# How can we conclude $h(x)$ is constant in Eisenstein's Criterion proof

I came across the following proof of Eisenstein's criterion from the book Galois Theory by Joseph Rotman.

Let $$f(x) = a_o +a_1x + ... +a_nx^n \in Z [x]$$. If there is a prime p dividing $$a_j$$ for all i < n, but with p not dividing $$a_n$$ and $$p^2$$ not dividing $$a_o$$, then $$f(x)$$ is irreducible in $$\mathbb Q[x]$$.

Let $$f(x) = g(x)h(x) = (bo + b_1x + ... + b_mx^m)(c_o + c_1x + ... + c_kx^k)$$ we may assume that both g and h lie in Z[x]. By hypothesis, $$p|a_0 = b_oc_o$$ so that $$p | b_0$$ or $$p | c_0$$, by Euclid's lemma in Z; since $$p^2$$ does not divide $$a_o$$, only one of them is divisible by $$p$$, say, $$p | c_0$$ but p does not divide $$b_o$$. The leading coefficient $$a_n = b_mc_k$$ is not divisible by p, so that p does not divide $$c_k$$ (or $$b_m$$). Let $$c_r$$ be the first coefficient not divisible by p (so p does divide $$c_o, ... , c_{r-1}$$). If r < n, then $$p|a_n$$ and $$b_oc_r = a_r - (b_1c_{r-1} + ... + b_rc_o)$$ is divisible by p; hence $$p | b_oc_n$$ contradicting Euclid's lemma (because p divides neither factor). It follows that r = n, hence k = 0, and $$h(x)$$ is constant. Therefore, f(x) is irreducible.

Shouldn't we conclude that $$g(x)$$ is constant from $$r=n$$, because $$h(x)$$ would have degree $$n$$ and therefore $$g(x)$$ would have degree 0, otherwise the degree of $$g(x)h(x)>n$$. How do we conclude $$k=0$$?

• If $k=0$ then the polynomial $h(x)=c_0+\ldots+c_kx^k$ amounts to $h(x)=c_0$ which is constant Commented May 3, 2021 at 22:22
• Yes, I know that but how do we conclude that $k = 0$? Commented May 3, 2021 at 22:23
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• The sum of the degrees must equal $n$, so $r+k=n$. Once you know $r=n$, you know $k=0$. Commented May 3, 2021 at 22:31
• I didn't know that about images, I will try to re-write the question. Commented May 3, 2021 at 22:33

It’s a typo. It should be either “$$m=0$$” (instead of $$k=0$$) or “$$k=n$$” (instead of $$k=0$$).
This follows because we know $$c_k$$ is not divisible by $$p$$, $$k\leq n$$, and $$r$$ is the smallest index such that $$c_r$$ is not divisible by $$p$$. Therefore, $$r\leq k\leq n$$. Since you just established that $$r=n$$, it follows that $$n=r\leq k\leq n$$, so $$k=n$$ and since $$m+k=n$$, then $$m=0$$.