Gelfond-Schneider Theorem

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

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 $\alpha,\beta\in\mathbb R$ and $\alpha>0$, assuming that $\exp(\beta\ln\alpha)$ 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 $(s_1, s_2)$ and $(s'_1, s'_2)$ 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 $0 \le s_1, s_2, s'_1, s'_2 < S$.

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 $L = \left({L_0 + 1}\right) \left({L_1 + 1}\right)$.

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:
 * $L_0 = \left \lfloor {S \ln S}\right \rfloor, L_1 = \left \lfloor {S / (2\ln S)}\right \rfloor$

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

Consider the matrix $M$ described as follows.

Consider some arrangement $\left({s_1 \left({i}\right), s_2 \left({i}\right)}\right)$ of the $S^2$ integral pairs $\left({s_1, s_2}\right)$ with $0 \le s_1 < S$ and $0 \le s_2 < S$.

Also, consider some arrangement $\left({u \left({j}\right), v \left({j}\right)}\right)$, with $1 \le j \le L$, of the integral pairs $\left({u, v}\right)$ where $0 \le u \le L_0$ and $0 \le v \le L_1$.

Then we define:
 * $M = \left[{ \left({s_1 \left({i}\right) + s_2 \left({i}\right) \beta}\right)^{u \left({j}\right)} \left({\alpha^{s_1 \left({i}\right) + s_2 \left({i}\right) \beta} }\right)^{v \left({j}\right)} }\right]$

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 $\ln \left|{\Delta}\right|$).


 * $(3): \quad$ Use Lemma 4 to motivate that if $\Delta \ne 0$, then $\left|{\Delta}\right| \ge B_2$ 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 $F \left({t}\right)$ as in Lemma 1 with less than $L$ roots but with $F \left({s_1 \left({i}\right) + s_2 \left({i}\right) \beta}\right) = 0$ for $1 \le i \le L$.


 * $(6): \quad$ Conclude that $\beta$ is rational as required.

Upper bound for $\boldsymbol{|\Delta|}$
We have not specified the arrangements defining $\left({s_1 \left({i}\right), s_2 \left({i}\right)}\right)$ and $\left({u \left({j}\right), v \left({j}\right)}\right)$.

Therefore it suffices to consider $\Delta = \det \left[{f_j \left({\zeta_i}\right)}\right]$ where:
 * $f_j \left({z}\right) = z^{u \left({j}\right)} \alpha^{v \left({j}\right) z} \quad (1 \le j \le L)$

and:
 * $\zeta_i = s_1 \left({i}\right) + s_2 \left({i}\right) \beta \quad (1 \le i \le L)$

Observe that $u \left({j}\right)$ is a non-negative integer for each $j$.

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

Hence, $f_j \left({z}\right)$ is uniquely defined.

Then $f_j \left({z}\right)$ represents an entire function for each $j$.

Observe that:
 * $\left|{e^{z_1 z_2}}\right| = e^{\Re \left({z_1 z_2}\right)} \le e^{\left|{z_1 z_2}\right|} = e^{\left|{z1}\right| \left|{z2}\right|}$

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

Hence, for any $R > 0$:
 * $\left|{f_j}\right|_R \le R^{u \left({j}\right)} e^{v \left({j}\right) R \left|{\ln \alpha}\right|}$

We use Lemma 3 with $r = S \left({1 + \left|{\beta}\right|}\right)$ and $R = e^2 r$.

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

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

Therefore, if $c$ is sufficiently large (that is,$ c \ge 4c_1$), then:
 * $\ln \left|{\Delta}\right| \le −\dfrac{L^2} 2$

Lower bound for $\boldsymbol{|\Delta|}$ if $\boldsymbol\Delta\neq 0$
Suppose now that $T'$ is a positive rational integer for which $T' \alpha, T' \beta$ and $T'\alpha^\beta$ are all algebraic integers.

Then $T = \left({T'}\right)^{L_0 + 2 S L_1}$ 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:
 * $\ge T^{−L} = \left({T'}\right)^{−L L_0 − 2 S L L_1}$

It may be reasonable to expect that a similar inequality might hold for $\left|{\Delta}\right|$ 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:
 * $(A) \qquad \ln \left|{\Delta}\right| \ge −c_2 \left({L L_0 \ln S + S L L_1}\right)$

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

Since $\Delta = \det \left[{f_j \left({\zeta_i}\right)}\right]$ as defined above, we get that the columns of $\left[{f_j \left({\zeta_i}\right)}\right]$ must be linearly dependent over the real numbers.

In other words, there exist real numbers $b_j$, not all $0$, such that:
 * $\displaystyle \sum_{j \mathop = 1}^L b_j f_j \left({\zeta_i}\right) = 0$ for $1 \le i \le L$

We now order the pairs $(u,v)$ in such a way that $(u,v)\le (u',v')$ if and only if $v<v'$, or $v=v'$ and $u\le u'$. We deduce that:
 * $\displaystyle \sum_{v \mathop = 0}^{L_1} \sum_{u \mathop = 0}^{L_0} b_{\left({L_0 + 1}\right) v + u + 1} \zeta_i^u \alpha^{v \zeta_i} = 0$ for $1 \le i \le L$

But
 * $\displaystyle \sum_{v \mathop = 0}^{L_1} \sum_{u \mathop = 0}^{L_0} b_{\left({L_0 + 1}\right) v + u + 1} \zeta_i^u \alpha^{v \zeta_i} = \sum_{v \mathop = 0}^{L_1} a_v \left({t}\right) e^{w_v t}$

where:
 * $\displaystyle a_v \left({t}\right) = \sum_{u \mathop = 0}^{L_0} b_{\left({L_0 + 1}\right) v + u + 1} t^u, w_v = v \ln \alpha$

and:
 * $t = \zeta_i = s_1 \left({i}\right) + s_2 \left({i}\right) \beta$

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

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

Therefore, two roots $\zeta_i$ must be the same, and we can conclude that:
 * $s_1 \left({i}\right) + s_2 \left({i}\right) \beta = s_1\left({i'}\right) + s_2 \left({i'}\right) \beta$

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

On the other hand, the pairs $\left({s_1 \left({i}\right), s_2 \left({i}\right)}\right)$ and $\left({s_1 \left({i'}\right) , s_2 \left({i'}\right)}\right)$ are necessarily distinct.

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

Proof of inequality $\boldsymbol{(A)}$
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 $\left\Vert{w}\right\Vert$ the maximum of the absolute value of a conjugate of $w$.

Observe that:


 * $\left\Vert{w+w'}\right\Vert \le \left\Vert{w}\right\Vert + \left\Vert{w'}\right\Vert$, $\left\Vert{ww'}\right\Vert \le \left\Vert{w}\right\Vert \left\Vert{w'}\right\Vert$, and $\left\Vert{\lambda w}\right\Vert = \left\vert{\lambda}\right\vert \left\Vert{w}\right\Vert$

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

Then we obtain that:
 * $\left\Vert{T^L \Delta}\right\Vert = T^L \left\Vert{\Delta}\right\Vert \le T^L L!S^{L_0 L} \left({1 + \left \Vert{ \beta }\right \Vert}\right)^{L_0 L} \left(1 + \left\Vert{\alpha}\right\Vert\right)^{S L_1 L} \left(1 + \left\Vert{\alpha^\beta}\right\Vert\right)^{S L_1 L}$

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

So we can deduce that $T^L \Delta$ is a root of an irreducible monic polynomial $g \left({x}\right)$ 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 $g \left({x}\right)$ is a conjugate of $T^L \Delta$.

Since:
 * the product of all the roots of $g \left({x}\right)$ has absolute value $\left|{g \left({0}\right)}\right| \ge 1$

and:
 * each root of $g \left({x}\right)$ has absolute value $\le \left\Vert{T^L \Delta}\right\Vert$

it follows that:

Hence:

Recall that $T = (T')^{L_0 + 2 S L_1}$ 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.