description/proof of that for polynomial over field with finite number of variables, if polynomial is constantly \(0\), coefficients are \(0\)
Topics
About: field
The table of contents of this article
Starting Context
- The reader knows a definition of polynomials ring over commutative ring.
- The reader admits the proposition that over any field, any n-degree polynomial has at most n roots.
Target Context
- The reader will have a description and a proof of the proposition that for any polynomial over any field (with more than the-polynomial-degree elements) with any finite number of variables, if the polynomial is constantly \(0\), the coefficients are \(0\).
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\) with more than \(m\) elements
\(p (x_1, ..., x_n)\): \(= \sum_{j_1 + ... + j_n = m} p_{j_1, ..., j_n} {x_1}^{j_1} ... {x_n}^{j_n} + ... + \sum_{j_1 + ... + j_n = 0} p_{j_1, ..., j_n} {x_1}^{j_1} ... {x_n}^{j_n}\) where \(j_l \in \mathbb{N}\), \(\in \{\text{ the polynomials over } F \text{ with variables, } x_1, ..., x_n\}\)
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Statements:
\(p (x_1, ..., x_n) \equiv 0\)
\(\implies\)
\(\forall l \in \{0, ..., m\} (\forall (j_1, ..., j_n) \text{ such that } j_1 + ... + j_n = l (p_{j_1, ..., j_n} = 0))\)
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2: Proof
Whole Strategy: Step 1: regard \(p (x_1, ..., x_n)\) as the polynomial of \(x_1\) with each fixed \((x_2, ..., x_n)\), and see that each coefficient is \(0\); Step 2: regard each coefficient of Step 1 as the polynomial of \(x_2\) with each fixed \((x_3, ..., x_n)\), and see that each coefficient is \(0\); Step 3: and so on.
Step 1:
\(p (x_1, ..., x_n) = p_m {x_1}^m + ... + p_0 {x_1}^0\), where \(p_j\) is a polynomial of \(x_2, ..., x_n\): \(p_m := p_{m, 0, ..., 0} {x_2}^0 ... {x_n}^0\), \(p_{m - 1} := \sum_{m - 1 + j_2 + ... + j_n = m} p_{m - 1, j_2, ..., j_n} {x_2}^{j_2} ... {x_n}^{j_n} + p_{m - 1, 0, ..., 0} {x_2}^0 ... {x_n}^0\), ....
Let \((x_2, ..., x_n)\) be fixed at any value.
Then, \(p (x_1) := p (x_1, ..., x_n)\) is a polynomial of \(x_1\) over \(F\).
If \(l\) was the largest such that \(p_l \neq 0\), \(p (x_1)\) would have at most \(l\) roots, by the proposition that over any field, any n-degree polynomial has at most n roots, a contradiction against that \(p (x_1, ..., x_n) \equiv 0\), which mean that \(p (x_1)\) has the whole \(F\) as the roots (\(F\) has more than \(m\) elements).
So, all the coefficients, \(p_j\), s are \(0\) with each fixed \((x_2, ..., x_n)\), so, \(p_j \equiv 0\).
Step 2:
Each \(p_{j_1} (x_2, ..., x_n)\) is a polynomial of \((x_2, ..., x_n)\) constantly \(0\).
\(p_{j_1} (x_2, ..., x_n) = p_{j_1, m - j_1} {x_2}^{m - j_1} + ... + p_{j_1, 0} {x_2}^0\), where \(p_{j_1, j}\) is a polynomial of \(x_3, ..., x_n\): \(p_{j_1, m - j_1} := p_{j_1, m - j_1, 0, ..., 0} {x_3}^0 ... {x_n}^0\), \(p_{j_1, m - j_1 - 1} := \sum_{j_1 + (m - j_1 - 1) + j_3 + ... + j_n = m} p_{j_1, m - j_1 - 1, j_3, ..., j_n} {x_3}^{j_3} ... {x_n}^{j_n} + p_{j_1, m - j_1 - 1, 0, ..., 0} {x_3}^0 ... {x_n}^0\), ....
Let \((x_3, ..., x_n)\) be fixed at any value.
Then, \(p_{j_1} (x_2) := p_{j_1} (x_1, ..., x_n)\) is a polynomial of \(x_2\) over \(F\).
If \(l\) was the largest such that \(p_{j_1, l} \neq 0\), \(p_{j_1} (x_2)\) would have at most \(l\) roots, as before, a contradiction against that \(p_{j_1} (x_2, ..., x_n) \equiv 0\), which mean that \(p_{j_1} (x_2)\) has the whole \(F\) as the roots.
So, all the coefficients, \(p_{j_1, l}\), s are \(0\) with each fixed \((x_3, ..., x_n)\), so, \(p_{j_1, l} \equiv 0\).
Step 3:
And so on, after all, we think of the polynomials of \(x_n\), whose coefficients are constants, and the coefficients are \(0\).
But each \(p_{j_1, ..., j_n}\) is one of those constant coefficients, because \(p_{j_1, ..., j_n} {x_1}^{j_1} ... {x_n}^{j_n}\) is put into \(p_{j_1}\) as \(p_{j_1, ..., j_n} {x_2}^{j_2} ... {x_n}^{j_n}\), which is put into \(p_{j_1, j_2}\) as \(p_{j_1, ..., j_n} {x_3}^{j_3} ... {x_n}^{j_n}\), ..., which becomes \(p_{j_1, ..., j_n}\) as a constant coefficient of \(p_{j_1, ..., j_{n - 1}}\).
So, \(p_{j_1, ..., j_n} = 0\).
