# Eisenstein's criteria irreducible polynomial - issue in proof of Gareth Jones's book.

I was reading the topic and was stuck on the last lines, where the author states that : -- " $a_i = b_0c_i + b_1c_{i-1}+ ... +b_{i-1}c-1 = b_ic_0$, with p dividing both $a_i$ (since $i \le s = n - t \lt n)$ and ... ". The image is shown below.

I have two questions : (i) how is $i \le s = n - t \lt n$.

(ii) How is it that there are all terms, starting from $b_0c_i$ to $b_ic_0$, in each of the $a_i$ term, I felt that there should be only one product term, i.e. $b_ic_i$. But, here each term is itself a sum of series of product of two terms. I feel confused, totally.

Addendum As shown in the image also, how can I prove that using the Eisenstein's criteria, that the polynomial $x^3 -4x +2$ is irreducible polynomial?

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

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

$p \nmid a_n$ => 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 $p \mid p^n => p \mid a_n$.

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

I have another question, does the prime to be used for testing, need be a root of the polynomial. It is obvious from the book's example that $2$ is not a root of $x^3 -4x +2$, as on substitution yields : $2$, a non-zero value.

If this is correct, then 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, before stating it to pass or fail a given polynomial in the irreducibility test. And how it is possible that there is only one value enough for pass/fail, means what is the logic behind it.

Also, as given above, the second criteria has failed. So, I feel that I am unable to comprehend the crux, and have not understood at all.

$p$ cannot divide each of $b_0,...,b_s,$ else $p$ would divide every co-efficient of $f(x).$ And $p$ divides $b_0$. So take the least $i\leq s$ such that $p$ does not divide $b_i.$ Then $p$ divides each of $b_0,...,b_{i-1}.$ Now $s+t=deg (g)+deg (h)=deg (gh)=deg(f)=n$ and $deg(g), deg (h)$ are both positive [ because we are supposing that $f(x)$ is reducible], so $s<n$, so $i\leq s<n.$

With $f(x)=x^3-4x+2=x^3+0x^2-4x+2=a_3x^3+a_2x^2+a_1x+a_0$, we have $(a_3,a_2,a_1,a_0)=(1,0,-4,2)$. The prime $2$ does not divide $a_3$, but divides $a_2,a_1$ and $a_0$, while $2^2$ does not divide $a_0$, so $f(x)$ is irreducible in $\Bbb Z [x]$ (the ring of polynomials with integer co-efficients).

Sometimes Eisenstein's Criterion can be applied after a change of variable: $f(x)=x^2+1$ does not meet it, but let $x=1+y)$, so $f(x)=y^2+2y+2.$ Then $f_1(y)$ meets Eisenstein's Criterion so it is irreducible. If $f(x)=g(x)h(x)$ then $f_1(y)=g(x)h(x)=f(1+y)=g(1+y)h(1+y)$, and since $f_1$ is irreducible, $deg (g)=0$ or $deg (h)=0$. So $f(x)$ is irreducible.

A theorem of Gauss, proved in a (somewhat) similar way is that if $f(x)\in \Bbb Z[x]$ is irreducible in $\Bbb Z[x]$ then $f(x)$ is irreducible in $\Bbb Q[x].$

• Thanks, but if it were clear theoretically that only one prime if passes the test, then the polynomial is irreducible. Also, how to choose the primes - in increasing order from 2. Third, if the polynomial fails the test at any prime, will it also fail at all the other primes. – jiten Dec 19 '17 at 7:39
• If $one$ prime passes the test then the polynomial is irreducble. So examine the prime divisors of the co-efficients to see whether any of them works....Your third point is wrong. – DanielWainfleet Dec 19 '17 at 8:03

(i) The degrees of $f,g,h$ are $n,s,t$ respectively, and $f(x)=g(x)h(x)$, so $n=s+t$.

(ii) Just think about how you multiply polynomials. For example, in $$(b_0+b_1x+b_2x^2)(c_0+c_1x+c_2x^2+c_3x^3)$$ the coefficient of $x^2$ is $$b_0c_2+b_1c_1+b_2c_0\ .$$

• Thanks a lot for that. – jiten Dec 18 '17 at 4:38
• Please see my edited OP above, and tell what all I have not understood. – jiten Dec 18 '17 at 6:27