This is the second part in a 3-part blog post in which we prove that is transcendental.
Recall that in step 1 we proved the following lemma.
Lemma 1. Suppose is a root of the polynomial . Let be a polynomial and . Then there exist such that , where .
Now we are ready pick a specific polynomial for Lemma 1. Since there are infinitely many primes, we can find a prime number greater than both and . Let
Our goal for this step is to show that is a nonzero integer. In fact we will show is that it is an integer not divisible by , which implies that it is nonzero.
We will accomplish this by proving that for , (and hence ) is an integer divisible by , but that is an integer that is not divisible by .
First we state but do not prove the following lemma. (This is a homework problem in Herstein.)
Lemma 2. If is a polynomial with integer coefficients and , then for , is a polynomial with integer coefficients, each of which is divisible by .
Applying Lemma 2 to our function we see that when , is a polynomial with integer coefficients, each of which is divisible by . In particular, when and is any integer, is an integer that is divisible by .
Now let us restrict our attention to . From the definition of we see that is a root of with multiplicity . The following lemma tells us that . (We leave the proof of this lemma as an exercise for the reader.)
Lemma 3. Suppose is a root of a polynomial with multiplicity . Then for .
We know from above that each of the terms is an integer divisible by . Thus is an integer divisible by .
Now let us consider . By the definition of , is a root with multiplicity . By Lemma 3, .
From the result above, we know that are all divisible by . What about ?
If we multiply out the terms in , we obtain
and hence . Since , does not divide . Therefore does not divide . Moreover, , so does not divide . We conclude that does not divide .
Finally this implies that is an integer not divisible by . In particular, it is a nonzero integer.
We are almost finished with the proof. In the next step we will show that , thus it cannot be a nonzero integer and we will obtain our sought-after contradiction.