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Let $S$ be a set of non-zero polynomials over a field $F$. If no two elements of $S$ have the same degree, show that $S$ is an independent subset of $F[x]$.

I tried to prove the statement by induction on the number of elements of any finite subset of $S$. Please let me know if my proof is correct or not. Besides, is there other method to prove the statement?

Thanks!

Let n be the number of elements in a finite subset of S.

Base case $n = 1$

$\alpha f = 0 \implies \alpha = 0$ because $f \ne 0$.

Assume the statement is true for $n=k$.

When $n = k+1$, suppose $$\sigma= \alpha_1 f_1+....+\alpha_k f_k+ \alpha_{k+1} f_{k+1}=0$$ WLOG, we may assume that $f_{k+1}$ has the highest degree among all $f_i${i=1,2,...,k+1}. Let $m= \deg f_{k+1}$

Denote the leading coefficient of $f_{k+1}$ by $\lambda$.

The coefficient of $x^{m}$ will be zero for all other $f_i$.

Note that the coefficient of $x^{m}$ in $\sigma$ will then be $$\alpha_{k+1}\lambda=0$$ By the definition of leading coefficient, $$\lambda \ne 0$$. So $$\alpha_{k+1}=0$$ This implies $$\alpha_1 f_1+....+\alpha_k f_k=0$$ By induction assumption, $$\alpha_1 = \alpha_2=...=\alpha_{k}=0$$ Hence, $f_1,f_2,....,f_{k+1}$ are linearly independent. By the principle of M.I., the statement is true for all natural numbers n.

Best regards, Michael

luimichael
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1 Answers1

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Here is my take.

Order the polynomials in $S$ by decreasing degree.

Write their coefficients in a matrix, from higher degree to lower degree.

The resulting matrix is in row echelon form and has no zero rows, and so has full rank.

Therefore, $S$ is linearly independent.

For example, if $S = \{1+x^3+x^4, x^2+x^6, x+x^7 \}$, then the matrix is $$ \pmatrix{ 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 \\ } $$

lhf
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  • Sorry, I can't follow your procedure. Take for example, if $f_1=1+x^3+x^4$, $f_2=x^2+x^6$ and $f_3=x+x^7$, then what will be the matrix? – luimichael Jul 18 '17 at 17:45
  • ,Oh! I see, you mean row echelon form.Very nice method. When I used induction to prove the statement I felt that I din't understand the reason why these elements are linearly independent. Now I totally understand the reasoning behind. Thanks a lot! – luimichael Jul 18 '17 at 18:07
  • that's probably the best way to approach the problem (+1)! – G Cab Jul 18 '17 at 18:26