Fundamental Theorem of Algebra: Difference between revisions

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There are also proofs that do not depend on [[complex analysis]], but they require more [[algebra|algebraic]] or [[topology|topological]] machinery.  
There are also proofs that do not depend on [[complex analysis]], but they require more [[algebra|algebraic]] or [[topology|topological]] machinery.  


We need to show that any algebraic extension of <math>\scriptstyle\mathbb{C}</math> has degree one. Since <math>\scriptstyle\mathbb{C} = \mathbb{R}[i]</math>, any such field extension also extends <math>\scriptstyle\mathbb{R}</math>. Now, any <math>\scriptstyle\alpha</math> algebraic over <math>\scriptstyle\mathbb{C}</math> must also be algebraic over <math>\scriptstyle\mathbb{R}</math>, but its minimal polynomial cannot be of odd degree, because any such polynomial must have a real root by the intermediate value theorem, so the splitting field of <math>\scriptstyle\alpha</math> over <math>\mathbb{R}</math> must have degree a power of 2. Its Galois group must have normal subgroup of index 2, but a generating element must already be in <math>\scriptstyle\mathbb{C}</math> by the quadratic formula. This shows that the extension has, at most, degree 2, but appealing once again to the quadratic formula, we see that <math>\scriptstyle\mathbb{C}</math> is closed under quadratic extensions, so <math>\scriptstyle\mathbb{C}</math> must itself be algebraically closed.
We need to show that any algebraic extension of <math>\scriptstyle\mathbb{C}</math> has degree one. Since <math>\scriptstyle\mathbb{C} = \mathbb{R}[i]</math>, any such field extension also extends <math>\scriptstyle\mathbb{R}</math>. Now, any <math>\scriptstyle\alpha</math> algebraic over <math>\scriptstyle\mathbb{C}</math> must also be algebraic over <math>\scriptstyle\mathbb{R}</math>, but its [[minimal polynomial]] cannot be of odd degree, because any such polynomial must have a real root by the [[intermediate value theorem]], so the [[splitting field]] of <math>\scriptstyle\alpha</math> over <math>\mathbb{R}</math> must have degree a power of 2. Its [[Galois group]] must have a [[normal subgroup]] of [[index of a subgroup|index]] 2, but a generating element must already be in <math>\scriptstyle\mathbb{C}</math> by the quadratic formula. This shows that the extension has, at most, degree 2, but appealing once again to the quadratic formula, we see that <math>\scriptstyle\mathbb{C}</math> is closed under quadratic extensions, so <math>\scriptstyle\mathbb{C}</math> must itself be algebraically closed.


===Using the fundamental group of the punctured plane <math>\mathbb{C}\setminus\{0\}</math>===
===Using the fundamental group of the punctured plane <math>\mathbb{C}\setminus\{0\}</math>===
Assume the contrary. W.l.og. the leading coefficient of <math>p(z)</math> is 1. By our assumption the maps <math>[0,1]\to\mathbb{C}\setminus\{0\}</math> given by <math>t\to p(r e^{2\pi i t})</math> are homotopic in <math>\mathbb{C}\setminus\{0\}</math> for all <math>r</math>, and hence they are all null homotopic there (take <math>r=0</math>). However, for large enough <math>r</math>, the maps <math>[0,1]\to\mathbb{C}\setminus\{0\}</math> given by <math>t\to p(r e^{2\pi i t})</math> and <math>t\to r e^{2\pi i t}</math> are homotopic, and
We can assume without loss of generality that the leading coefficient of <math>p(z)</math> is 1. Now assume that <math>p</math> has no roots. Then, the maps <math>[0,1]\to\mathbb{C}\setminus\{0\}</math> given by <math>t\mapsto p(r e^{2\pi i t})</math> are [[homotopy|homotopic]] in <math>\mathbb{C}\setminus\{0\}</math> for all <math>r</math>, and hence they are all null homotopic there (take <math>r=0</math>). However, for large enough <math>r</math>, the maps <math>[0,1]\to\mathbb{C}\setminus\{0\}</math> given by <math>t\mapsto p(r e^{2\pi i t})</math> and <math>t\mapsto r^d e^{2d\pi i t}</math> (where <math>d</math> is the degree of the polynomial <math>p</math>) are homotopic, and thus <math>t\mapsto p(r e^{2\pi i t})</math> is not null homotopic.
<math>t\to p(r e^{2\pi i t})</math> is not null homotopic.


===Using the second homotopy group of the Riemann sphere <math>\hat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}\cong S^2</math>===
===Using the second homotopy group of the Riemann sphere <math>\hat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}\cong S^2</math>===
W.l.o.g the leading coefficient of <math>p(z)</math> is 1. We consider <math>z\to p(z)</math> as a map from the Riemann sphere to itself (taking infinity to infinity). By considering the homotopy <math> t z^n+(1-t) p(z)</math>, where <math>t\in [0,1]</math>, this map is homotopic to the map <math>z\to z^n</math>. Hence it suffices to show that the map <math>z\to z^n</math> is not null-homotopic. However, in the homotopy group <math>\pi_2(S^2)</math> we have <math>[z\to z^n] = [z\to z][z\to z^{n-1}]</math>, and so it suffices to show that <math>z\to z</math> is not null homotpic, which is equivalent to the fact that the sphere is not contractible.
Without loss of generality the leading coefficient of <math>p(z)</math> is 1. We consider <math>z\mapsto p(z)</math> as a map from the [[Riemann sphere]] to itself (taking infinity to infinity). By considering the homotopy <math> t z^n+(1-t) p(z)</math>, where <math>t\in [0,1]</math>, this map is homotopic to the map <math>z\mapsto z^n</math>. Hence it suffices to show that the map <math>z\mapsto z^n</math> is not null homotopic. However, in the homotopy group <math>\pi_2(S^2)</math> we have <math>[z\mapsto z^n] = [z\mapsto z][z\mapsto z^{n-1}]</math>, and so it suffices to show that <math>z\mapsto z</math> is not null homotpic, which is equivalent to the fact that the sphere is not contractible.


==Further reading==
==Further reading==
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|publisher = Springer-Verlag
|publisher = Springer-Verlag
|date = 1997
|date = 1997
|isbn = 0-387-94657-8 }}
|isbn = 0-387-94657-8 }}[[Category:Suggestion Bot Tag]]

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The Fundamental Theorem of Algebra is a mathematical theorem stating that every nonconstant polynomial whose coefficients are complex numbers has at least one complex number as a root. In other words, given any polynomial

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) = c_d x^d + c_{d-1} x^{d-1} + \cdots + c_2 x^2 + c_1 x + c_0}

(where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d} is any positive integer), we can find a complex number Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t} so that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_d t^d + c_{d-1} t^{d-1} + \cdots + c_2 t^2 + c_1 t + c_0 = 0.}

One important case of the Fundamental Theorem of Algebra is that every nonconstant polynomial with real coefficients must have at least one complex root. Since it is not true that every such polynomial has to have at least one real root (as the example Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) = x^2+1} demonstrates), many mathematicians feel that the complex numbers form the most natural setting for working with polynomials.

In fact, a stronger version of the Fundamental Theorem of Algebra is also true: a polynomial of degree Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d} can be factored completely into a product of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d} linear polynomials:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) = c_d (x-t_1) (x-t_2) \cdots (x-t_d). }

A way of saying this is that every polynomial of degree Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d} has exactly Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d} complex roots, "counting multiplicity".

Carl Friedrich Gauss is generally credited with the first satisfactory proof of this theorem, his proof being the principal result in his Ph.D. thesis finished in 1799.

Proving the Fundamental Theorem of Algebra

Using complex analysis

A startlingly simple proof is based on Liouville's theorem: If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(z)} is a polynomial function of a complex variable then both Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(z)} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1/p(z)} will be holomorphic in any domain where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle p(z) \ne 0} . But, by the triangle inequality, we know that outside a neighborhood of the origin Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |p(z)| > |p(0)|} , so if there is no Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z_0 } such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(z_0) = 0} , we know that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1/p(z)} is a bounded entire (i.e., holomorphic in all of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{C}} ) function. By Liouville's theorem, it must be constant, so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(z)} must also be constant.

Using algebra (and a bit of real analysis)

There are also proofs that do not depend on complex analysis, but they require more algebraic or topological machinery.

We need to show that any algebraic extension of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle\mathbb{C}} has degree one. Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle\mathbb{C} = \mathbb{R}[i]} , any such field extension also extends Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle\mathbb{R}} . Now, any Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle\alpha} algebraic over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle\mathbb{C}} must also be algebraic over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle\mathbb{R}} , but its minimal polynomial cannot be of odd degree, because any such polynomial must have a real root by the intermediate value theorem, so the splitting field of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle\alpha} over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}} must have degree a power of 2. Its Galois group must have a normal subgroup of index 2, but a generating element must already be in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle\mathbb{C}} by the quadratic formula. This shows that the extension has, at most, degree 2, but appealing once again to the quadratic formula, we see that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle\mathbb{C}} is closed under quadratic extensions, so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \scriptstyle\mathbb{C}} must itself be algebraically closed.

Using the fundamental group of the punctured plane Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{C}\setminus\{0\}}

We can assume without loss of generality that the leading coefficient of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(z)} is 1. Now assume that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} has no roots. Then, the maps Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0,1]\to\mathbb{C}\setminus\{0\}} given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t\mapsto p(r e^{2\pi i t})} are homotopic in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{C}\setminus\{0\}} for all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} , and hence they are all null homotopic there (take Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=0} ). However, for large enough Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} , the maps Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0,1]\to\mathbb{C}\setminus\{0\}} given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t\mapsto p(r e^{2\pi i t})} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t\mapsto r^d e^{2d\pi i t}} (where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d} is the degree of the polynomial Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} ) are homotopic, and thus Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t\mapsto p(r e^{2\pi i t})} is not null homotopic.

Using the second homotopy group of the Riemann sphere Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}\cong S^2}

Without loss of generality the leading coefficient of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(z)} is 1. We consider Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z\mapsto p(z)} as a map from the Riemann sphere to itself (taking infinity to infinity). By considering the homotopy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t z^n+(1-t) p(z)} , where , this map is homotopic to the map . Hence it suffices to show that the map is not null homotopic. However, in the homotopy group we have , and so it suffices to show that is not null homotpic, which is equivalent to the fact that the sphere is not contractible.

Further reading

  • Fine, Benjamin; Rosenberger, Gerhard (1997). The Fundamental Theorem of Algebra. Springer-Verlag. ISBN 0-387-94657-8. 
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