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Why is it that the Eisenstein Criterion would work when substituting $x$ with $x + 1$? Why is it OK to do this for polynomials in $\mathbb{Q}[x]$?

Thank you

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2 Answers 2

Well, here is a proof.

Now if $f(x+1) = q(x)r(x)$, then $f(x) = f((x-1)+1) = q(x-1)r(x-1)$.

And if $f(x) = q(x)r(x)$, then $f(x+1) = q(x+1)r(x+1)$.

Notice also that $g(x)$, $g(x-1)$ and $g(x+1)$ always have the same degree. Thus we may conclude by above that $f(x)$ is irreducible if and only if $f(x+1)$ is irreducible.

With the same proof you can generalize this to prove that for all $c \in \mathbb{Z}$, $f(x)$ is irreducible if and only if $f(x+c)$ is irreducible.

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I think the trick you are referring to is the fact that a polynomial $f(x)\in \mathbb{Q}[x]$ is irreducible if and only if the polynomial $f(x+1)$ is irreducible in $\mathbb{Q}[x]$. This can make applying the Eisenstein criterion possible when it is not initially, which is why it is useful to do this.

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