Gelfond-Schneider Theorem

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Theorem

Let $\alpha$ and $\beta$ be algebraic numbers (possibly complex) such that $\alpha \notin \set {0, 1}$.

Let $\beta$ be irrational.


Then any value of $\alpha^\beta$ is transcendental.


Proof

Let $\alpha$ be an algebraic number such that $\alpha \ne 0$ and $\alpha \ne 1$.

Let $\beta$ be an algebraic number such that some value of $\alpha^\beta$ is algebraic.

The result will follow if we can show that $\beta \in \Q$.


We treat only the special case that $\alpha, \beta \in \R$ and $\alpha > 0$, assuming that $\map \exp {\beta \ln \alpha}$ is algebraic for the real value of $\ln \alpha$.

Observe that $\alpha^{s_1 + s_2 \beta}$ is an algebraic number for all integers $s_1$ and $s_2$.

To establish the result, it is enough to show that there are two distinct pairs of integers $\tuple {s_1, s_2}$ and $\tuple {s'_1, s'_2}$ for which:

$s_1 + s_2 \beta = s'_1 + s'_2 \beta$

since that implies:

$\beta = \dfrac{s_1 - s'_1} {s'_2 - s_2} \in \Q$

We will choose $S$ sufficiently large and show such pairs exist with $0 \le s_1, s_2, s'_1, s'_2 < S$.


Lemma 1

Let $a_1 \left({t}\right), \ldots, a_n \left({t}\right)$ be non-zero polynomials in $\R \left[{t}\right]$ of degrees $d_1, \ldots, d_n$ respectively.

Let $w_1, \ldots, w_n$ be pairwise distinct real numbers.


Then:

$\displaystyle F \left({t}\right) = \sum_{j \mathop = 1}^n a_j \left({t}\right) e^{w_j t}$

has at most $d_1 + \cdots + d_n + n − 1$ real roots (counting multiplicities).


Lemma 2

Let $\map f z$ be an analytic function in the disk $D \subseteq \C: D = \set {z : \size z < R}$ for some real number $R > 0$.

Let $f$ also be continuous on the closure of $D$, that is, on $D^- = \set {z : \size z \le R}$.


Then:

$\forall z \in D^-: \size {\map f z} \le \size f_R$, where we set $\size f_R = \max_{\size z = R} \size {\map f z}$.


Lemma 3

Let $r$ and $R$ be two real numbers such that $0 < r \le R$.

Let $f_1 \left({z}\right), f_2 \left({z}\right), \ldots, f_L \left({z}\right)$ be:

analytic in $D \subseteq \C: D = \left\{{z : \left|{z}\right| < R}\right\}$
continuous on the closure $D$, that is, $D^- = \left\{{z : \left \vert{z}\right \vert \le R}\right\}$.

Let $\zeta_1, \ldots, \zeta_L$ be complex numbers such that:

$\forall j \in \left\{{1, 2, \ldots, L}\right\}: \left|{\zeta_j}\right| \le r$


Then the determinant:

$\Delta = \det \begin{bmatrix} f_1 \left({\zeta_1}\right) & \cdots & f_L \left({\zeta_1}\right) \\ \vdots & \ddots & \vdots \\ f_1 \left({\zeta_L}\right) & \cdots & f_L \left({\zeta_L}\right) \end{bmatrix}$

satisfies:

$\displaystyle \left|{\Delta}\right| \le \left({\frac R r}\right)^{−L \left({L−1}\right) / 2} L! \prod_{\lambda \mathop = 1}^L \left|{f_λ}\right|_R$


Lemma 4

Let:

$\Delta = \det \left[{\alpha_{i, j}}\right]_{L \times L}$

where the $\alpha_{i,j}$ are algebraic numbers.

Suppose that $T$ is a positive (rational) integer for which $T \alpha_{i, j}$ is an algebraic integer for every $i, j \in \left\{{1, 2, \ldots, L}\right\}$.

Also, suppose that $\Delta \ne 0$.


Then there is a conjugate of $\Delta$ with absolute value $\ge T^{−L}$.


Construction of the matrix $\boldsymbol M$ and outline of the proof

Let $c$ be a sufficiently large real number (which will be specified in due course).

Consider integers $L_0, L_1, S$ each $\ge 2$.

Let $L = \paren {L_0 + 1} \paren {L_1 + 1}$.

Observe that we can find such $L_0, L_1, S$ (and we do so) with:

$c L_0 \ln S \le L$, $c L_1 S \le L$, $L \le S^2$

For example, take $S$ large, and:

$L_0 = \floor {S \ln S}, L_1 = \floor {\dfrac S {2 \ln S} }$

Note that we could take $c = \ln \ln S$.


Consider the matrix $M$ described as follows.

Consider some arrangement $\tuple {\map {s_1} i, \map {s_2} i}$ of the $S^2$ integer pairs $\tuple {s_1, s_2}$ with $0 \le s_1 < S$ and $0 \le s_2 < S$.

Also, consider some arrangement $\tuple {\map u j, \map v j}$, with $1 \le j \le L$, of the integer pairs $\tuple {u, v}$ where $0 \le u \le L_0$ and $0 \le v \le L_1$.

Then we define:

$M = \sqbrk {\paren {\map {s_1} i + \map {s_2} i \beta}^{\map u j} \paren {\alpha^{\map {s_1} i + \map {s_2} i \beta} }^{\map v j} }$

so that $M$ is a $S^2 \times L$ matrix.


The idea of the proof is to:

$(1): \quad$ Consider the determinant $\Delta$ of an arbitrary $L \times L$ submatrix of $M$ (any one will do).
$(2): \quad$ Use Lemma 3 to obtain an upper bound $B_1$ for the absolute value of $\Delta$ (or, more specifically, an upper bound for $\ln \size \Delta$).
$(3): \quad$ Use Lemma 4 to motivate that if $\Delta \ne 0$, then $\size \Delta \ge B_2$ for some $B_2 > B_1$.
$(4): \quad$ Conclude that $\Delta$ must be $0$ and, hence, the rank of $M$ is less than $L$.
$(5): \quad$ Take a linear combination of the columns of $M$ to obtain an $\map F t$ as in Lemma 1 with less than $L$ roots but with $\map F {\map {s_1} i + \map {s_2} i \beta} = 0$ for $1 \le i \le L$.
$(6): \quad$ Conclude that $\beta$ is rational as required.


Upper bound for $\size \Delta$

We have not specified the arrangements defining $\tuple {\map {s_1} i, \map {s_2} i}$ and $\tuple {\map u j, \map v j}$.

Therefore it suffices to consider $\Delta = \det \sqbrk {\map {f_j} {\zeta_i} }$ where:

$\map {f_j} z = z^{\map u j} \alpha^{\map v j z} \quad \paren {1 \le j \le L}$

and:

$\zeta_i = \map {s_1} i + \map {s_2} i \beta \quad \paren {1 \le i \le L}$

Observe that $\map u j$ is a non-negative integer for each $j$.

Also, $\alpha^{\map v j z} = \map \exp {\map v j z \ln \alpha}$, where $\ln \alpha$ denotes the real value of the logarithm.

Hence, $\map {f_j} z$ is uniquely defined.

Then $\map {f_j} z$ represents an entire function for each $j$.

Observe that:

$\size {e^{z_1 z_2} } = e^{\map \Re {z_1 z_2} } \le e^{\size {z_1 z_2} } = e^{\size {z_1} \size {z_2} }$

for all complex numbers $z_1$ and $z_2$.



Hence, for any $R > 0$:

$\size {f_j}_R \le R^{\map u j} e^{\map v j R \size {\ln \alpha} }$

We use Lemma 3 with $r = S \paren {1 + \size {\beta} }$ and $R = e^2 r$.

Then for some constant $c_1 > 0$, we obtain:

\(\displaystyle \ln \size \Delta\) \(\le\) \(\displaystyle -L \paren {L - 1} + \ln L! + L \max_{1 \mathop \le j \mathop \le L} \set {\ln \size {f_j}_R}\)
\(\displaystyle \) \(\le\) \(\displaystyle -L \paren {L - 1} + L \ln L + L L_0 \ln R + L L_1 R \size {\ln \alpha}\)
\(\displaystyle \) \(\le\) \(\displaystyle -L^2 + c_1 \paren {L L_0 \ln S + L L_1 S}\)

The constant $c_1$ above is independent of $c$.

Therefore, if $c$ is sufficiently large (that is, $c \ge 4 c_1$), then:

$\ln \size \Delta \le -\dfrac {L^2} 2$


Lower bound for $\size \Delta$ if $\Delta \ne 0$

Suppose now that $T'$ is a positive rational integer for which $T' \alpha, T' \beta$ and $T'\alpha^\beta$ are all algebraic integers.

Then $T = \paren {T'}^{L_0 + 2 S L_1}$ has the property that $T$ times any element of $M$ (and, hence, $T$ times any element of the matrix defining $\Delta$) is an algebraic integer.

Therefore, by Lemma 4, if $\Delta \ne 0$, then there is a conjugate of $\Delta$ with absolute value $\ge T^{-L} = \paren {T'}^{-L L_0 - 2 S L L_1}$.

It may be reasonable to expect that a similar inequality might hold for $\size \Delta$ itself (rather than for the absolute value of a conjugate of $\Delta$).

It will be shown later that if $\Delta \ne 0$, then there is indeed a constant $c_2$ (independent of $c$) for which:

$(A) \quad \ln \size \Delta \ge -c_2 \paren {L L_0 \ln S + S L L_1}$


Conclusion

We see that, for $c$ sufficiently large ($c \ge 8 c_2$ will do), our upper bound for $\ln \size \Delta$ would contradict $(A)$, hence we obtain that $\Delta = 0$.

Since $\Delta = \det \sqbrk {\map {f_j} {\zeta_i} }$ as defined above, we get that the columns of $\sqbrk {\map {f_j} {\zeta_i} }$ must be linearly dependent over the real numbers.

In other words, there exist real numbers $b_j$, not all $0$, such that:

$\displaystyle \sum_{j \mathop = 1}^L b_j \map {f_j} {\zeta_i} = 0$ for $1 \le i \le L$

We now order the pairs $\tuple {u, v}$ in such a way that $\tuple {u, v} \le \tuple {u', v'}$ if and only if $v < v'$, or $v = v'$ and $u \le u'$.

We deduce that:

$\displaystyle \sum_{v \mathop = 0}^{L_1} \sum_{u \mathop = 0}^{L_0} b_{\paren {L_0 + 1} v + u + 1} \zeta_i^u \alpha^{v \zeta_i} = 0$ for $1 \le i \le L$

But:

$\displaystyle \sum_{v \mathop = 0}^{L_1} \sum_{u \mathop = 0}^{L_0} b_{\paren {L_0 + 1} v + u + 1} \zeta_i^u \alpha^{v \zeta_i} = \sum_{v \mathop = 0}^{L_1} \map {a_v} t e^{w_v t}$

where:

$\displaystyle \map {a_v} t = \sum_{u \mathop = 0}^{L_0} b_{\paren {L_0 + 1} v + u + 1} t^u, w_v = v \ln \alpha$

and:

$t = \zeta_i = \map {s_1} i + \map {s_2} i \beta$

Each of the $L$ values of $\zeta_i$ is a root of the function $\displaystyle \sum_{v \mathop = 0}^{L_1} \map {a_v} t e^{w_v t}$.

Since some $b_j \ne 0$, we deduce from Lemma 1 that there are at most $L_0 \paren {L_1 + 1} + \paren {L_1 + 1} - 1 < L$ distinct real roots.

Therefore, two roots $\zeta_i$ must be the same, and we can conclude that:

$\map {s_1} i + \map {s_2} i \beta = \map {s_1} {i'} + \map {s_2} {i'} \beta$

for some $i, i'$ with $1 \le i < i' \le L$.

On the other hand, the pairs $\tuple {\map {s_1} i, \map {s_2} i}$ and $\tuple {\map {s_1} {i'}, \map {s_2} {i'} }$ are necessarily distinct.

So we can conclude that $\beta$ is rational, completing the proof.


Proof of inequality $(A)$

To finish the proof, all we need to do is to show that if $\Delta \ne 0$, then $(A)$ holds.

Assume $\Delta \ne 0$ and recall that $T^L \Delta$ is an algebraic integer, where $T=(T')^{L_0+2SL_1}$ and $T'$ is a positive integer (depending only on $\alpha$, $\beta$, and $\alpha^\beta$).

For an algebraic number $w$, we denote by $\norm w$ the maximum of the absolute value of a conjugate of $w$.

Observe that:

$\norm {w + w'} \le \norm w + \norm {w'}$, $\norm {w w'} \le \norm w \norm {w'}$

and:

$\norm {\lambda w} = \size \lambda \norm w$

whenever $w, w'$ are algebraic numbers and $\lambda \in \Q$.

Then we obtain that:

$\norm {T^L \Delta} = T^L \norm \Delta \le T^L L! S^{L_0 L} \paren {1 + \norm \beta}^{L_0 L} \paren {1 + \norm \alpha}^{S L_1 L} \paren {1 + \norm {\alpha^\beta} }^{S L_1 L}$

We have that $T^L \Delta$ is an algebraic integer in $\map Q {\alpha, \beta, \alpha^\beta}$.

So we can deduce that $T^L \Delta$ is a root of an irreducible monic polynomial $\map g x$ of degree $N$ where $N$ is the product of the degrees of the minimal polynomials for $\alpha, \beta, \alpha^\beta$.

Note that each root of $\map g x$ is a conjugate of $T^L \Delta$.

Since:

the product of all the roots of $\map g x$ has absolute value $\size {\map g 0} \ge 1$

and:

each root of $\map g x$ has absolute value $\le \norm {T^L \Delta}$

it follows that:

\(\displaystyle \size {T^L \Delta}\) \(\ge\) \(\displaystyle \dfrac {\size {\map g 0} } {\norm {T^L \Delta}^{N - 1} }\)
\(\displaystyle \) \(\ge\) \(\displaystyle T^{-\paren {N - 1} L} \paren {L!}^{-N} S^{-N L_0 L} \paren {1 + \norm \beta}^{-N L_0 L} \paren {1 + \norm \alpha}^{-N S L_1 L} \paren {1 + \norm {\alpha^\beta} }^{-N S L_1 L}\)

Hence:

\(\displaystyle \ln \size \Delta\) \(\ge\) \(\displaystyle -N L \ln T - N L \ln L - N L_0 L \ln S - N L_0 L \map \ln {1 + \norm \beta}\)
\(\displaystyle \) \(\) \(\displaystyle - N S L_1 L \map \ln {1 + \norm \alpha} - N S L_1 L \map \ln {1 + \norm {\alpha^\beta} }\)

Recall that $T = \paren {T'}^{L_0 + 2 S L_1}$ and that $T'$ and $N$ are constants depending only on $\alpha$ and $\beta$.

It follows that $(A)$ holds, and the proof for $\alpha > 0$ and $\beta$ real follows.

$\blacksquare$


Source of Name

This entry was named for Alexander Osipovich Gelfond and Theodor Schneider.


Historical Note

The Gelfond-Schneider Theorem was proved in $\text {1934}$ – $\text {1935}$ by Alexander Osipovich Gelfond and Theodor Schneider.


Hilbert $23$

This problem is no. $7$ in the Hilbert $23$.


Sources