https://artofproblemsolving.com/wiki/index.php?title=Minimal_polynomial&feed=atom&action=historyMinimal polynomial - Revision history2024-03-28T21:13:02ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Minimal_polynomial&diff=92590&oldid=prevMakar: /* Proof of existence/uniqueness */2018-03-02T03:02:51Z<p><span dir="auto"><span class="autocomment">Proof of existence/uniqueness</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Proof of existence/uniqueness ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Proof of existence/uniqueness ==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>First note that as <math>\alpha</math> is algebraic over <math>F</math>, there do exist polynomials <math>f(x)\in F[x<del class="diffchange diffchange-inline">}</del></math> with <math>f(\alpha)=0</math>, and hence there must exist at least one such polynomial, say <math>g(x)</math>, of minimum degree. Now multiplying a polynomial by a scalar does not change it's roots, so we can find some nonzero <math>a\in F</math> such that <math>m(x) = ag(x)</math> is monic. Now by definition it follows that <math>m(x)</math> is ''a'' minimal polynomial for <math>\alpha</math> over <math>F</math>. We now show that is is the only one.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>First note that as <math>\alpha</math> is algebraic over <math>F</math>, there do exist polynomials <math>f(x)\in F[x<ins class="diffchange diffchange-inline">]</ins></math> with <math>f(\alpha)=0</math>, and hence there must exist at least one such polynomial, say <math>g(x)</math>, of minimum degree. Now multiplying a polynomial by a scalar does not change it's roots, so we can find some nonzero <math>a\in F</math> such that <math>m(x) = ag(x)</math> is monic. Now by definition it follows that <math>m(x)</math> is ''a'' minimal polynomial for <math>\alpha</math> over <math>F</math>. We now show that is is the only one.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Assume that there is some other monic polynomial <math>m'(x)\in F[x]</math> such that <math>m'(\alpha)=0</math> and <math>\deg m' = \deg m</math>. By the division algorithm there must exist polynomials <math>q(x),r(x)\in F[x]</math> with <math>\deg r<\deg m</math> such that <math>m(x) = m'(x)q(x)+r(x)</math>. But now we have <math>r(\alpha) = m(\alpha)-m'(\alpha)q(\alpha) = 0</math>, which contradicts the minimality of <math>m(x)</math> unless <math>r(x) = 0</math>. It now follows that <math>m(x) = q(x)m'(x)</math>. And now, as <math>m(x)</math> and <math>m'(x)</math> are both monic polynomials of the same degree, it is easy to verify that <math>q(x)=1</math>, and hence <math>m(x) = m'(x)</math>. So indeed, <math>m(x)</math> is the ''only'' minimal polynomial for <math>\alpha</math> over <math>F</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Assume that there is some other monic polynomial <math>m'(x)\in F[x]</math> such that <math>m'(\alpha)=0</math> and <math>\deg m' = \deg m</math>. By the division algorithm there must exist polynomials <math>q(x),r(x)\in F[x]</math> with <math>\deg r<\deg m</math> such that <math>m(x) = m'(x)q(x)+r(x)</math>. But now we have <math>r(\alpha) = m(\alpha)-m'(\alpha)q(\alpha) = 0</math>, which contradicts the minimality of <math>m(x)</math> unless <math>r(x) = 0</math>. It now follows that <math>m(x) = q(x)m'(x)</math>. And now, as <math>m(x)</math> and <math>m'(x)</math> are both monic polynomials of the same degree, it is easy to verify that <math>q(x)=1</math>, and hence <math>m(x) = m'(x)</math>. So indeed, <math>m(x)</math> is the ''only'' minimal polynomial for <math>\alpha</math> over <math>F</math>.</div></td></tr>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Field theory]]</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Field theory]]</div></td></tr>
</table>Makarhttps://artofproblemsolving.com/wiki/index.php?title=Minimal_polynomial&diff=32699&oldid=prevJam: definition and existance/uniqueness2009-08-21T01:34:36Z<p>definition and existance/uniqueness</p>
<p><b>New page</b></p><div>Given a [[field extension]] <math>F\subseteq K</math>, if <math>\alpha\in K</math> is [[algebraic]] over <math>F</math> then the '''minimal polynomial''' of <math>\alpha</math> over <math>F</math> is defined the [[monic]] [[polynomial]] <math>f(x)\in F[x]</math> of smallest degree such that <math>f(\alpha)=0</math>. This polynomial is often denoted by <math>m_{\alpha,F}(x)</math>, or simply by <math>m_\alpha(x)</math> if <math>F</math> is clear from context.<br />
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== Proof of existence/uniqueness ==<br />
First note that as <math>\alpha</math> is algebraic over <math>F</math>, there do exist polynomials <math>f(x)\in F[x}</math> with <math>f(\alpha)=0</math>, and hence there must exist at least one such polynomial, say <math>g(x)</math>, of minimum degree. Now multiplying a polynomial by a scalar does not change it's roots, so we can find some nonzero <math>a\in F</math> such that <math>m(x) = ag(x)</math> is monic. Now by definition it follows that <math>m(x)</math> is ''a'' minimal polynomial for <math>\alpha</math> over <math>F</math>. We now show that is is the only one.<br />
<br />
Assume that there is some other monic polynomial <math>m'(x)\in F[x]</math> such that <math>m'(\alpha)=0</math> and <math>\deg m' = \deg m</math>. By the division algorithm there must exist polynomials <math>q(x),r(x)\in F[x]</math> with <math>\deg r<\deg m</math> such that <math>m(x) = m'(x)q(x)+r(x)</math>. But now we have <math>r(\alpha) = m(\alpha)-m'(\alpha)q(\alpha) = 0</math>, which contradicts the minimality of <math>m(x)</math> unless <math>r(x) = 0</math>. It now follows that <math>m(x) = q(x)m'(x)</math>. And now, as <math>m(x)</math> and <math>m'(x)</math> are both monic polynomials of the same degree, it is easy to verify that <math>q(x)=1</math>, and hence <math>m(x) = m'(x)</math>. So indeed, <math>m(x)</math> is the ''only'' minimal polynomial for <math>\alpha</math> over <math>F</math>.<br />
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[[Category:Field theory]]</div>Jam