# Math Help - Laurent polynomials

1. ## Laurent polynomials

Notations: Let $R=\mathbb{C}[x,x^{-1}]$ be the set of Laurent polynomials, i.e. every element of $R$ is in the form $a_mx^m+ \cdots + a_nx^n,$ where $m,n \in \mathbb{Z}, \ a_j \in \mathbb{C}.$ For any $y \in R,$ we can consider

$\mathbb{C}[y],$ which is just the set of ordinary polynomials in $y,$ i.e. every element of $\mathbb{C}[y]$ is in the form $b_0 + b_1y + \cdots + b_ny^n,$ where $n \geq 0$ and $b_j \in \mathbb{C}.$ Finally, by $y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y],$ where

$y,y_1, \cdots , y_n \in R,$ we obviously mean the set of all elements in the form $y_1z_1 + \cdots + y_nz_n,$ where $z_j \in \mathbb{C}[y].$ Clearly we have $y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y] \subseteq R.$

And here's the problem that anyone, even with no knowledge of abstract algebra, can attack it:

[ $\star \star \star$] True or False: There exist an inetger $n \geq 1$ and $y, y_1, \cdots , y_n \in R$ such that $R=y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$.

2. I'll give it a try.

Spoiler:

False.

Proof:
First, let us prove a short lemma:

For any $y_1, y_2 \in R$: $y_1 = y_2$ if and only if $y_1$ and $y_2$ have the same powers of of $y$ and same coefficients.
The proof is simple:
Let $a$ be the absolute value of the most negative power of $y$ in $y_1 + y_2$ ( $a=0$ if there are no negative powers). For example, when $y_1 = y^{-2}+1+y$ and $y_2 = 3y$ we have $a=2$. Thus $y^a y_1, y^a y_2 \in \mathbb{C}[y]$, and therefore the powers and coefficients of $y^a y_1, y^a y_2$ and the same (since we assumed $y_1 = y_2$ for all $y$). The result follows.

Back to the original problem:
Let $n \in \mathbb{N}$ and $y_1, y_2 \cdots , y_n \in R$. As before, let $a$ be the absolute value of the most negative power of $y$ in $y_1 + y_1 + \cdots + y_n$, and $a=0$ if no negative powers exist. Thus, the most negative power in $y_1 z_1 + \cdots + y_n z_n$ for any $z_j \in \mathbb{C}[y]$ is no lower than $-a$. Therefore, the most negative power of $y$ in the set $y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$ is $-a$. From the lemma above and from the fact that the powers of $y$ in Laurent polynomials are not bounded, we conclude that there exist a Laurent polynomial $b$ for which $b \not\in y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$, so $y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y] \subset R$ for any choose of $n \in \mathbb{N}$ and $y_j \in \mathbb{C}[y]$. In other words, it is false that there exist an inetger $n \geq 1$ and $y, y_1, \cdots , y_n \in R$ such that $R=y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$.

3. Originally Posted by Unbeatable0
I'll give it a try.

Spoiler:

False.

Proof:
First, let us prove a short lemma:

For any $y_1, y_2 \in R$: $y_1 = y_2$ if and only if $y_1$ and $y_2$ have the same powers of of $y$ and same coefficients.
The proof is simple:
Let $a$ be the absolute value of the most negative power of $y$ in $y_1 + y_2$ ( $a=0$ if there are no negative powers). For example, when $y_1 = y^{-2}+1+y$ and $y_2 = 3y$ we have $a=2$. Thus $y^a y_1, y^a y_2 \in \mathbb{C}[y]$, and therefore the powers and coefficients of $y^a y_1, y^a y_2$ and the same (since we assumed $y_1 = y_2$ for all $y$). The result follows.

Back to the original problem:
Let $n \in \mathbb{N}$ and $y_1, y_2 \cdots , y_n \in R$. As before, let $a$ be the absolute value of the most negative power of $y$ in $y_1 + y_1 + \cdots + y_n$, and $a=0$ if no negative powers exist. Thus, the most negative power in $y_1 z_1 + \cdots + y_n z_n$ for any $z_j \in \mathbb{C}[y]$ is no lower than $-a$. Therefore, the most negative power of $y$ in the set $y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$ is $-a$. From the lemma above and from the fact that the powers of $y$ in Laurent polynomials are not bounded, we conclude that there exist a Laurent polynomial $b$ for which $b \not\in y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$, so $y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y] \subset R$ for any choose of $n \in \mathbb{N}$ and $y_j \in \mathbb{C}[y]$. In other words, it is false that there exist an inetger $n \geq 1$ and $y, y_1, \cdots , y_n \in R$ such that $R=y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$.
i think you're confusing $y,$ which can be any element of $R=\mathbb{C}[x,x^{-1}],$ with $x.$ note that if the least power of $x$ in $y$ is negative, then the set of least powers of $x$ in $y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$

will be unbounded. it is obvious that, if the equality $R=y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$ holds, then $y=a_mx^m + a_{m+1}x^{m+1} + \cdots a_nx^n,$ where $m < 0, \ n > 0.$ but would this condition be enough?

4. Originally Posted by NonCommAlg
i think you're confusing $y,$ which can be any element of $R=\mathbb{C}[x,x^{-1}],$ with $x.$ note that if the least power of $x$ in $y$ is negative, then the set of least powers of $x$ in $y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$

will be unbounded. it is obvious that, if the equality $R=y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$ holds, then $y=a_mx^m + a_{m+1}x^{m+1} + \cdots a_nx^n,$ where $m < 0, \ n > 0.$ but would this condition be enough?
I see my mistake here. I indeed confused $x$ and $y$. I suspected something was wrong, because you posted it as a 3-stars question, and my proof was very simple and short, but I couldn't find any mistake in my proof.
Thanks for the correction! I may try again soon.