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I have a question that, for lack of familiarity or understanding of the relevant fields, I'm not quite sure how to formulate, so I'll just start off with an example and list some questions as I go. As the title suggests, it has to do with factorizing an otherwise irreducible expression into "polynomials" of algebraic degree.

Consider the binomial $x+y$. If I'm reading the definition correctly, then this is certainly an irreducible polynomial since we can't decompose it into polynomial pieces. We can $(\spadesuit)$, however, factorize it to obtain $$x+y=\left(x^{1/2}\right)^2-\left(i y^{1/2}\right)^2=\left(x^{1/2}+iy^{1/2}\right)\left(x^{1/2}-iy^{1/2}\right)$$

$(1)$ Under what circumstances is $(\spadesuit)$ a "legal move"? Do I need to be specific about what $\left(a^{1/n}\right)^n$ might mean?

$(2)$ Is there a name for this kind of "reducibility"?

We can continue in this manner to obtain the next iteration, $$x+y=\left(x^{1/4}\color{red}\pm i^{3/2}y^{1/4}\right)\left(x^{1/4}\color{red}\pm i^{1/2}y^{1/4}\right)$$ followed by $$x+y=\left(x^{1/8}\color{red}\pm i^{7/4}y^{1/8}\right)\left(x^{1/8}\color{red}\pm i^{5/4}y^{1/8}\right)\left(x^{1/8}\color{red}\pm i^{3/4}y^{1/8}\right)\left(x^{1/8}\color{red}\pm i^{1/4}y^{1/8}\right)$$ and so on, where I use $\color{red}\pm$ to mean $(a\color{red}\pm b)=(a+b)(a-b)$ just to save space.

$(3)$ Same as $(1)$ but for successive iterations.

Barring any errors thus far, we can arrive at a compact form relying on product notation: $$\large \prod_{\substack{n\in\mathbb{N}\\[.5ex]1\le r\le 2^{n-1}}}\left(x^{2^{-n}}\color{red}\pm i^{(2r-1)2^{-(n-1)}}y^{2^{-n}}\right)$$

$(4)$ Is there anything about this kind of manipulation that is blatantly wrong or otherwise doesn't hold up?

$(5)$ What fields of math, if any, would be interested in studying this kind of factorization?

In addition to being unsure how to frame this question properly, I'm also confuddled as to how to tag this question. Any suggestions/edits would be appreciated.

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    $\begingroup$ I think you're basically showing that the ring $\bigcup_{n=1}^\infty \mathbb{C}[x^{1/n}, y^{1/n}]$ is not noetherian. $\endgroup$ – André 3000 Jul 7 '16 at 2:11
  • $\begingroup$ @SpamIAm I'd replace $n$ by $2^n$ in order to be sure that the union forms a ring. $\endgroup$ – user26857 Jul 7 '16 at 6:09
  • $\begingroup$ @user26857 Yeah, you're right. Really I should probably say the direct limit $\varinjlim_{n} \mathbb{C}[x^{1/n}, y^{1/n}]$ where $\mathbb{C}[x^{1/m}, y^{1/m}] \hookrightarrow \mathbb{C}[x^{1/n}, y^{1/n}]$ if $m \mid n$. $\endgroup$ – André 3000 Jul 7 '16 at 14:30
  • $\begingroup$ All this notation is making my head spin... But I gather ring theory is something I should be looking into. $\endgroup$ – user170231 Jul 7 '16 at 14:58
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Essentially you are asking about roots of unity. That is, for any positive integer $n>0$ there is a factorization $$ x^n-1 = \prod_{k=0}^{n-1} (x-\zeta_n^k) \tag{1}$$ where $\,\{\zeta_n^0,\zeta_n^1,\dots,\zeta_n^{n-1}\}$ are the $n$-th roots of unity. Since $(x-1)(x+1)=x^2-1,$ we have a similar factorization $$ x^n+1 = \prod_{k=0}^{n-1} (x-\zeta_{2n}^{2k+1}) \tag{2}$$ where $\,\{\zeta_{2n}^1,\zeta_{2n}^3,\dots,\zeta_{2n}^{2n-1}\}\,$ are the $2n$-th roots of unity which are not $n$-th roots of unity. Now $$ x+y = \prod_{k=0}^{n-1} (x^{1/n}-\zeta_{2n}^{2k+1}y^{1/n}). \tag{3}$$

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The trick here is that an expression containing powers of the variable which are not nonnegative integers is not a polynomial. Polynomials, by definition, at least in a single variable, are expressions of the form

$$a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0$$

with $a_n \ne 0$ (unless $n = 0$), and $n$ is some nonnegative integer, and called the polynomial's degree. Thus, given that the powers are only nonnegative integers from $0$ to $n$ and $n$ itself is such, no polynomial can contain any power that is not a nonnegative integer, which includes both negative powers and fractional powers. In particular,

$$\frac{1}{x}$$

is not a polynomial, nor is

$$x^{1/2}$$

and thus neither is the expression in terms of the "factors" you give. It is a valid "factorization" among a more general class of expressions than polynomials - but not the polynomials themselves. This is similar to how that a number like $3$ has no non-trivial factorization in terms of other integers, but if we move to more general forms of number like rational and real numbers, then it can admit such, and in fact infinitely many, factorizations.

One possible generalization of a regular polynomial, which still retains a lot of their structured nature, is what could be called a "Puiseux series" and its finite form, the Puiseux polynomial, in which in addition to the degree $n$ we also have a denominator $N$ in the exponent and define it as

$$a_n x^{n/N} + a_{n-1} x^{(n-1)/N} + \cdots + a_2 x^{2/N} + a_1 x^{1/N} + a_0$$

with same conditions for the coefficients, and likewise for more than one variable by adding suitable terms. In this case, $x + y$ is a Puiseux polynomial of two variables with $N = 1$, and it factors into Puiseux polynomials with $N = 2$.

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