^2 = xz$ -- (eq. 2). Since the second one also satisfies the required condition.

But I can't do that manually for every point.

Yes, that is what I want.

So basically my question is establishing both ways equality between parametrised equation and cartesian coordinates curve, how do I ensure they are indeed equal.

I can sure write it simply as $y = x^2$ which would be just one equation to represent it. But that is not sufficient.

Let $x = t$, $y = t^2$ and $z = t^3$ where $t$ is a parameter in space. How do you eliminate $t$ from it?

@SteamyRoot got it.

How do the functions (or expressions) $\exp(\frac{x}{2})$, and $\sqrt{\exp(x)}$ over the complex numbers differ?

I posted the question now on site itself math.stackexchange.com/questions/2400454/…

There is one paper that discusses the Weierstrass substitution in symbolic integration but "not" specifically in terms of Risch algorithm, this one apmaths.uwo.ca/~djeffrey/Offprints/toms1994.ps

Does anyone know of a manuscript that discusses "Weierstrass substitution"(also called tangent half angle substitution) in terms of the Risch algorithm?

Anyone who can help me out with Symbolic Integration problem?

$F$ is overall a field here, made by building up the extension, similar to the tower of extensions.

Is $\theta = \exp(\frac{1}{\log(x)})$ an element of $F = C(x)(t_1, t_2, ..., t_n)$ where $C$ is the constant field and $t_i$'s are monomials over $C(x)(t_1, ... t_{i - 1})$, for each $i$.

@TobiasKildetoft if you think I need to improve upon my question asking way, I had be happy to hear if that is your opinion.

@TobiasKildetoft what I think rational solution space means here is: just the rational linear combination of $f_i$'s i.e $\sum_{i=1}^{n} r_i f_i$ where $r_i \in \mathbb{Q}$ (i.e for all rational values of $r_i$'s).

@TobiasKildetoft that rational relation is $\sum_{i=1}^{n} r_i f_i = 0$ (I missed equals zero in that).

@TobiasKildetoft if you think that the comment is not complete in some sense, I can tell you for what the missing thing means.

@TobiasKildetoft This is a line from a comment I received in a discussion: "The task is to find those linear relations where the coefficients $​r_i$​​ are rational numbers. The computation depends on the existence of some basis over $\mathbb{Q}$ but the result should be a matrix of rational entries ($r_i$'s) such that the rational relations ($\sum_{i=1}^{n} r_i f_i$) are exactly the elements of the rational solution space of the matrix."

What is the meaning of "solution space of a matrix"?

That $f_i$'s are elements of the type $\frac{a(x)}{b(x)}$ where $a(x)$, $b(x)$ are polynomials in variable $x$ where coefficients of $x$ are elements of $K$?

What does it mean by $f_i$'s a are fractions over an algebraic number field $K$?

Where $K$ is a field.

@Danu okay. When I say linear relations over K, it would mean "linear relation among the elements of $K$"?

Seems like everyone is out for the weekend :)

Till now I have often come across the terms "over a field" in case of algebraic structures like fields. Can you point me to some link where I can see its usage and meaning?

@EricStucky and that is because the transcendence degree of $\pi$ is $\infty$?

Basis for $\mathbb{Q}(\pi)$ over $\mathbb{Q}$ is of infinite cardinality?

@LeakyNun thanks.

@LeakyNun wait a moment please, I am still searching for chatJax enabling.

@AlessandroCodenotti BTW how do see the latex written in here?

That would be $$\mathbb{C}$$.

Are there any computer algebra systems that currently do these type of computations?

And now consider a field $$\mathbb{Q}(\sqrt 2, \sqrt3)$$ which is equivalent to $$\mathbb{Q}(\sqrt 2 + \sqrt3)$$ (is it?), and the extension $$\mathbb{Q}(\sqrt 2 + \sqrt3)$$ is a simple extension, so its equivalent extension should also be simple.

I am considering the base of the extension to be $$\mathbb{Q}$$.

Is every algebraic extension a simple extension?

To be true, I didn't thought it to find in something for permutations. But after you sent me the link, I thought the term "cyclic" should have made me to look for it, even in the permutations related search results.

Thank you very @MartinSleziak especially for the link.

So for the set would be the union of cyclic conjugates of all elements.

It seems quite intuitive.

May be the definition for a set of "words" is something similar to that of definition for individual "words".

Yes, this is about "words" precisely, to repeat I meant: the elements of set are "words" in generators and their inverses.

Seems like nobody in the room right now. Ping me if anyone can help me with the definition of this.

In particular I want to find the "cyclic conjugate" of a set of union of relators and their inverses of a finitely presented group.

Perhaps something I am studying about the "Coset Enumeration", though probably "cyclic conjugate" in group theory seems to be a much general term.