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elde edilir. Burada sadece toplamdaki indeksin ''k'' = 0 olduğu durumda integralin değeri 0 olmaz. Bu bir çelişkidir. O yüzden, ''A'''nın özdeğeri vardır.
===Topolojik kanıtlar===
Let ''z''<sub>0</sub> ∈ '''C''' be such that the minimum of |''p''(''z'')| on the whole complex plane is achieved at ''z''<sub>0</sub>; it was seen at the proof which uses Liouville's theorem that such a number must exist. We can write ''p''(''z'') as a polynomial in ''z'' − ''z''<sub>0</sub>: there is some natural number ''k'' and there are some complex numbers ''c<sub>k</sub>'', ''c''<sub>''k'' + 1</sub>, ..., ''c<sub>n</sub>'' such that ''c<sub>k</sub>'' ≠ 0 and that
:<math>p(z)=p(z_0)+c_k(z-z_0)^k+c_{k+1}(z-z_0)^{k+1}+ \cdots +c_n(z-z_0)^n.</math>
It follows that if ''a'' is a ''k''<sup>th</sup> root of −''p''(''z''<sub>0</sub>)/''c<sub>k</sub>'' and if ''t'' is positive and sufficiently small, then |''p''(''z''<sub>0</sub> + ''ta'')| < |''p''(''z''<sub>0</sub>)|, which is impossible, since |''p''(''z''<sub>0</sub>)| is the minimum of |''p''| on ''D''.
For another topological proof by contradiction, suppose that ''p''(''z'') has no zeros. Choose a large positive number ''R'' such that, for |''z''| = ''R'', the leading term ''z<sup>n</sup>'' of ''p''(''z'') dominates all other terms combined; in other words, such that |''z''|<sup>''n''</sup> > |''a''<sub>''n'' − 1</sub>''z''<sup>''n'' −1</sup> + ··· + ''a''<sub>0</sub>|. As ''z'' traverses the circle given by the equation |''z''| = ''R'' once counter-clockwise, ''p''(''z''), like ''z<sup>n</sup>'', winds ''n'' times counter-clockwise around 0. At the other extreme, with |''z''| = 0, the “curve” ''p''(''z'') is simply the single (nonzero) point ''p''(0), whose [[winding number]] is clearly 0. If the loop followed by ''z'' is [[homotopy|continuously deformed]] between these extremes, the path of ''p''(''z'') also deforms continuously. We can explicitly write such a deformation as <math>H(Re^{i\theta},t)=p((1-t)Re^{i\theta})</math> where ''t'' is greater than or equal to 0 and less than or equal to 1. If one views the variable ''t'' as time, then at time zero the curve is ''p(z)'' and at time one the curve is ''p(0)''. Clearly at every point ''t'', ''p(z)'' cannot be zero by the original assumption, therefore during the deformation, the curve never crosses zero. Therefore the winding number of the curve around zero should never change. However, given that the winding number started as ''n'' and ended as 0, this is absurd. Therefore, ''p''(''z'') has at least one zero.
===Cebirsel kanıtlar===
These proofs use two facts about real numbers that require only a small amount of analysis (more precisely, the [[intermediate value theorem]]):
* every polynomial with odd degree and real coefficients has some real root;
* every non-negative real number has a square root.
The second fact, together with the [[quadratic formula]], implies the theorem for real quadratic polynomials. In other words, algebraic proofs of the fundamental theorem actually show that if ''R'' is any [[real-closed field]], then its extension <math>\scriptstyle C = R ( \sqrt{-1} )</math> is algebraically closed.
As mentioned above, it suffices to check the statement “every non-constant polynomial ''p''(''z'') with real coefficients has a complex root”. This statement can be proved by induction on the greatest non-negative integer ''k'' such that 2<sup>''k''</sup> divides the degree ''n'' of ''p''(''z''). Let ''a'' be the coefficient of ''z<sup>n</sup>'' in ''p''(''z'') and let ''F'' be a [[splitting field]] of ''p''(''z'') over ''C''; in other words, the field ''F'' contains ''C'' and there are elements ''z''<sub>1</sub>, ''z''<sub>2</sub>, ..., ''z''<sub>n</sub> in ''F'' such that
:<math>p(z)=a(z-z_1)(z-z_2) \cdots (z-z_n).</math>
If ''k'' = 0, then ''n'' is odd, and therefore ''p''(''z'') has a real root. Now, suppose that ''n'' = 2''<sup>k</sup>m'' (with ''m'' odd and ''k'' > 0) and that the theorem is already proved when the degree of the polynomial has the form 2<sup>''k'' − 1</sup>''m''′ with ''m''′ odd. For a real number ''t'', define:
:<math>q_t(z)=\prod_{1\le i<j\le n}\left(z-z_i-z_j-tz_iz_j\right).\,</math>
Then the coefficients of ''q<sub>t</sub>''(''z'') are [[symmetric polynomial]]s in the ''z<sub>i</sub>'''s with real coefficients. Therefore, they can be expressed as polynomials with real coefficients in the [[elementary symmetric polynomial]]s, that is, in −''a''<sub>1</sub>, ''a''<sub>2</sub>, ..., (−1)''<sup>n</sup>a<sub>n</sub>''. So ''q<sub>t</sub>''(''z'') has in fact ''real'' coefficients. Furthermore, the degree of ''q<sub>t</sub>''(''z'') is ''n''(''n'' − 1)/2 = 2<sup>''k'' − 1</sup>''m''(''n'' − 1), and ''m''(''n'' − 1) is an odd number. So, using the induction hypothesis, ''q<sub>t</sub>'' has at least one complex root; in other words, ''z<sub>i</sub>'' + ''z<sub>j</sub>'' + ''tz<sub>i</sub>z<sub>j</sub>'' is complex for two distinct elements ''i'' and ''j'' from {1,...,''n''}. Since there are more real numbers than pairs (''i'',''j''), one can find distinct real numbers ''t'' and ''s'' such that ''z<sub>i</sub>'' + ''z<sub>j</sub>'' + ''tz<sub>i</sub>z<sub>j</sub>'' and ''z<sub>i</sub>'' + ''z<sub>j</sub>'' + ''sz<sub>i</sub>z<sub>j</sub>'' are complex (for the same ''i'' and ''j''). So, both ''z<sub>i</sub>'' + ''z<sub>j</sub>'' and ''z<sub>i</sub>z<sub>j</sub>'' are complex numbers. It is easy to check that every complex number has a complex square root, thus every complex polynomial of degree 2 has a complex root by the quadratic formula. It follows that ''z<sub>i</sub>'' and ''z<sub>j</sub>'' are complex numbers, since they are roots of the quadratic polynomial ''z''<sup>2</sup> − (''z<sub>i</sub>'' + ''z<sub>j</sub>'')''z'' + ''z<sub>i</sub>z<sub>j</sub>''.
J. Shipman showed in 2007 that the assumption that odd degree polynomials have roots is stronger than necessary; any field in which polynomials of prime degree have roots is algebraically closed (so "odd" can be replaced by "odd prime" and furthermore this holds for fields of all characteristics). This is the best possible, as there are counterexamples if a single prime is excluded.
Another algebraic proof of the fundamental theorem can be given using [[Galois theory]]. It suffices to show that '''C''' has no proper finite [[field extension]].<ref>A proof of the fact that this suffices can be seen [[Algebraically closed field#The field has no proper finite extension|here]].</ref> Let ''K''/'''C''' be a finite extension. Since the [[Normal extension#Normal closure|normal closure]] of ''K'' over '''R''' still has a finite degree over '''C''' (or '''R'''), we may assume [[without loss of generality]] that ''K'' is a [[normal extension]] of '''R''' (hence it is a [[Galois extension]], as every algebraic extension of a field of [[characteristic (algebra)|characteristic]] 0 is [[separable extension|separable]]). Let ''G'' be the [[Galois group]] of this extension, and let ''H'' be a [[Sylow theorems|Sylow]] 2-group of ''G'', so that the [[order (group theory)|order]] of ''H'' is a power of 2, and the [[index of a subgroup|index]] of ''H'' in ''G'' is odd. By the [[fundamental theorem of Galois theory]], there exists a subextension ''L'' of ''K''/'''R''' such that Gal(''K''/''L'') = ''H''. As [''L'':'''R'''] = [''G'':''H''] is odd, and there are no nonlinear irreducible real polynomials of odd degree, we must have ''L'' = '''R''', thus [''K'':'''R'''] and [''K'':'''C'''] are powers of 2. Assuming for contradiction [''K'':'''C'''] > 1, the [[p-group|2-group]] Gal(''K''/'''C''') contains a subgroup of index 2, thus there exists a subextension ''M'' of '''C''' of degree 2. However, '''C''' has no extension of degree 2, because every quadratic complex polynomial has a complex root, as mentioned above.
== Notlar ==
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