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The formulation of Eisenstein's criterion in my text is:

Let $a(x) = a_0 + a_1x+\cdots+a_nx^n$ be a polynomial with integer coefficients. Suppose there is a prime number $p$ which divides every coefficient of $a(x)$ except the leading coefficient $a_n$; suppose $p$ does not divide $a_n$ and $p^2$ does not divide $a_0$. Then $a(x)$ is irreducible over $\mathbb{Q}$.

I want to understand the proof of this given on Wikipedia. The proof outline is the following. Assume by contradiction that a polynomial $Q=q_n x^n + \cdots + q_0$ satisfying the conditions for some $p$ is reducible such that $Q=G*H=(g_mx^m+\cdots+g_0)(h_kx^k+\cdots+h_0)$. Then send both sides of $Q=G*H$ to $\mathbb{Z}_p[x]$ (using the homomorphism induced by the natural mapping from $\mathbb{Z} \rightarrow \mathbb{Z_p})$ and you are necessarily left with $\bar{q_n}x^n = (\bar{g_m}x^ m)( \bar{h_k}x^k)$. This implies that $g_0$ and $h_0$ are multiples of $p$, contradicting the assumption that $p^2\nmid q_0$.

Why is the bolded statement above true? How can I prove that there aren't other polynomials $G',H' \in \mathbb{Z}_p[x]$ with more terms than just the leading terms $\bar{g_m}x^m, \bar{h_k}x^k$, respectively, such that $\bar{q_n}x^n = G'*H'$ ?

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    $\begingroup$ $\mathbb{Z}_{p}[x]$ is a unique factorization domain, so the situation you describe in particular is impossible. $\endgroup$ Mar 12, 2016 at 1:03
  • $\begingroup$ Because you know that $p$ divides each $q_i$ for $i \leq n-1$, so only the leading coefficient on the left can remain, and if you expand the RHS, only that term will have the matching degree. I am not sure what your second question is asking, since there are of course many polynomials $G,H$ such that $GH = \bar q_nx^n$ $\endgroup$
    – user208649
    Mar 12, 2016 at 1:07

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Let $i$ be the smallest index such that $\bar{g}_i\neq 0$, and $j$ the smallest index such that $\bar{h}_j\neq 0$.

Then the smallest degree term of $\overline{G}\cdot\overline{H}$ is $\bar{g}_i\bar{h}_jx^{i+j}$. On the other hand, we know that $\overline{G}\cdot\overline{H}=\bar{q}_nx^n$, hence $i+j=n$. Since $i\leq m$ and $j\leq k$, and $m+k=n$, this is only possible if $i=m$ and $j=k$. Therefore all the lower degree terms must be zero, as claimed.

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