Difference between revisions of "Fundamental Theorem of Algebra"

Latest revision as of 13:22, 11 January 2024

The fundamental theorem of algebra states that every nonconstant polynomial with complex coefficients has a complex root. In fact, every known proof of this theorem involves some analysis, since the result depends on certain properties of the complex numbers that are most naturally described in topological terms.

It follows from the division algorithm that every complex polynomial of degree $n$ has $n$ complex roots, counting multiplicities. In other words, every polynomial over $\mathbb{C}$ splits over $\mathbb{C}$ , or decomposes into linear factors.

Proofs

Proof by Liouville's Theorem

We use Liouville's Boundedness Theorem of complex analysis, which says that every bounded entire function is constant.

Suppose that $P(z)$ is a complex polynomial of degree $n$ with no complex roots; without loss of generality, suppose that $P$ is monic. Then $1/P(z)$ is an entire function; we wish to show that it is bounded. It is clearly bounded when $n=0$ ; we now consider the case when $n>0$ .

Let $R$ be the sum of absolute values of the coefficients of $P$ , so that $R \ge 1$ . Then for $\lvert z \rvert \ge R$ , $\lvert P(z) \rvert \ge \lvert z^n \rvert - (R-1) \lvert z^{n-1} \rvert = \lvert z^{n-1} \rvert \cdot \bigl[ \lvert z \rvert - (R-1) \bigr] \ge R^{n-1} .$ It follows that $1/P(z)$ is a bounded entire function for $\lvert z \rvert > R$ . On the other hand, by the Heine-Borel Theorem, the set of $z$ for which $\lvert z \rvert \le R$ is a compact set so its image under $1/P$ is also compact; in particular, it is bounded. Therefore the function $1/P(z)$ is bounded on the entire complex plane when $n>0$ .

Now we apply Liouville's theorem and see that $1/P(z)$ is constant, so $P(z)$ is a constant polynomial. The theorem then follows. $\blacksquare$

Algebraic Proof

Let $P(x)$ be a polynomial with complex coefficients. Since $F(x) = P(x) \overline{P(x)}$ is a polynomial with real coefficients such that the roots of $P$ are also roots of $F$ , it suffices to show that every polynomial with real coefficients has a complex root. To this end, let the degree of $F$ be $d = 2^n q$ , where $q$ is odd. We induct on the quantity $n$ .

For $n=0$ , we note that for sufficiently large negative real numbers $x$ , $F(x) < 0$ ; for sufficiently large positive real numbers $x$ , $F(x) > 0$ . It follows from the Intermediate Value Theorem that $F(x)$ has a real root.

Now suppose that $n > 0$ . Let $C$ be a splitting field of $F$ over $\mathbb{C}$ , and let $x_1, \dotsc, x_d$ be the roots of $F$ in $C$ .

Let $c$ be an arbitrary real number, and let $y_{c,i,j} = x_i + x_j + cx_ix_j$ for $1 \le i \le j \le d$ . Let $G_c(x) = \prod_{1 \le i \le j \le d} (x-y_{c,i,j}) .$ The coefficients of $G$ are symmetric in $x_1, \dotsc, x_d$ . Therefore they can be expressed as linear combinations of real numbers times the elementary symmetric polynomials in $x_1, \dotsc, x_n$ ; thus they are real numbers. Since the degree of $G_c$ is $\binom{d+1}{2} = 2^{n-1}q(d+1)$ , it follows by inductive hypothesis that $G_c$ has a complex root; that is, $y_{c,i(c),j(c)} \in \mathbb{C}$ for some $1 \le i(c) \le j(c) \le d$ .

Now, since there are infinitely many real numbers but only finitely many integer pairs $(i,j)$ with $1 \le i \le j \le d$ , it follows that for two distinct numbers $c,c'$ , $(i(c),j(c)) = (i(c'),j(c')) = (i,j)$ . It follows that $x_i + x_j$ and $x_ix_j$ are both complex numbers, so $x_i$ and $x_j$ satisfy a quadratic equation with complex coefficients. Hence they are complex numbers. Therefore $F$ has a complex root, as desired. $\blacksquare$

References

Samuel, Pierre (trans. A. Silberger), Algebraic Theory of Numbers, Dover 1970, ISBN 978-0-486-46666-8 .
Proofs of the Fundamental Theorem of Algebra on Cut the Knot

@@ Line 1: / Line 1: @@
-== Introduction ==
+The '''fundamental theorem of algebra''' states that every [[nonconstant]] [[polynomial]] with [[complex number|complex]] [[coefficient]]s has a complex [[root]]. In fact, every known proof of this theorem involves some [[analysis]], since the result depends on certain properties of the complex numbers that are most naturally described in [[topology | topological]] terms.
-The '''fundamental theorem of algebra''' states that every [[nonconstant]] [[polynomial]] with [[complex number|complex]] coefficients has a complex [[root]]. In fact, every known proof of this theorem involves a bit of [[analysis]], since a purely algebraic construction of the complex numbers is very hard to work with.
-A corollary to the Fundamental Theorem of Algebra also states that for a polynomial of degree <math>n</math>, there are <math>n</math> roots, counting multiplicities.
+It follows from the [[division algorithm]] that every complex polynomial of degree <math>n</math> has <math>n</math> complex roots, counting multiplicities.  In other words, every polynomial over <math>\mathbb{C}</math> splits over <math>\mathbb{C}</math>, or decomposes into linear factors.
-== Proof ==
+== Proofs ==
-One of the shortest proofs of the fundamental theorem of algebra uses [[Liouville's Theorem]] of [[complex analysis]], which says that a [[bounded]] [[entire function]] is [[constant]]. Suppose that <math>p(x)</math> were a complex polynomial with no complex roots. Then <math>1/p(x)</math> would be an entire function. To see that it is bounded, we must show that <math>|p(x)|\ge c>0</math> for some <math>c</math>. It can easily be shown that outside a sufficiently large [[disc]], <math>p(x)</math> is dominated by its [[leading term]], which gets large in [[absolute value]], so it suffices to work inside a disc of radius <math>R</math> for some <math>R</math> to find the minimum value. But the (closed) disc of radius <math>R</math> is a [[compact set]], so <math>|p(x)|</math> must achieve a minimum inside the disc. This minimum value is <math>c</math>. Hence <math>\left|\frac{1}{p(x)}\right|\le\frac{1}{c}</math>, so <math>1/p(x)</math> would be a bounded entire function and hence constant. This proves the theorem.
+=== Proof by Liouville's Theorem ===
+We use [[Liouville's Boundedness Theorem]] of [[complex analysis]], which says that every [[bounded]] [[entire function]] is [[constant]].
+Suppose that <math>P(z)</math> is a complex polynomial of degree <math>n</math> with no complex roots; without loss of generality, suppose that <math>P</math> is [[monic]].  Then <math>1/P(z)</math> is an [[entire]] function; we wish to show that it is bounded.  It is clearly bounded when <math>n=0</math>; we now consider the case when <math>n>0</math>.
+Let <math>R</math> be the sum of absolute values of the coefficients of <math>P</math>, so that <math>R \ge 1</math>.  Then for <math>\lvert z \rvert \ge R</math>,
+<cmath> \lvert P(z) \rvert \ge \lvert z^n \rvert - (R-1) \lvert z^{n-1} \rvert
+= \lvert z^{n-1} \rvert \cdot \bigl[ \lvert z \rvert - (R-1) \bigr]
+\ge R^{n-1} . </cmath>
+It follows that <math>1/P(z)</math> is a bounded entire function for <math>\lvert z \rvert > R</math>.  On the other hand, by the [[Heine-Borel Theorem]], the set of <math>z</math> for which <math>\lvert z \rvert \le R</math> is a [[compact set]] so its image under <math>1/P</math> is also compact; in particular, it is bounded.  Therefore the function <math>1/P(z)</math> is bounded on the entire complex plane when <math>n>0</math>.
+Now we apply Liouville's theorem and see that <math>1/P(z)</math> is constant, so <math>P(z)</math> is a constant polynomial.  The theorem then follows.  <math>\blacksquare</math>
+=== Algebraic Proof ===
+Let <math>P(x)</math> be a polynomial with complex coefficients.  Since <math>F(x) = P(x) \overline{P(x)}</math> is a polynomial with real coefficients such that the roots of <math>P</math> are also roots of <math>F</math>, it suffices to show that every polynomial with ''real'' coefficients has a complex root.  To this end, let the degree of <math>F</math> be <math>d = 2^n q</math>, where <math>q</math> is odd.  We induct on the quantity <math>n</math>.
+For <math>n=0</math>, we note that for sufficiently large negative real numbers <math>x</math>, <math>F(x) < 0</math>; for sufficiently large positive real numbers <math>x</math>, <math>F(x) > 0</math>.  It follows from the [[Intermediate Value Theorem]] that <math>F(x)</math> has a real root.
+Now suppose that <math>n > 0</math>.  Let <math>C</math> be a [[splitting field]] of <math>F</math> over <math>\mathbb{C}</math>, and let <math>x_1, \dotsc, x_d</math> be the roots of <math>F</math> in <math>C</math>.
+Let <math>c</math> be an arbitrary real number, and let <math>y_{c,i,j} = x_i + x_j + cx_ix_j</math> for <math>1 \le i \le j \le d</math>.  Let
+<cmath> G_c(x) = \prod_{1 \le i \le j \le d} (x-y_{c,i,j}) . </cmath>
+The coefficients of <math>G</math> are symmetric in <math>x_1, \dotsc, x_d</math>.  Therefore they can be expressed as linear combinations of real numbers times the [[elementary symmetric polynomial]]s in <math>x_1, \dotsc, x_n</math>; thus they are real numbers.  Since the degree of <math>G_c</math> is <math>\binom{d+1}{2} = 2^{n-1}q(d+1)</math>, it follows by inductive hypothesis that <math>G_c</math> has a complex root; that is, <math>y_{c,i(c),j(c)} \in \mathbb{C}</math> for some <math>1 \le i(c) \le j(c) \le d</math>.
+Now, since there are infinitely many real numbers but only finitely many integer pairs <math>(i,j)</math> with <math>1 \le i \le j \le d</math>, it follows that for two distinct numbers <math>c,c'</math>, <math>(i(c),j(c)) = (i(c'),j(c')) = (i,j)</math>.  It follows that <math>x_i + x_j</math> and <math>x_ix_j</math> are both complex numbers, so <math>x_i</math> and <math>x_j</math> satisfy a quadratic equation with complex coefficients.  Hence they are complex numbers.  Therefore <math>F</math> has a complex root, as desired.  <math>\blacksquare</math>
+== References ==
+* Samuel, Pierre (trans. A. Silberger), ''Algebraic Theory of Numbers'', Dover 1970, ISBN 978-0-486-46666-8 .
+* [http://www.cut-the-knot.org/fta/analytic.shtml Proofs of the Fundamental Theorem of Algebra on Cut the Knot]
 == See also ==
 * [[Algebra]]
-(This ought to be cleaned up.)
-There is another proof that uses [[group theory]] and [[Galois theory]] that is very nice, and someone (possibly me) should write it here at some point. The advantage of this proof is that it uses only a very small amount of analysis (just the [[Intermediate Value Theorem]]), and the rest is all actually [[algebra]].
+[[Category:Algebra]]
+[{Category:Complex analysis]]
+[[Category:Polynomials]]
+[[Category:Theorems]]

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