# Eisenstein criterion's condition

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The question has already some answer here: Irreducibility check for polynomials not satisfying Eisenstein Criterion. In addition, there are some results on how many monic polynomials in $\mathbb{Z}[x]$ can be shown to be irreducible by Eisenstein's crtiterion. For example, less that $1\%$ of the polynomials with at least seven non-zero coefficients are irreducible by Eisenstein (A. Dubickas, 2003).

Edit: To the new question. No, if the constant term is $\pm 1$, we cannot apply Eisenstein directly, and also not in general after a substitution $x\mapsto x+a$ (e.g., $x^3+x+1$). And yes, there are certain rules, called Eisenstein shifts, when you can attempt a substitution. See the article "On shifted Eisenstein polynomials" of R Heyman, I. Shparlinski (2013).

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### forlorn

Hi, help me improve my IQ in mathematics.

Updated on August 01, 2022

• forlorn 3 months

Proposition: If the constant term is $$1$$ or $$-1$$, then we can't use the Eisenstein criterion to determine whether the polynomial is irreducible over $$Q$$.

Is it right?

### Edit

Since directly use is not right. So the proposition is right or not right?

And further question is, When should I try some substitutions to use Eisenstein criterion? Any guidelines and rules?

For example，I found two examples：

$$x^7+7x+1$$

Eisenstein's criterion requires that the constant term be a multiple of some prime number $p$. The numbers $\pm 1$ are the only ones that do not have a prime factor.
• Tobias Kildetoft about 9 years
Yes, if the constant term is $1$ then we cannot use Eisenstein directly to prove irreducibility, since there is no prime we could apply it with.
However, as the wikipedia link @lhf put forward points out under Examples - Cyclotomic polynomials, substituting $x$ for $(x-1)$ or $(x+1)$ and expanding the brackets, might just make a suitable prime pop out on the other end. Any other substitute $(x+a)$ for an integer $a$ might also work, so while it might be long and tedious to calculate, it might just be worth it in the end.
Sometimes (but rarely) a substitution helps, e.g. $f(x)=x^4+4x^3+10x^2+12x+7$ becomes, after $x\mapsto x+1$, $g(x)=x^4+4x^2+2$, which is irreducible by Eisenstein.
Yes, for example with $x^3+x+1$ a substitution is useless, see the answer in math.stackexchange.com/questions/458802/….