Monomials of Polynomial Ring are Linearly Independent/One Variable
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Theorem
Let $R$ be a commutative ring with unity.
Let $R \sqbrk X$ be a polynomial ring in one variable $X$ over $R$.
Then the set of monomials $\set {X^k : k \in \N}$ is linearly independent.
Proof
We consider polynomial ring over sequences in $R$:
Let $\struct {R^{\left({\N}\right)}, \oplus, \odot}$ be the underlying ring of sequences of finite support.
Recall:
- $X = \sequence {0, 1, 0, 0, \ldots}$
Observe:
| \(\ds X^2\) | \(=\) | \(\ds \sequence {0, 0, 1, 0, 0, \ldots}\) | ||||||||||||
| \(\ds X^3\) | \(=\) | \(\ds \sequence {0, 0, 0, 1, 0, \ldots}\) | ||||||||||||
| \(\ds \) | \(\cdots\) | \(\ds \) | ||||||||||||
| \(\ds X^k\) | \(=\) | \(\ds \sequence {\underbrace {0, \ldots, 0}_k, 1, 0, \ldots}\) | ||||||||||||
| \(\ds \) | \(\cdots\) | \(\ds \) |
More generally, for all $r \in R$ and $k \in \N$:
| \(\ds r X^k\) | \(=\) | \(\ds \map \iota r \odot \sequence {\underbrace{0,\ldots,0}_k,1,0,\ldots}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sequence {r,0,\ldots} \odot \sequence {\underbrace{0,\ldots,0}_k,1,0,\ldots}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sequence {\underbrace{0,\ldots,0}_k,r,0,\ldots}\) |
Now let $n \in \N$ and $r_0,\ldots,r_n \in R$ be such that:
- $r_0 + r_1 X + r_2 X^2 + \cdots + r_n X^n = 0$
That is:
| \(\ds \) | \(\) | \(\ds \sequence {r_0,0,\ldots} \oplus \sequence {0,r_1,0,\ldots} \oplus \sequence {0,0,r_2,0,\ldots} \oplus \cdots \oplus \sequence {0,\ldots,0,r_n,0,\ldots}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sequence {r_0,r_1,r_2,\ldots,r_n,0,\ldots}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sequence {0,0,\ldots}\) |
Therefore:
- $r_0 = \cdots = r_n = 0$
$\blacksquare$