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












3












$begingroup$


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











$endgroup$








  • 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
    42 mins ago
















3












$begingroup$


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











$endgroup$








  • 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
    42 mins ago














3












3








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











$endgroup$




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




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 44 mins ago









Calum Gilhooley

4,577629




4,577629










asked 3 hours ago









GnumbertesterGnumbertester

582113




582113








  • 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
    42 mins ago














  • 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
    42 mins ago








1




1




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




$begingroup$
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 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
42 mins 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
42 mins ago










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









$endgroup$













  • $begingroup$
    The proof is incomplete. You need to prove that it has exactly $n$ roots.
    $endgroup$
    – Bill Dubuque
    1 hour 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
    29 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
    26 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
    22 mins ago













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1 Answer
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1 Answer
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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









$endgroup$













  • $begingroup$
    The proof is incomplete. You need to prove that it has exactly $n$ roots.
    $endgroup$
    – Bill Dubuque
    1 hour 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
    29 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
    26 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
    22 mins ago


















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









$endgroup$













  • $begingroup$
    The proof is incomplete. You need to prove that it has exactly $n$ roots.
    $endgroup$
    – Bill Dubuque
    1 hour 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
    29 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
    26 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
    22 mins ago
















6












6








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









$endgroup$



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












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









lulululu

42.3k25080




42.3k25080












  • $begingroup$
    The proof is incomplete. You need to prove that it has exactly $n$ roots.
    $endgroup$
    – Bill Dubuque
    1 hour 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
    29 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
    26 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
    22 mins ago




















  • $begingroup$
    The proof is incomplete. You need to prove that it has exactly $n$ roots.
    $endgroup$
    – Bill Dubuque
    1 hour 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
    29 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
    26 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
    22 mins ago


















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






$begingroup$
The proof is incomplete. You need to prove that it has exactly $n$ roots.
$endgroup$
– Bill Dubuque
1 hour 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
29 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
29 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
26 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
26 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
22 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
22 mins ago




















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