# Proving irreducible polynomial using Eisenstein's criteria.

How can I prove that using the Eisenstein's criteria, that the polynomial : $x^3 -4x +2$ is irreducible polynomial for prime $2$?

I am new to this topic, and while attempting would try to get the logic of the criteria too.

My attempt is : for $x=2$, the value of $a_0 = 2, a_1= -8, a_2 = 0, a_3 =8$.

To apply Eisenstein's criterion there should be a prime $p$ such that

(i) $p\mid a_0, a_1,..., a_{n-1}$ => passes as $2 \mid 2, -8, 0$

(ii) $p \nmid a_n$ => fails as $2 \mid 8$, this also means that the highest term (let, of the power $n$) must be multiplied by a suitable integer multiplier (let, $m$), so that $p \nmid m.p^n$. I am not aware of its algebraic proof, or even its significance, or ramifications; but it is just an observation. It also means that for monic polynomials it must always fail, as $p \mid p^n => p \mid a_n$.

(iii) $p^2 \nmid a_0$ => passes as $4 \nmid 2$, this also means that $c$ term must be non-zero and composite for this condition to be true.

As given above, the second criteria has failed, so where is the flaw in my attempt?

I have another question, how will one test for any prime fitting in for test, i.e. there must be some restrictions on the choice of primes to be tested. And how it is possible that there is only one value enough for pass/fail.

• Please give reason for down-voting. I made a very serious attempt, and it is nowhere told that lack of knowledge is not appreciated, particularly in the face of attempting to get it. There was no mention of such a simple way to check - online or offline, and so asked rather than sit and mug it. – jiten Dec 18 '17 at 12:41
• I did not downvote this, but I assume that those who did found the question unclear. Eisenstein directs you to perform a test on the coefficients of the polynomial, not on the values. You can read about the method and why it works here. I agree that the test is somewhat mysterious on first glance...that's why it gets so much attention. But the proof that it works is not terribly difficult and is well worth studying. – lulu Dec 18 '17 at 12:47
• @lulu Thanks for the link, it is also given below in one of the answers for the "Basic Proof" section. I will try to find the sources also that lead to the article. – jiten Dec 18 '17 at 12:54

Because $0$, $4$ and $2$ they are divisible by $2$, but the last coefficient $2$ is not divisible by $4$.

For irreducibility it's enough to use the Eisenstein's criteria for one prime number.

• I request why, what is the logic for that? Also, will it be enough for failing too. Or do you mean that to clear the test, only one prime is enough, and any number of failure cases of primes are not enough. If so, then please answer my question in the OP.. – jiten Dec 18 '17 at 12:06
• In this the Eisenstein's criteria. To check it for some prime number. It's not always works. – Michael Rozenberg Dec 18 '17 at 12:11
• Please elaborate your statement. I am a beginner, and need more details. – jiten Dec 18 '17 at 12:14
• For example, $x^2+2x+4$ is irreducible above $\mathbb Z$. – Michael Rozenberg Dec 18 '17 at 12:43
• Yes, of course! – Michael Rozenberg Dec 18 '17 at 13:00

Your flaw is in the $a_{i}$'s. These are the coefficients of the polynomial, i.e. in your case

$$p(x)=a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}$$

$$\Longrightarrow a_{3}=1,\:a_{2}=0,\:a_{1}=-4,\:a_{0}=2$$

and not the value of each monomial evaluated at a specific $x$.

• Thanks, but I feel that there should be some explanation also, as to why only the coefficients $a_j$ are being taken, i.e. no substitution for the value of $x$ is needed. – jiten Dec 18 '17 at 11:57
• @jiten The key here is that a polynomial is an entirely abstract entity. $x$ is not a variable, but an element of the ring $\mathbb{Z}\left[x\right]$. The statement of irreducibility is for the polynomial as an abstract element. – eranreches Dec 18 '17 at 12:06
• Thanks for that, but if there were an equivalent statement to the abstract language statement: "an element of the ring $\mathbb{Z}[x]$". If it means that :" the element $x$ is not a variable, but is a particular value taken for arriving at a particular curve given by its substitution". It also means that the polynomial is taken as a parametric equation, and the parameter is not an issue, rather the form of the parametric equation matters, the order, and the coefficients. --- If I am correct, then please state where to find literature that explains in such way. – jiten Dec 18 '17 at 12:11
• @jiten I don't understand what are you getting at. The notion of polynomials as curves is not good when dealing with abstract algebra. – eranreches Dec 18 '17 at 12:15
• But, if state to myself the explanation that: "$x$ is an element of the ring $\mathbb {Z}[x]$", then I would try to see the logic too. I request you to elaborate in a simpler context, as I am not aware of the abstract algebra concepts. – jiten Dec 18 '17 at 12:18