Did You Know? -> Gelfond–Schneider theorem

Gelfond–Schneider theorem, DYK Fact #583

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JaneFairfax	Posted: 15:40 Wednesday 05 August 2009
The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007	Did you know? The Gelfond–Schneider theorem states if α is an algebraic number not equal to 0 or 1 and β is an irrational algebraic number, then α^β is transcendental. A proof of this theorem is included in the chapter on transcendental numbers in the book A Course in Number Theory by H.E. Rose.

algebraic topology	Posted: 20:28 Wednesday 05 August 2009
Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007	Did you know? This theorem is a resolution of part of the seventh of the mathematical problems famously proposed by David Hilbert in 1900.

Ebudae	Posted: 20:44 Wednesday 05 August 2009
The Enlightenment Group: Admin Posts: 935 Member No.: 1 Joined: 02 Jan 2007	Did you know? Schneider is the German word for a tailor. -------------------- *Ebudæ*

algebraic topology	Posted: 21:48 Wednesday 05 August 2009
Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007	Did you know? The theorem was first proved by Aleksandr Gelfond in 1934; Theodor Schneider refined the proof in 1935.

JaneFairfax	Posted: 12:20 Thursday 06 August 2009
The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007	Did you know? The proof of the Gelfond–Schneider theorem given in the book by Rose goes by assuming to the contraray that α, β and α^β all belong to some algebraic-number field K and then constructing a member of K with an inconsistent norm.

algebraic topology	Posted: 11:49 Thursday 27 August 2009
Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007	Did you know? If F is a subfield of E, a number α in E is said to be algebraic over F iff α is a root of a nonzero polynomial with coefficients in F; α is said to be transcendental over F iff it is not algebraic over F. The Gelfond–Schneider theorem is a result concerning complex transcendental numbers over the rationals.

JaneFairfax	Posted: 17:00 Thursday 27 August 2009
The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007	Did you know? A complex number is a root of a polynomial with rational coefficients if and only if it is a root of a polynomial with integer coefficients. This is Gauß’s lemma.

algebraic topology	Posted: 18:19 Thursday 27 August 2009
Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007	Did you know? If R is an integral domain with field of fractions F, an element in some field containing F is a root of a polynomial with coefficients in F if and only if it is a root of a polynomial with coefficients in R. For if α is a root of $f_0+f_1x+\cdots+f_nx^n\in{F}[x]$ where $f_i=r_i/s_i$ , $r_i,s_i\in{R}$ and $s_i\ne0_R$ , for $i=0,1,\ldots,n$ , then, if $s=\prod_{i\,=\,0}^ns_i$ , we have that α is a root of the polynomial $g_0+g_1x+\cdots+g_nx^n\in{R}[x]$ where $g_i=sf_i$ for $i=0,1,\ldots,n$ . The other implication is trivial.

JaneFairfax	Posted: 09:58 Friday 28 August 2009
The Enlightenment Group: Moderators Posts: 673 Member No.: 20 Joined: 03 Mar 2007	Did you know? If F is a subfield of E and $\alpha\in{E}$ is algebraic over F, the (unique) monic polynomial in F[x] of least degree for which α is a root is called the minimal polynomial of α over F. It is a principal-ideal generator of the kernel of the ring homomorphism $\varphi_\alpha:F[x]\to{E}$ , $\varphi_\alpha(f_0+f_1x+\cdots+f_nx^n)=f_0+f_1\alpha+\cdots+f_n\alpha^n$ .

algebraic topology	Posted: 16:29 Friday 28 August 2009
Renaissance Group: Friends Posts: 143 Member No.: 14 Joined: 20 Feb 2007	Did you know? If F is a subfield of E, the elements of E which are algebraic over F form a subfield of E containing F.

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