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Algebraic proof that two statements of the fundamental theorem of algebra are equivalent

A question on the Fundamental Theorem of AlgebraFundamental Theorem of AlgebraFundamental theorem of algebra simple proof for rewriting with rootsIs the fundamental theorem of algebra valid with polynomial terms like $bar{z}$ and $Re (z)$?Degenerate zeroes, fundamental theorem of algebra.Is this an equivalent statement to the Fundamental Theorem of Algebra?Derivation of the Fundamental Theorem of AlgebraWhy fundamental theorem of algebra is stated less intuitively in college math?Fundamental theorem of algebra - connecting equivalent formulationsEvery complex polynomial is a product of first degree polynomials, alternative proofs?

3

1

$begingroup$

Students studying the fundamental theorem of algebra in high school are probably familiar with the statement that goes something like the following.

Every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ complex roots.

However, another statement I often see (and comes first in the Wikipedia article) is the following.

The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root.

These statements are equivalent and Wikipedia says that this can be proven by successive polynomial long division. However, Wikipedia does not give this proof. It is absent from Wolfram MathWorld as well.

My question is, what is the proof that these statements are equivalent? I know that the fundamental theorem of algebra cannot be proven algebraically, but can this equivalence be proven with only algebra, i.e., polynomial division?

real-analysis complex-analysis algebra-precalculus polynomials proof-explanation

share|cite|improve this question

edited 27 mins ago

Calum Gilhooley

4,577629

asked 3 hours ago

GnumbertesterGnumbertester

582113

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I prefer the second statement, which is really the bulk of the question.
  $endgroup$
  – egreg
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The first statement is a corollary of the second. The second is also simpler to grasp: "If $p$ is a polynomial, then $p$ has at least one root."
  $endgroup$
  – Bernard Massé
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The real-analysis tag seems inappropriate, but I'm not confident enough of this judgement to delete it. I've added a couple more tags that do seem appropriate.
  $endgroup$
  – Calum Gilhooley
  25 mins ago
  

  
  
  
  

add a comment | 

3

1

$begingroup$

Students studying the fundamental theorem of algebra in high school are probably familiar with the statement that goes something like the following.

Every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ complex roots.

However, another statement I often see (and comes first in the Wikipedia article) is the following.

The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root.

These statements are equivalent and Wikipedia says that this can be proven by successive polynomial long division. However, Wikipedia does not give this proof. It is absent from Wolfram MathWorld as well.

My question is, what is the proof that these statements are equivalent? I know that the fundamental theorem of algebra cannot be proven algebraically, but can this equivalence be proven with only algebra, i.e., polynomial division?

real-analysis complex-analysis algebra-precalculus polynomials proof-explanation

share|cite|improve this question

edited 27 mins ago

Calum Gilhooley

4,577629

asked 3 hours ago

GnumbertesterGnumbertester

582113

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I prefer the second statement, which is really the bulk of the question.
  $endgroup$
  – egreg
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The first statement is a corollary of the second. The second is also simpler to grasp: "If $p$ is a polynomial, then $p$ has at least one root."
  $endgroup$
  – Bernard Massé
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The real-analysis tag seems inappropriate, but I'm not confident enough of this judgement to delete it. I've added a couple more tags that do seem appropriate.
  $endgroup$
  – Calum Gilhooley
  25 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Students studying the fundamental theorem of algebra in high school are probably familiar with the statement that goes something like the following.

Every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ complex roots.

However, another statement I often see (and comes first in the Wikipedia article) is the following.

The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root.

These statements are equivalent and Wikipedia says that this can be proven by successive polynomial long division. However, Wikipedia does not give this proof. It is absent from Wolfram MathWorld as well.

My question is, what is the proof that these statements are equivalent? I know that the fundamental theorem of algebra cannot be proven algebraically, but can this equivalence be proven with only algebra, i.e., polynomial division?

real-analysis complex-analysis algebra-precalculus polynomials proof-explanation

share|cite|improve this question

edited 27 mins ago

Calum Gilhooley

4,577629

asked 3 hours ago

GnumbertesterGnumbertester

582113

$endgroup$

share|cite|improve this question

edited 27 mins ago

Calum Gilhooley

4,577629

asked 3 hours ago

GnumbertesterGnumbertester

582113

share|cite|improve this question

edited 27 mins ago

Calum Gilhooley

4,577629

asked 3 hours ago

GnumbertesterGnumbertester

582113

edited 27 mins ago

Calum Gilhooley

4,577629

edited 27 mins ago

Calum Gilhooley

4,577629

asked 3 hours ago

GnumbertesterGnumbertester

582113

asked 3 hours ago

GnumbertesterGnumbertester

582113

1 Answer
1

active

oldest

votes

6

$begingroup$

Induction (on the degree of the polynomial) suffices.

As it's clear that the first implies the second, we need only argue that the second implies the first.

This is clear for degree $1$.

Inductively suppose it for degree $n-1$.

Let $P(x)$ have degree $n$. By the second definition it has at least one root, $alpha$. Then, by standard polynomial division we may write $P(x)=(x-alpha)times Q(x)$ where $Q(x)$ has degree $n-1$. Applying the inductive hypothesis to $Q(x)$ shows that the second definition implies the first.

share|cite|improve this answer

answered 2 hours ago

lulululu

42.3k25080

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The proof is incomplete. You need to prove that it has exactly $n$ roots.
  $endgroup$
  – Bill Dubuque
  58 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For example, over the ring $,Bbb Z/8 = $ integers $bmod 8,$ both $x-1$ and $x+1$ have exacty $1$ root, but their product $,x^2-1,$ has $4$ roots $,pm1,pm3. $
  $endgroup$
  – Bill Dubuque
  12 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BillDubuque but over $mathbb{Z}/8$ we have things like $x^2+1$ has no roots, 'violating' the second statement of the fundamental theorem of $mathbb{Z}/8$-algebra, so that's not a counterexample (because you can't prove the first statement using the second statement if the second statement isn't true). (I do agree with your initial comment.)
  $endgroup$
  – boboquack
  8 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @boboquack My point was to show how additivity of number of roots fails in general rings, so it requires proof in this special case. Your remark has nothing to do with that so it may confuse readers.
  $endgroup$
  – Bill Dubuque
  5 mins ago
  
  
  

  
  
  
  

add a comment | 

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6

$begingroup$

Induction (on the degree of the polynomial) suffices.

As it's clear that the first implies the second, we need only argue that the second implies the first.

This is clear for degree $1$.

Inductively suppose it for degree $n-1$.

Let $P(x)$ have degree $n$. By the second definition it has at least one root, $alpha$. Then, by standard polynomial division we may write $P(x)=(x-alpha)times Q(x)$ where $Q(x)$ has degree $n-1$. Applying the inductive hypothesis to $Q(x)$ shows that the second definition implies the first.

share|cite|improve this answer

answered 2 hours ago

lulululu

42.3k25080

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The proof is incomplete. You need to prove that it has exactly $n$ roots.
  $endgroup$
  – Bill Dubuque
  58 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For example, over the ring $,Bbb Z/8 = $ integers $bmod 8,$ both $x-1$ and $x+1$ have exacty $1$ root, but their product $,x^2-1,$ has $4$ roots $,pm1,pm3. $
  $endgroup$
  – Bill Dubuque
  12 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BillDubuque but over $mathbb{Z}/8$ we have things like $x^2+1$ has no roots, 'violating' the second statement of the fundamental theorem of $mathbb{Z}/8$-algebra, so that's not a counterexample (because you can't prove the first statement using the second statement if the second statement isn't true). (I do agree with your initial comment.)
  $endgroup$
  – boboquack
  8 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @boboquack My point was to show how additivity of number of roots fails in general rings, so it requires proof in this special case. Your remark has nothing to do with that so it may confuse readers.
  $endgroup$
  – Bill Dubuque
  5 mins ago
  
  
  

  
  
  
  

add a comment | 

6

$begingroup$

Induction (on the degree of the polynomial) suffices.

As it's clear that the first implies the second, we need only argue that the second implies the first.

This is clear for degree $1$.

Inductively suppose it for degree $n-1$.

Let $P(x)$ have degree $n$. By the second definition it has at least one root, $alpha$. Then, by standard polynomial division we may write $P(x)=(x-alpha)times Q(x)$ where $Q(x)$ has degree $n-1$. Applying the inductive hypothesis to $Q(x)$ shows that the second definition implies the first.

share|cite|improve this answer

answered 2 hours ago

lulululu

42.3k25080

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The proof is incomplete. You need to prove that it has exactly $n$ roots.
  $endgroup$
  – Bill Dubuque
  58 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For example, over the ring $,Bbb Z/8 = $ integers $bmod 8,$ both $x-1$ and $x+1$ have exacty $1$ root, but their product $,x^2-1,$ has $4$ roots $,pm1,pm3. $
  $endgroup$
  – Bill Dubuque
  12 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BillDubuque but over $mathbb{Z}/8$ we have things like $x^2+1$ has no roots, 'violating' the second statement of the fundamental theorem of $mathbb{Z}/8$-algebra, so that's not a counterexample (because you can't prove the first statement using the second statement if the second statement isn't true). (I do agree with your initial comment.)
  $endgroup$
  – boboquack
  8 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @boboquack My point was to show how additivity of number of roots fails in general rings, so it requires proof in this special case. Your remark has nothing to do with that so it may confuse readers.
  $endgroup$
  – Bill Dubuque
  5 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Induction (on the degree of the polynomial) suffices.

As it's clear that the first implies the second, we need only argue that the second implies the first.

This is clear for degree $1$.

Inductively suppose it for degree $n-1$.

Let $P(x)$ have degree $n$. By the second definition it has at least one root, $alpha$. Then, by standard polynomial division we may write $P(x)=(x-alpha)times Q(x)$ where $Q(x)$ has degree $n-1$. Applying the inductive hypothesis to $Q(x)$ shows that the second definition implies the first.

share|cite|improve this answer

answered 2 hours ago

lulululu

42.3k25080

$endgroup$

share|cite|improve this answer

answered 2 hours ago

lulululu

42.3k25080

answered 2 hours ago

lulululu

42.3k25080

answered 2 hours ago

lulululu

42.3k25080