Factorization of a polynomial into linear and quadratic factors
$begingroup$
It is known that every non-constant polynomial with real coefficients admits a factorization in terms of real and quadratic factors. The proof normally uses the Fundamental Theorem of Algebra. Is there an elementary proof of the above which does not involve complex numbers at all?
polynomials factoring alternative-proof
edited Dec 3 '18 at 14:20
José Carlos Santos
154k22124227
asked Feb 16 '18 at 17:56
VpdVpd
877
$endgroup$
add a comment |
$begingroup$
It is known that every non-constant polynomial with real coefficients admits a factorization in terms of real and quadratic factors. The proof normally uses the Fundamental Theorem of Algebra. Is there an elementary proof of the above which does not involve complex numbers at all?
polynomials factoring alternative-proof
edited Dec 3 '18 at 14:20
José Carlos Santos
154k22124227
asked Feb 16 '18 at 17:56
VpdVpd
877
$endgroup$
add a comment |
$begingroup$
It is known that every non-constant polynomial with real coefficients admits a factorization in terms of real and quadratic factors. The proof normally uses the Fundamental Theorem of Algebra. Is there an elementary proof of the above which does not involve complex numbers at all?
polynomials factoring alternative-proof
edited Dec 3 '18 at 14:20
José Carlos Santos
154k22124227
asked Feb 16 '18 at 17:56
VpdVpd
877
$endgroup$
It is known that every non-constant polynomial with real coefficients admits a factorization in terms of real and quadratic factors. The proof normally uses the Fundamental Theorem of Algebra. Is there an elementary proof of the above which does not involve complex numbers at all?
polynomials factoring alternative-proof
polynomials factoring alternative-proof
edited Dec 3 '18 at 14:20
José Carlos Santos
154k22124227
asked Feb 16 '18 at 17:56
VpdVpd
877
edited Dec 3 '18 at 14:20
José Carlos Santos
154k22124227
asked Feb 16 '18 at 17:56
VpdVpd
877
edited Dec 3 '18 at 14:20
José Carlos Santos
154k22124227
edited Dec 3 '18 at 14:20
José Carlos Santos
154k22124227
edited Dec 3 '18 at 14:20
José Carlos Santos
154k22124227
154k22124227
asked Feb 16 '18 at 17:56
VpdVpd
877
asked Feb 16 '18 at 17:56
VpdVpd
877
asked Feb 16 '18 at 17:56
VpdVpd
877
877
add a comment |
add a comment |
1 Answer
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oldest
votes
$begingroup$
I published such a proof (see my article Another Proof of the Fundamental Theorem of Algebra, American Mathematical Monthly 112 (1), 2005, pp. 76–78), although I doubt that you'll find it elementary.
So, let $p(x)inmathbb{R}[x]$ be an irreducible polynomial. If I prove that its degree is $1$ or $2$, I will have proved that every polynomial in $mathbb{R}[x]$ can be written as a product of linear and quadratic polynomials. Let $n=deg p(x)$ and assume that $n>1$. The idea is to prove that $n=2$. Note that $mathbb{R}[x]/bigllangle p(x)bigrrangle$ is a field which is an extension of $mathbb R$ and whose dimension as a $mathbb R$-vector space is $n$. So, all that is needed is to prove that if $K$ is such an extension of $mathbb R$, then $n=2$.
In order to prove that, I proved that there is a norm $|cdot|$ on $K$ such that$$(forall x,yin K):|x.y|leqslant|x|.|y|.$$This allows us to define the exponential function$$begin{array}{rccc}expcolon&K&longrightarrow&K\&x&mapsto&displaystylesum_{n=0}^inftyfrac{x^n}{n!}.end{array}$$It turns out that $(forall x,yin K):exp(x+y)=exp(x)exp(y)$. That is, $exp$ is a group homomorphism from $(K,+)$ into $(Ksetminus{0},.)$. It can be proved that $exp(K)$ is both an open and a closed set of $Ksetminus{0}$. Since $n>1$, $Ksetminus{0}$ is connected and therefore $exp$ is surjective.
It can also be proved that $kerexp$ is a discrete subgroup of $(K,+)$. This means that either $kerexp={0}$ or that there are $k$ linearly independent vectors $v_1,ldots,v_kin K$ such that $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. It is not hard to prove that $exp$ induces a homeomorphism between $K/kerexp$ and $Ksetminus{0}$. But then there are two possibilites:
- $kerexp={0}$: this is impossible, because $mathbb{R}^n$ and $mathbb{R}^nsetminus{0}$ are not homeomorphic.
- $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. Then $K/kerexp$ is homeomorphic to $(S^1)^ktimesmathbb{R}^{n-k}$, which is not simply connected. But $mathbb{R}^nsetminus{0}$ is simply connected when $n>2$. Therefore, $n=2$ and this completes my (not that much elementary) proof.
answered Feb 18 '18 at 12:21
José Carlos SantosJosé Carlos Santos
154k22124227
$endgroup$
$begingroup$
Well done for actually publishing a proof. I want to publish something in the following years, whether it be a proof, theorem, conjecture, or improvement on a certain discovery (like reducing a well-known bound or something), etc.
$endgroup$
– user477343
Feb 18 '18 at 12:24
add a comment |
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$begingroup$
I published such a proof (see my article Another Proof of the Fundamental Theorem of Algebra, American Mathematical Monthly 112 (1), 2005, pp. 76–78), although I doubt that you'll find it elementary.
So, let $p(x)inmathbb{R}[x]$ be an irreducible polynomial. If I prove that its degree is $1$ or $2$, I will have proved that every polynomial in $mathbb{R}[x]$ can be written as a product of linear and quadratic polynomials. Let $n=deg p(x)$ and assume that $n>1$. The idea is to prove that $n=2$. Note that $mathbb{R}[x]/bigllangle p(x)bigrrangle$ is a field which is an extension of $mathbb R$ and whose dimension as a $mathbb R$-vector space is $n$. So, all that is needed is to prove that if $K$ is such an extension of $mathbb R$, then $n=2$.
In order to prove that, I proved that there is a norm $|cdot|$ on $K$ such that$$(forall x,yin K):|x.y|leqslant|x|.|y|.$$This allows us to define the exponential function$$begin{array}{rccc}expcolon&K&longrightarrow&K\&x&mapsto&displaystylesum_{n=0}^inftyfrac{x^n}{n!}.end{array}$$It turns out that $(forall x,yin K):exp(x+y)=exp(x)exp(y)$. That is, $exp$ is a group homomorphism from $(K,+)$ into $(Ksetminus{0},.)$. It can be proved that $exp(K)$ is both an open and a closed set of $Ksetminus{0}$. Since $n>1$, $Ksetminus{0}$ is connected and therefore $exp$ is surjective.
It can also be proved that $kerexp$ is a discrete subgroup of $(K,+)$. This means that either $kerexp={0}$ or that there are $k$ linearly independent vectors $v_1,ldots,v_kin K$ such that $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. It is not hard to prove that $exp$ induces a homeomorphism between $K/kerexp$ and $Ksetminus{0}$. But then there are two possibilites:
- $kerexp={0}$: this is impossible, because $mathbb{R}^n$ and $mathbb{R}^nsetminus{0}$ are not homeomorphic.
- $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. Then $K/kerexp$ is homeomorphic to $(S^1)^ktimesmathbb{R}^{n-k}$, which is not simply connected. But $mathbb{R}^nsetminus{0}$ is simply connected when $n>2$. Therefore, $n=2$ and this completes my (not that much elementary) proof.
answered Feb 18 '18 at 12:21
José Carlos SantosJosé Carlos Santos
154k22124227
$endgroup$
$begingroup$
Well done for actually publishing a proof. I want to publish something in the following years, whether it be a proof, theorem, conjecture, or improvement on a certain discovery (like reducing a well-known bound or something), etc.
$endgroup$
– user477343
Feb 18 '18 at 12:24
add a comment |
$begingroup$
I published such a proof (see my article Another Proof of the Fundamental Theorem of Algebra, American Mathematical Monthly 112 (1), 2005, pp. 76–78), although I doubt that you'll find it elementary.
So, let $p(x)inmathbb{R}[x]$ be an irreducible polynomial. If I prove that its degree is $1$ or $2$, I will have proved that every polynomial in $mathbb{R}[x]$ can be written as a product of linear and quadratic polynomials. Let $n=deg p(x)$ and assume that $n>1$. The idea is to prove that $n=2$. Note that $mathbb{R}[x]/bigllangle p(x)bigrrangle$ is a field which is an extension of $mathbb R$ and whose dimension as a $mathbb R$-vector space is $n$. So, all that is needed is to prove that if $K$ is such an extension of $mathbb R$, then $n=2$.
In order to prove that, I proved that there is a norm $|cdot|$ on $K$ such that$$(forall x,yin K):|x.y|leqslant|x|.|y|.$$This allows us to define the exponential function$$begin{array}{rccc}expcolon&K&longrightarrow&K\&x&mapsto&displaystylesum_{n=0}^inftyfrac{x^n}{n!}.end{array}$$It turns out that $(forall x,yin K):exp(x+y)=exp(x)exp(y)$. That is, $exp$ is a group homomorphism from $(K,+)$ into $(Ksetminus{0},.)$. It can be proved that $exp(K)$ is both an open and a closed set of $Ksetminus{0}$. Since $n>1$, $Ksetminus{0}$ is connected and therefore $exp$ is surjective.
It can also be proved that $kerexp$ is a discrete subgroup of $(K,+)$. This means that either $kerexp={0}$ or that there are $k$ linearly independent vectors $v_1,ldots,v_kin K$ such that $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. It is not hard to prove that $exp$ induces a homeomorphism between $K/kerexp$ and $Ksetminus{0}$. But then there are two possibilites:
- $kerexp={0}$: this is impossible, because $mathbb{R}^n$ and $mathbb{R}^nsetminus{0}$ are not homeomorphic.
- $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. Then $K/kerexp$ is homeomorphic to $(S^1)^ktimesmathbb{R}^{n-k}$, which is not simply connected. But $mathbb{R}^nsetminus{0}$ is simply connected when $n>2$. Therefore, $n=2$ and this completes my (not that much elementary) proof.
answered Feb 18 '18 at 12:21
José Carlos SantosJosé Carlos Santos
154k22124227
$endgroup$
$begingroup$
Well done for actually publishing a proof. I want to publish something in the following years, whether it be a proof, theorem, conjecture, or improvement on a certain discovery (like reducing a well-known bound or something), etc.
$endgroup$
– user477343
Feb 18 '18 at 12:24
add a comment |
$begingroup$
I published such a proof (see my article Another Proof of the Fundamental Theorem of Algebra, American Mathematical Monthly 112 (1), 2005, pp. 76–78), although I doubt that you'll find it elementary.
So, let $p(x)inmathbb{R}[x]$ be an irreducible polynomial. If I prove that its degree is $1$ or $2$, I will have proved that every polynomial in $mathbb{R}[x]$ can be written as a product of linear and quadratic polynomials. Let $n=deg p(x)$ and assume that $n>1$. The idea is to prove that $n=2$. Note that $mathbb{R}[x]/bigllangle p(x)bigrrangle$ is a field which is an extension of $mathbb R$ and whose dimension as a $mathbb R$-vector space is $n$. So, all that is needed is to prove that if $K$ is such an extension of $mathbb R$, then $n=2$.
In order to prove that, I proved that there is a norm $|cdot|$ on $K$ such that$$(forall x,yin K):|x.y|leqslant|x|.|y|.$$This allows us to define the exponential function$$begin{array}{rccc}expcolon&K&longrightarrow&K\&x&mapsto&displaystylesum_{n=0}^inftyfrac{x^n}{n!}.end{array}$$It turns out that $(forall x,yin K):exp(x+y)=exp(x)exp(y)$. That is, $exp$ is a group homomorphism from $(K,+)$ into $(Ksetminus{0},.)$. It can be proved that $exp(K)$ is both an open and a closed set of $Ksetminus{0}$. Since $n>1$, $Ksetminus{0}$ is connected and therefore $exp$ is surjective.
It can also be proved that $kerexp$ is a discrete subgroup of $(K,+)$. This means that either $kerexp={0}$ or that there are $k$ linearly independent vectors $v_1,ldots,v_kin K$ such that $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. It is not hard to prove that $exp$ induces a homeomorphism between $K/kerexp$ and $Ksetminus{0}$. But then there are two possibilites:
- $kerexp={0}$: this is impossible, because $mathbb{R}^n$ and $mathbb{R}^nsetminus{0}$ are not homeomorphic.
- $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. Then $K/kerexp$ is homeomorphic to $(S^1)^ktimesmathbb{R}^{n-k}$, which is not simply connected. But $mathbb{R}^nsetminus{0}$ is simply connected when $n>2$. Therefore, $n=2$ and this completes my (not that much elementary) proof.
answered Feb 18 '18 at 12:21
José Carlos SantosJosé Carlos Santos
154k22124227
$endgroup$
I published such a proof (see my article Another Proof of the Fundamental Theorem of Algebra, American Mathematical Monthly 112 (1), 2005, pp. 76–78), although I doubt that you'll find it elementary.
So, let $p(x)inmathbb{R}[x]$ be an irreducible polynomial. If I prove that its degree is $1$ or $2$, I will have proved that every polynomial in $mathbb{R}[x]$ can be written as a product of linear and quadratic polynomials. Let $n=deg p(x)$ and assume that $n>1$. The idea is to prove that $n=2$. Note that $mathbb{R}[x]/bigllangle p(x)bigrrangle$ is a field which is an extension of $mathbb R$ and whose dimension as a $mathbb R$-vector space is $n$. So, all that is needed is to prove that if $K$ is such an extension of $mathbb R$, then $n=2$.
In order to prove that, I proved that there is a norm $|cdot|$ on $K$ such that$$(forall x,yin K):|x.y|leqslant|x|.|y|.$$This allows us to define the exponential function$$begin{array}{rccc}expcolon&K&longrightarrow&K\&x&mapsto&displaystylesum_{n=0}^inftyfrac{x^n}{n!}.end{array}$$It turns out that $(forall x,yin K):exp(x+y)=exp(x)exp(y)$. That is, $exp$ is a group homomorphism from $(K,+)$ into $(Ksetminus{0},.)$. It can be proved that $exp(K)$ is both an open and a closed set of $Ksetminus{0}$. Since $n>1$, $Ksetminus{0}$ is connected and therefore $exp$ is surjective.
It can also be proved that $kerexp$ is a discrete subgroup of $(K,+)$. This means that either $kerexp={0}$ or that there are $k$ linearly independent vectors $v_1,ldots,v_kin K$ such that $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. It is not hard to prove that $exp$ induces a homeomorphism between $K/kerexp$ and $Ksetminus{0}$. But then there are two possibilites:
- $kerexp={0}$: this is impossible, because $mathbb{R}^n$ and $mathbb{R}^nsetminus{0}$ are not homeomorphic.
- $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. Then $K/kerexp$ is homeomorphic to $(S^1)^ktimesmathbb{R}^{n-k}$, which is not simply connected. But $mathbb{R}^nsetminus{0}$ is simply connected when $n>2$. Therefore, $n=2$ and this completes my (not that much elementary) proof.
answered Feb 18 '18 at 12:21
José Carlos SantosJosé Carlos Santos
154k22124227
edited Feb 19 '18 at 10:33
answered Feb 18 '18 at 12:21
José Carlos SantosJosé Carlos Santos
154k22124227
answered Feb 18 '18 at 12:21
José Carlos SantosJosé Carlos Santos
154k22124227
answered Feb 18 '18 at 12:21
José Carlos SantosJosé Carlos Santos
154k22124227
154k22124227
$begingroup$
Well done for actually publishing a proof. I want to publish something in the following years, whether it be a proof, theorem, conjecture, or improvement on a certain discovery (like reducing a well-known bound or something), etc.
$endgroup$
– user477343
Feb 18 '18 at 12:24
add a comment |
$begingroup$
Well done for actually publishing a proof. I want to publish something in the following years, whether it be a proof, theorem, conjecture, or improvement on a certain discovery (like reducing a well-known bound or something), etc.
$endgroup$
– user477343
Feb 18 '18 at 12:24
$begingroup$
Well done for actually publishing a proof. I want to publish something in the following years, whether it be a proof, theorem, conjecture, or improvement on a certain discovery (like reducing a well-known bound or something), etc.
$endgroup$
– user477343
Feb 18 '18 at 12:24
$begingroup$
Well done for actually publishing a proof. I want to publish something in the following years, whether it be a proof, theorem, conjecture, or improvement on a certain discovery (like reducing a well-known bound or something), etc.
$endgroup$
– user477343
Feb 18 '18 at 12:24
add a comment |
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