Eisenstein's criterion proof

[Hill]

Homework Statement

Prove Eisenstein's criterion, see below

Relevant Equations

$a_n X^n + ... + a_0 = (b_k X^k + ...+ b_0)(c_m X^m+ ... +c_0)$

My solution:
Assume it is reducible, i.e., $a_n X^n + ... + a_0 = (b_k X^k + ...+ b_0)(c_m X^m+ ... +c_0)$.
$a_0=b_0 c_0$. Since $p \mid a_0$, either $p \mid b_0$ or $p \mid c_0$, but not both, because $p^2 \nmid a_0$. Assume $p \mid b_0, p \nmid c_0$.
$a_1=b_0 c_1+b_1 c_0$.
$p \mid a_1, p \mid b_0, p \nmid c_0 \Rightarrow p \mid b_1$.
Continuing the same steps up the powers, we get $p \mid b_k$. But since $a_n=b_k c_m$ it makes $p \mid a_n$, which contradicts the statement, $p \nmid a_n$.

The book hints to reduce the coefficients $a_i, b_i, c_i$ modulo p and to consider the relation between the reduced polynomials. I can convert my solution using this hint as follows:
After the reduction modulo p, we get $\bar a_n X^n = (\bar b_k X^k + ...+ \bar b_0)(\bar c_m X^m+ ... +\bar c_0)$.
$0=\bar b_0 \bar c_0$. Then, either $\bar b_0=0$ or $\bar c_0=0$, but not both, because $p^2 \nmid a_0$. Assume $\bar b_0=0 , \bar c_0 \neq 0$.
$0=\bar b_1 \bar c_0 \Rightarrow \bar b_1=0$. Etc. until $\bar a_n=0$ which leads to contradiction.

I wonder if the book's suggestion allows for another, better proof that I don't see.
??

 

[WWGD]

Slightly OT: Your title mentions Einstein, not Eisenstein.

 

[fresh_42]

Homework Statement: Prove Eisenstein's criterion, see below
Relevant Equations: $a_n X^n + ... + a_0 = (b_k X^k + ...+ b_0)(c_m X^m+ ... +c_0)$

View attachment 354698
My solution:
Assume it is reducible, i.e., $a_n X^n + ... + a_0 = (b_k X^k + ...+ b_0)(c_m X^m+ ... +c_0)$.
$a_0=b_0 c_0$. Since $p \mid a_0$, either $p \mid b_0$ or $p \mid c_0$, but not both, because $p^2 \nmid a_0$. Assume $p \mid b_0, p \nmid c_0$.
$a_1=b_0 c_1+b_1 c_0$.
$p \mid a_1, p \mid b_0, p \nmid c_0 \Rightarrow p \mid b_1$.
Continuing the same steps up the powers, we get $p \mid b_k$. But since $a_n=b_k c_m$ it makes $p \mid a_n$, which contradicts the statement, $p \nmid a_n$.

The book hints to reduce the coefficients $a_i, b_i, c_i$ modulo p and to consider the relation between the reduced polynomials. I can convert my solution using this hint as follows:
After the reduction modulo p, we get $\bar a_n X^n = (\bar b_k X^k + ...+ \bar b_0)(\bar c_m X^m+ ... +\bar c_0)$.
$0=\bar b_0 \bar c_0$. Then, either $\bar b_0=0$ or $\bar c_0=0$, but not both, because $p^2 \nmid a_0$. Assume $\bar b_0=0 , \bar c_0 \neq 0$.
$0=\bar b_1 \bar c_0 \Rightarrow \bar b_1=0$. Etc. until $\bar a_n=0$ which leads to contradiction.

I wonder if the book's suggestion allows for another, better proof that I don't see.
??

That's essentially the proof van der Waerden gives in his book. Whether you use the language modulo p or the language of divisibility is not really a difference, just words.

 

[fresh_42]

Slightly OT: Your title mentions Einstein, not Eisenstein.

At least both had a Jewish background, although Eisenstein's parents converted to Protestantism (before 1823) and were born in Germany, Eisenstein 1823 in Berlin, Einstein 1879 in Ulm.

Let's call it a draw:

 

[Frabjous]

Slightly OT: Your title mentions Einstein, not Eisenstein.

Somebody changed the title. Earlier it was Eisenstein.

 

[Hill]

Somebody changed the title. Earlier it was Eisenstein.

Yes, it was.

 

[fresh_42]

Yes, it was.

Auto-correct by ignorance I guess.

 

[WWGD]

Somebody changed the title. Earlier it was Eisenstein.

Sure it wasn't Einstein. This thread isn't _ that_ old.

 

[fresh_42]

Sure it wasn't Einstein. This thread isn't _ that_ old.

But Eisenstein was older! And my book by van der Waerden is pretty old, too. I remember that I once almost had written a PM to John Baez who once visited us and began with "Hello Joan!" I recognized it in time, but it was close.

 

[WWGD]

But Eisenstein was older! And my book by van der Waerden is pretty old, too. I remember that I once almost had written a PM to John Baez who once visited us and began with "Hello Joan!" I recognized it in time, but it was close.

He in return gave you Diamonds and Rust. Right? Edit: The two are likely related. It must have been a treat to have someone of John Baez' caliber as a guest. His Math writeups are better than those of many Mathematicians.

 

[Frabjous]

Sure it wasn't Einstein. This thread isn't _ that_ old.

Yes, because I was wondering if they had mispelled Einstein when I saw the title.

 

[Hill]

Here is part (c) of the same exercise, i.e., related to the Eisenstein's criterion; part (a) is in the post #1 above.

If it said $f(X)=p^{-1} X^n + pX +1$, I would multiply it by $p$ and apply the criterion. However, as it is written I cannot solve it with or without the criterion. Is it a typo?

 

[WWGD]

Here is part (c) of the same exercise, i.e., related to the Eisenstein's criterion; part (a) is in the post #1 above.

View attachment 354706

If it said $f(X)=p^{-1} X^n + pX +1$, I would multiply it by $p$ and apply the criterion. However, as it is written I cannot solve it with or without the criterion. Is it a typo?

Positive Integer 1, then you would get $P^0X^1+ PX+1$. I think you're using the case for $n=0$, which isn't positive.

 

[fresh_42]

Here is part (c) of the same exercise, i.e., related to the Eisenstein's criterion; part (a) is in the post #1 above.

View attachment 354706

If it said $f(X)=p^{-1} X^n + pX +1$, I would multiply it by $p$ and apply the criterion. However, as it is written I cannot solve it with or without the criterion. Is it a typo?

Assume $f(X)$ is reducible and consider $p\cdot f(X/p).$