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Math Help - Laurent polynomials

  1. #1
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    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].
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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 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].
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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 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?
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  4. #4
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    Quote Originally Posted by NonCommAlg View Post
    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.
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