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Thread: Laurent polynomials

  1. #1
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    Laurent polynomials

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

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

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


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


    [$\displaystyle \star \star \star$] True or False: There exist an inetger $\displaystyle n \geq 1$ and $\displaystyle y, y_1, \cdots , y_n \in R$ such that $\displaystyle R=y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$.
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  2. #2
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    I'll give it a try.

    Spoiler:

    False.

    Proof:
    First, let us prove a short lemma:

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

    Back to the original problem:
    Let $\displaystyle n \in \mathbb{N}$ and $\displaystyle y_1, y_2 \cdots , y_n \in R$. As before, let $\displaystyle a$ be the absolute value of the most negative power of $\displaystyle y$ in $\displaystyle y_1 + y_1 + \cdots + y_n$, and $\displaystyle a=0$ if no negative powers exist. Thus, the most negative power in $\displaystyle y_1 z_1 + \cdots + y_n z_n$ for any $\displaystyle z_j \in \mathbb{C}[y]$ is no lower than $\displaystyle -a$. Therefore, the most negative power of $\displaystyle y$ in the set $\displaystyle y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$ is $\displaystyle -a$. From the lemma above and from the fact that the powers of $\displaystyle y$ in Laurent polynomials are not bounded, we conclude that there exist a Laurent polynomial $\displaystyle b$ for which $\displaystyle b \not\in y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$, so $\displaystyle y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y] \subset R$ for any choose of $\displaystyle n \in \mathbb{N}$ and $\displaystyle y_j \in \mathbb{C}[y]$. In other words, it is false that there exist an inetger $\displaystyle n \geq 1$ and $\displaystyle y, y_1, \cdots , y_n \in R$ such that $\displaystyle R=y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$.
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  3. #3
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    Quote Originally Posted by Unbeatable0 View Post
    I'll give it a try.

    Spoiler:

    False.

    Proof:
    First, let us prove a short lemma:

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

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

    will be unbounded. it is obvious that, if the equality $\displaystyle R=y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$ holds, then $\displaystyle y=a_mx^m + a_{m+1}x^{m+1} + \cdots a_nx^n,$ where $\displaystyle m < 0, \ n > 0.$ but would this condition be enough?
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  4. #4
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    Quote Originally Posted by NonCommAlg View Post
    i think you're confusing $\displaystyle y,$ which can be any element of $\displaystyle R=\mathbb{C}[x,x^{-1}],$ with $\displaystyle x.$ note that if the least power of $\displaystyle x$ in $\displaystyle y$ is negative, then the set of least powers of $\displaystyle x$ in $\displaystyle y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$

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