Polynomials and Irreducible polynomials
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So I'm trying to prove this Proposition:
Every polynomial of degree greater than $0$ is divisible by an irreducible polynomial.
Proof:
Consider a polynomial $P in mathbb R [x]$ of degree $geq 1$
Let $S$ denote the set of non-trivial factors of $P$.
Suppose that $S$ does not contain any irreducible polynomials.
By the well ordering principle let $P_0$ denote the polynomial of least degree in $S$
By our assumption $P_0$ is not irreducible $therefore P_0=P_1P_2$ for some
$P_1,P_2 in mathbb R[x]$
This contradicts the definition of $P_0$ being the polynomial of least degree to be a factor of $P$
Thus we are forced to conclude the polynomial of least degree is irreducible and the claim follows.
$blacksquare$
My main issue is can I use the well ordering principle in this fashion?
thanks
polynomials
asked Dec 26 '18 at 11:16
PolynomialCPolynomialC
826
$endgroup$
add a comment |
$begingroup$
So I'm trying to prove this Proposition:
Every polynomial of degree greater than $0$ is divisible by an irreducible polynomial.
Proof:
Consider a polynomial $P in mathbb R [x]$ of degree $geq 1$
Let $S$ denote the set of non-trivial factors of $P$.
Suppose that $S$ does not contain any irreducible polynomials.
By the well ordering principle let $P_0$ denote the polynomial of least degree in $S$
By our assumption $P_0$ is not irreducible $therefore P_0=P_1P_2$ for some
$P_1,P_2 in mathbb R[x]$
This contradicts the definition of $P_0$ being the polynomial of least degree to be a factor of $P$
Thus we are forced to conclude the polynomial of least degree is irreducible and the claim follows.
$blacksquare$
My main issue is can I use the well ordering principle in this fashion?
thanks
polynomials
asked Dec 26 '18 at 11:16
PolynomialCPolynomialC
826
$endgroup$
$begingroup$
$S$ must be the set of monic (non-trivial factors of $P$). With this, you can say that $P_1,P_2$ are monic (and have degree $geq1$)
$endgroup$
– Martín Vacas Vignolo
Dec 26 '18 at 11:26
$begingroup$
Why must the factors be monic?
$endgroup$
– PolynomialC
Dec 26 '18 at 11:28
$begingroup$
No. $S$ is a subset of set of all factors. The key is that in your proof you can not say that $deg P_i geq 1$
$endgroup$
– Martín Vacas Vignolo
Dec 26 '18 at 11:30
$begingroup$
Is this to make sure the set is of finite size and thus bounded below? Cheers for the input.
$endgroup$
– PolynomialC
Dec 26 '18 at 11:33
add a comment |
$begingroup$
So I'm trying to prove this Proposition:
Every polynomial of degree greater than $0$ is divisible by an irreducible polynomial.
Proof:
Consider a polynomial $P in mathbb R [x]$ of degree $geq 1$
Let $S$ denote the set of non-trivial factors of $P$.
Suppose that $S$ does not contain any irreducible polynomials.
By the well ordering principle let $P_0$ denote the polynomial of least degree in $S$
By our assumption $P_0$ is not irreducible $therefore P_0=P_1P_2$ for some
$P_1,P_2 in mathbb R[x]$
This contradicts the definition of $P_0$ being the polynomial of least degree to be a factor of $P$
Thus we are forced to conclude the polynomial of least degree is irreducible and the claim follows.
$blacksquare$
My main issue is can I use the well ordering principle in this fashion?
thanks
polynomials
asked Dec 26 '18 at 11:16
PolynomialCPolynomialC
826
$endgroup$
So I'm trying to prove this Proposition:
Every polynomial of degree greater than $0$ is divisible by an irreducible polynomial.
Proof:
Consider a polynomial $P in mathbb R [x]$ of degree $geq 1$
Let $S$ denote the set of non-trivial factors of $P$.
Suppose that $S$ does not contain any irreducible polynomials.
By the well ordering principle let $P_0$ denote the polynomial of least degree in $S$
By our assumption $P_0$ is not irreducible $therefore P_0=P_1P_2$ for some
$P_1,P_2 in mathbb R[x]$
This contradicts the definition of $P_0$ being the polynomial of least degree to be a factor of $P$
Thus we are forced to conclude the polynomial of least degree is irreducible and the claim follows.
$blacksquare$
My main issue is can I use the well ordering principle in this fashion?
thanks
polynomials
polynomials
asked Dec 26 '18 at 11:16
PolynomialCPolynomialC
826
asked Dec 26 '18 at 11:16
PolynomialCPolynomialC
826
edited Dec 26 '18 at 11:23
PolynomialC
asked Dec 26 '18 at 11:16
PolynomialCPolynomialC
826
asked Dec 26 '18 at 11:16
PolynomialCPolynomialC
826
asked Dec 26 '18 at 11:16
PolynomialCPolynomialC
826
826
$begingroup$
$S$ must be the set of monic (non-trivial factors of $P$). With this, you can say that $P_1,P_2$ are monic (and have degree $geq1$)
$endgroup$
– Martín Vacas Vignolo
Dec 26 '18 at 11:26
$begingroup$
Why must the factors be monic?
$endgroup$
– PolynomialC
Dec 26 '18 at 11:28
$begingroup$
No. $S$ is a subset of set of all factors. The key is that in your proof you can not say that $deg P_i geq 1$
$endgroup$
– Martín Vacas Vignolo
Dec 26 '18 at 11:30
$begingroup$
Is this to make sure the set is of finite size and thus bounded below? Cheers for the input.
$endgroup$
– PolynomialC
Dec 26 '18 at 11:33
add a comment |
$begingroup$
$S$ must be the set of monic (non-trivial factors of $P$). With this, you can say that $P_1,P_2$ are monic (and have degree $geq1$)
$endgroup$
– Martín Vacas Vignolo
Dec 26 '18 at 11:26
$begingroup$
Why must the factors be monic?
$endgroup$
– PolynomialC
Dec 26 '18 at 11:28
$begingroup$
No. $S$ is a subset of set of all factors. The key is that in your proof you can not say that $deg P_i geq 1$
$endgroup$
– Martín Vacas Vignolo
Dec 26 '18 at 11:30
$begingroup$
Is this to make sure the set is of finite size and thus bounded below? Cheers for the input.
$endgroup$
– PolynomialC
Dec 26 '18 at 11:33
$begingroup$
$S$ must be the set of monic (non-trivial factors of $P$). With this, you can say that $P_1,P_2$ are monic (and have degree $geq1$)
$endgroup$
– Martín Vacas Vignolo
Dec 26 '18 at 11:26
$begingroup$
$S$ must be the set of monic (non-trivial factors of $P$). With this, you can say that $P_1,P_2$ are monic (and have degree $geq1$)
$endgroup$
– Martín Vacas Vignolo
Dec 26 '18 at 11:26
$begingroup$
Why must the factors be monic?
$endgroup$
– PolynomialC
Dec 26 '18 at 11:28
$begingroup$
Why must the factors be monic?
$endgroup$
– PolynomialC
Dec 26 '18 at 11:28
$begingroup$
No. $S$ is a subset of set of all factors. The key is that in your proof you can not say that $deg P_i geq 1$
$endgroup$
– Martín Vacas Vignolo
Dec 26 '18 at 11:30
$begingroup$
No. $S$ is a subset of set of all factors. The key is that in your proof you can not say that $deg P_i geq 1$
$endgroup$
– Martín Vacas Vignolo
Dec 26 '18 at 11:30
$begingroup$
Is this to make sure the set is of finite size and thus bounded below? Cheers for the input.
$endgroup$
– PolynomialC
Dec 26 '18 at 11:33
$begingroup$
Is this to make sure the set is of finite size and thus bounded below? Cheers for the input.
$endgroup$
– PolynomialC
Dec 26 '18 at 11:33
add a comment |
1 Answer
1
active
oldest
votes
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If you want to be precise, you need to apply the well-ordering principle to a non-empty set of $mathbb N$.
For that, define $D = { n in mathbb N : text{there is a factor of $P$ of degree $n$} }$.
Then $D$ is not empty because $P$ is a factor of $P$.
Let $m = min D$. Then there is a polynomial $Q$ such that $Q$ is a factor of $P$ of degree $m$. The minimality of $m$ implies that $Q$ is irreducible.
answered Dec 26 '18 at 11:28
lhflhf
165k10171396
$endgroup$
add a comment |
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$begingroup$
If you want to be precise, you need to apply the well-ordering principle to a non-empty set of $mathbb N$.
For that, define $D = { n in mathbb N : text{there is a factor of $P$ of degree $n$} }$.
Then $D$ is not empty because $P$ is a factor of $P$.
Let $m = min D$. Then there is a polynomial $Q$ such that $Q$ is a factor of $P$ of degree $m$. The minimality of $m$ implies that $Q$ is irreducible.
answered Dec 26 '18 at 11:28
lhflhf
165k10171396
$endgroup$
add a comment |
$begingroup$
If you want to be precise, you need to apply the well-ordering principle to a non-empty set of $mathbb N$.
For that, define $D = { n in mathbb N : text{there is a factor of $P$ of degree $n$} }$.
Then $D$ is not empty because $P$ is a factor of $P$.
Let $m = min D$. Then there is a polynomial $Q$ such that $Q$ is a factor of $P$ of degree $m$. The minimality of $m$ implies that $Q$ is irreducible.
answered Dec 26 '18 at 11:28
lhflhf
165k10171396
$endgroup$
add a comment |
$begingroup$
If you want to be precise, you need to apply the well-ordering principle to a non-empty set of $mathbb N$.
For that, define $D = { n in mathbb N : text{there is a factor of $P$ of degree $n$} }$.
Then $D$ is not empty because $P$ is a factor of $P$.
Let $m = min D$. Then there is a polynomial $Q$ such that $Q$ is a factor of $P$ of degree $m$. The minimality of $m$ implies that $Q$ is irreducible.
answered Dec 26 '18 at 11:28
lhflhf
165k10171396
$endgroup$
If you want to be precise, you need to apply the well-ordering principle to a non-empty set of $mathbb N$.
For that, define $D = { n in mathbb N : text{there is a factor of $P$ of degree $n$} }$.
Then $D$ is not empty because $P$ is a factor of $P$.
Let $m = min D$. Then there is a polynomial $Q$ such that $Q$ is a factor of $P$ of degree $m$. The minimality of $m$ implies that $Q$ is irreducible.
answered Dec 26 '18 at 11:28
lhflhf
165k10171396
edited Dec 26 '18 at 11:46
answered Dec 26 '18 at 11:28
lhflhf
165k10171396
answered Dec 26 '18 at 11:28
lhflhf
165k10171396
answered Dec 26 '18 at 11:28
lhflhf
165k10171396
165k10171396
add a comment |
add a comment |
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$begingroup$
$S$ must be the set of monic (non-trivial factors of $P$). With this, you can say that $P_1,P_2$ are monic (and have degree $geq1$)
$endgroup$
– Martín Vacas Vignolo
Dec 26 '18 at 11:26
$begingroup$
Why must the factors be monic?
$endgroup$
– PolynomialC
Dec 26 '18 at 11:28
$begingroup$
No. $S$ is a subset of set of all factors. The key is that in your proof you can not say that $deg P_i geq 1$
$endgroup$
– Martín Vacas Vignolo
Dec 26 '18 at 11:30
$begingroup$
Is this to make sure the set is of finite size and thus bounded below? Cheers for the input.
$endgroup$
– PolynomialC
Dec 26 '18 at 11:33