Why do we conclude that p|exactly one of b0,c0?

[evinda]

Hi! :)

I am looking at the proof of the Eisenstein's criterion
(Let $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0 \in \mathbb{Z}$ and let's suppose that there is a prime $p$ such that
$p \nmid a_n$ & $p \mid a_i \forall i=0,1,...,n-1$ & $p^2 \nmid a_0$.
Then $f(x)$ is unfactorable into the product of non constant polynomials with rational coefficients):

We suppose that the conditions: $p \nmid a_n$ & $p \mid a_i \forall i=0,1,...,n-1$ & $p^2 \nmid a_0$ are satisfied.Also,we suppose that $f(x)$ is not irreducible.
Then,from Gauss's lemma, it will be like that: $f(x)=(b_mx^m+...+b_1x+b_0) \cdot (c_kx^k+...+c_1x+c_0) , b_i,b_j \in \mathbb{Z}, b_mc_k \neq 0, m+k=n$

So,we get:

$(1) b_0c_0=a_0$
$(2) b_1c_0+c_1b_0=a_1$
$(3) c_0b_2+c_1b_1+c_2b_0=a_2$
.
.
.
.
$(n) b_mc_k=a_n$

$p \mid a_0 \Rightarrow p \mid b_0c_0 \Rightarrow p \mid$ at least one of $b_0,c_0$

Then, because of the fact that $p^2 \nmid a_0 \Rightarrow$ $p$ does not divide both $b_0,c_0 \Rightarrow p \mid$ exactly one of $b_0,c_0$

Why is it like that?? I haven't understood it..Isn't it possible that,for example, $p^2 \mid b_0$ ?? (Thinking)

 

[Evgeny.Makarov]

$p \mid a_0 \Rightarrow p \mid b_0c_0 \Rightarrow p \mid$ at least one of $b_0,c_0$

Then, because of the fact that $p^2 \nmid a_0 \Rightarrow$ $p$ does not divide both $b_0,c_0 \Rightarrow p \mid$ exactly one of $b_0,c_0$

Why is it like that?? I haven't understood it..Isn't it possible that,for example, $p^2 \mid b_0$ ??

The fact that $p^2\nmid b_0$ is true, but it is not stated in the first two paragraphs of the quote above. They only claim something about the divisibility by $p$ (not $p^2$), namely, that
\[
(p\mid b_0\lor p\mid c_0)\land\neg(p\mid b_0\land p\mid c_0)\qquad(*)
\]
First, (*) is true because if $p\mid b_0$ and $p\mid c_0$, then there exist integers $m,n$ such that $b_0=pm$ and $c_0=pn$, so $a_0=b_0c_0=p^2mn$, i.e., $p^2\mid a_0$, contrary to the assumption. Second, you are right that $p^2\nmid b_0$ and $p^2\nmid c_0$. If $p^2\mid b_0$, then $p^2\mid b_0c_0=a_0$.

 

[Deveno]

Primes have the property that if $p|ab$ then either $p|a$ or $p|b$ (or both).

Now, here if $p|b_0c_0$, by the above, either $p|b_0$ or $p|c_0$. Could it be both?

No, because then we would have $p^2|b_0c_0 = a_0$ contrary to our original condition on $p$.

So it has to be one or the other, since both can't happen.

Now, if $p|b_0$ and $p^2|b_0$ then $p^2|a_0$ (which cannot be the case). So we only have ONE factor of $p$ in the product $b_0c_0$, and it has to be in $b_0$ or $c_0$.

 

[evinda]

The fact that $p^2\nmid b_0$ is true, but it is not stated in the first two paragraphs of the quote above. They only claim something about the divisibility by $p$ (not $p^2$), namely, that
\[
(p\mid b_0\lor p\mid c_0)\land\neg(p\mid b_0\land p\mid c_0)\qquad(*)
\]
First, (*) is true because if $p\mid b_0$ and $p\mid c_0$, then there exist integers $m,n$ such that $b_0=pm$ and $c_0=pn$, so $a_0=b_0c_0=p^2mn$, i.e., $p^2\mid a_0$, contrary to the assumption. Second, you are right that $p^2\nmid b_0$ and $p^2\nmid c_0$. If $p^2\mid b_0$, then $p^2\mid b_0c_0=a_0$.

Primes have the property that if $p|ab$ then either $p|a$ or $p|b$ (or both).

Now, here if $p|b_0c_0$, by the above, either $p|b_0$ or $p|c_0$. Could it be both?

No, because then we would have $p^2|b_0c_0 = a_0$ contrary to our original condition on $p$.

So it has to be one or the other, since both can't happen.

Now, if $p|b_0$ and $p^2|b_0$ then $p^2|a_0$ (which cannot be the case). So we only have ONE factor of $p$ in the product $b_0c_0$, and it has to be in $b_0$ or $c_0$.

I understand..Thank you very much! :D

 

[blue_raver22]

Hello there! I can explain why we conclude that p|exactly one of b0,c0. In the Eisenstein's criterion, we are looking at the conditions for a polynomial to be irreducible. This means that it cannot be factored into the product of non-constant polynomials with rational coefficients.

In the given proof, we are assuming that the polynomial f(x) is not irreducible and can be factored into two polynomials, which we will call g(x) and h(x). We also know from Gauss's lemma that the coefficients of g(x) and h(x) must be integers.

Now, let's look at the coefficients of f(x). We know that p does not divide the leading coefficient, a_n, and it divides all the other coefficients, a_i. This means that p cannot divide both b_0 and c_0, as that would contradict the fact that p does not divide a_n.

So, we can conclude that p divides either b_0 or c_0, but not both. This is because if p divided both b_0 and c_0, it would also divide their product, b_0c_0, which is equal to a_0. But we know that p^2 does not divide a_0, so this is not possible.

Therefore, we can conclude that p must divide exactly one of b_0 and c_0. This is important because if p divided both b_0 and c_0, then we would have a factorization of f(x) that does not satisfy the conditions of Eisenstein's criterion.

I hope this helps clarify why we conclude that p|exactly one of b0,c0. Let me know if you have any other questions!