Are my conjecture and proof below correct and well formulated?

How can they be improved?

It’s a new conjecture. It’s an extension of Lin’s theorem in (Lin 1983) which is used in (Chow 1999) – see both references at the bottom of this page.

The proof is very very simple because the preconditions of my conjecture contain the preconditions of Lin’s theorem except irreducibility. The irreducibility follows directly from very simple arguments from elementary algebra. But I need qualified confirmations.

$mathbb{L}$ are the elementary numbers, or Liouvillian numbers:

http://mathworld.wolfram.com/LiouvillianNumber.html, https://en.wikipedia.org/wiki/Elementary_number, http://mathworld.wolfram.com/ElementaryNumber.html, (Chow 1999).

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**Theorem** (Lin 1983):

If Schanuel’s conjecture is true and $tilde{P}(X,Y)inoverline{mathbb{Q}}(X,Y)$ is an irreducible polynomial involving both $X$ and $Y$ and $tilde{P}left(alpha,e^{alpha}right)=0$ for some non-zero $alpha$ in $mathbb{C}$, then $alpha$ is not in $mathbb{L}$.

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**Conjecture:**

Let

$f$ be an elementary function,

$ninmathbb{N}_{ge 1}$,

$P(x,y)inoverline{mathbb{Q}}(x,y)setminus(overline{mathbb{Q}}(x)cupoverline{mathbb{Q}}(y))$ irreducible over $mathbb{C}$,

$Q(x,y)inoverline{mathbb{Q}}(x,y)$ not zero,

$p(x)inoverline{mathbb{Q}}(x)$,

$q(x)inoverline{mathbb{Q}}(x)$ not zero,

$R(x,y)=frac{P(x,y)}{Q(x,y)cdot q(x)^n}$,

$r(x)=frac{p(x)}{q(x)}$ not constant

so that

$P(x,y)$ and $Q(x,y)$ coprime over $mathbb{C}$,

$p(x)$ and $q(x)$ coprime over $mathbb{C}$.

Suppose Schanuel’s conjecture is true.

If a $tilde{P}(x,y)inoverline{mathbb{Q}}(x,y)setminus(overline{mathbb{Q}}(x) cup overline{mathbb{Q}}(y))$ of $degree_x=n$ with $frac{P(x,y)}{{q(x)}^n}=tilde{P}(r(x),y)$ exists and $R(f(z_0),e^{r(f(z_0))})=0$ for $z_0inmathbb{C}$ and $r(f(z_0))neq 0$, then $z_0$ is not an elementary number.

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**Proof draft:**

We will prove our conjecture using Lin’s theorem. Lin’s theorem makes statements about equations of the form $tilde{P}(alpha,e^alpha)=0$, wherein $alphainmathbb{C}$ and $tilde{P}(x,y)$ is irreducible over $mathbb{C}$. We consider the equation $R(f(z_0),e^{r(f(z_0))})=0$ from our conjecture and will show that under the preconditions of our conjecture there exists an equation equivalent to it in the form from Lin’s theorem.

Setting $f(z_0)=x$ and $e^{r(f(z_0))}=y$ gives the equation $R(x, y)=0$. We consider it in the form $frac{P(x, y)}{Q(x, y)cdot q(x)^n}=0$ from the preconditions of our conjecture. Because $P(x,y)$ and $Q(x,y)$ are coprime over $mathbb{C}$ by precondition, we can multiply both sides of the equation by $Q(x,y)$ and get the equation $frac{P(x,y)}{q(x)^n}=0$ equivalent to it. By precondition, $frac{P(x,y)}{{q(x)}^n}=tilde{P}(r(x),y)$ holds. So our equation becomes $tilde{P}(r(x),y)=0$. Backsubstitution of $x$ and $y$ gives $tilde{P}(r(f(z_0)),e^r(f(z_0)))=0$.

We have $forall z_0inmathbb{C}colon r(f(z_0))inmathbb{C}$. We can therefore set $r(f(z_0))=alpha$, which gives the form $tilde{P}(alpha,e^alpha)=0$ from Lin’s theorem. To fulfill all preconditions from Lin’s theorem, we still have to show that $tilde{P}(x,y)$ is irreducible over $mathbb {C}$.

If $tilde{P}(x,y)$ is reducible over $mathbb{C}$, then

$n_1inmathbb{N}_{ge 1}$,

$n_2inmathbb{N}_0$,

$a_0,…,a_{n_1}inmathbb{C}(y)$,

$b_0,…,b_{n_2}inmathbb{C}(y)$ not all constant

exist so that

$a_{n_1},b_{n_2}neq 0$,

$tilde{P}(x,y)=(a_0+a_1x^1+…+a_{n_1}x^{n_1}) cdot(b_0+b_1x^1+…+b_{n_2}x^{n_2})$.

The highest degree regarding $x$ of the polynomial product on the right-hand side of the equation is $n_1+n_2$. This must be equal to the highest degree regarding $x$ of the polynomial $tilde{P}(x,y)$ on the left-hand side of the equation. This was $n$ according to the preconditions of our conjecture. We have therefore $n_1+n_2=n$.

By substitution of $x$ by $r(x)$, we get

$$tilde{P}(r(x),y))=(a_0+a_1r(x)^1+…+a_{n_1}r(x)^{n_1})cdot(b_0+b_1r(x)^1+…+b_{n_2}r(x)^{n_2}).$$

Because, according to the preconditions of our conjecture, $r(x)=frac{p(x)}{q(x)}$, and $p(x)$ and $q(x)$ are coprime over $mathbb{C}$, we have

$$tilde{P}(r(x),y))=(a_0+a_1left(frac{p(x)}{q(x)}right)^1+…+a_{n_1}left(frac{p(x)}{q(x)}right)^{n_1})cdot(b_0+b_1left(frac{p(x)}{q(x)}right)^1+…+b_{n_2}left(frac{p(x)}{q(x)}right)^{n_2})$$

$$=(a_0+a_1frac{p(x)^1}{q(x)^1}+…+a_{n_1}frac{p(x)^{n_1}}{q(x)^{n_1}})cdot(b_0+b_1frac{p(x)^1}{q(x)^1}+…+b_{n_2}frac{p(x)^{n_2}}{q(x)^{n_2}})$$

$$=(a_0+frac{a_1p(x)^1}{q(x)^1}+…+frac{a_{n_1}p(x)^{n_1}}{q(x)^{n_1}})cdot(b_0+frac{b_1p(x)^1}{q(x)^1}+…+frac{b_{n_2}p(x)^{n_2}}{q(x)^{n_2}})$$

$$=frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+…+a_{n_1}p(x)^{n_1})}{q(x)^{n_1}}cdotfrac{(b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+…+b_{n_2}p(x)^{n_2})}{q(x)^{n_2}}$$

$$=frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+…+a_{n_1}p(x)^{n_1})cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+…+b_{n_2}p(x)^{n_2})}{q(x)^{n_1+n_2}}.$$

Because, as stated above, $n_1+n_2=n$, we have

$$tilde{P}(r(x),y))=frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+…+a_{n_1}p(x)^{n_1})cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+…+b_{n_2}p(x)^{n_2})}{q(x)^n}.$$

According to the preconditions of our conjecture, $tilde{P}(r(x),y))=frac{P(x,y)}{q(x)^n}$. We have therefore

$$frac{P(x,y)}{q(x)^n}=frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+…+a_{n_1}p(x)^{n_1})cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+…+b_{n_2}p(x)^{n_2})}{q(x)^n}.$$

Multiplying by $q(x)^n$ yields

$$P(x,y)=(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+…+a_{n_1}p(x)^{n_1})cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+…+b_{n_2}p(x)^{n_2}).$$

If $tilde{P}(x,y)$ is reducible over $mathbb{C}$, then $P(x,y)$ is accordingly reducible over $mathbb{C}$. Because, according to the preconditions of our conjecture, $P(x,y)$ is irreduible over $mathbb{C}$, $tilde{P}(x,y)$ is irreducible over $mathbb{C}$.

Thereby all preconditions from Lin’s theorem are fulfilled and we can apply it to our equation. According to Lin’s theorem, $alpha$ is not in $mathbb{L}$. Because $alpha=r(f(z_0))$, $r(f(z_0))notinmathbb{L}$. Because $r$ and $f$ are elementary functions, $r(f(z))inmathbb{L}$ holds for all $zcolon zin dom(rcirc f) land zinmathbb{L}$. But because, as just stated, $r(f(z_0))notinmathbb{L}$, $z_0notinmathbb{L}$, so $z_0$ is not an elementary number.

q.e.d.

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(Lin 1983) Ferng-Ching Lin: Schanuel’s Conjecture Implies Ritt’s Conjectures. Chin. J. Math. 11 (1983) (1) 41-50

(Chow 1999) Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448