MHB 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