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In solution it said "If we apply Eisenstein criteria for $p=5$, we can check that is irreducible over $\mathbb{Q}$." But I don't understand why should that be. According to the Eisenstein criteria for the polynomial $a_nx^n+...+a_1x+a_0$, that thing should be "$5\vert a_i (0\leq i \leq n-1)$, $5$ does not divide $a_n$ and $25$ does not divide $a_0$". But the solution case, $5$ dose not divde $a_0(=1)$. So How could I apply this for $20x^4+15x^3-10x+1$ ?

kechang lee
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    Note that the dual polynomial, $x^4-10x^3+15x +20$ satisfies Eisenstein $\pmod 5$. – lulu Sep 17 '22 at 01:36
  • As explained in the linked dupes, it follow immediately by applying Eisenstein to the reversed (reciprocal) polynomial. – Bill Dubuque Sep 17 '22 at 01:42

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If we write $u = \frac{1}{x}$, then $$20 x^4 - 15 x^3 - 10 x + 1 = \frac{u^4 - 10 u^3 - 15 u + 20}{u^4},$$ so $20 x^4 - 15 x^3 - 10 x + 1$ is irreducible if its reciprocal polynomial, $u^4 - 10 u^3 - 15 u + 20$, is, and the latter polynomial is Eisenstein at $5$.

Travis Willse
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  • Wow! Then when I consider general case "f(x) is irreducible over $\mathbb{Q}$ $\iff$ $g(x)$ is irreducible over $\mathbb{Q}$ " holds? (Here $g(x)$ is reciprocal polynomial for $f$) – kechang lee Sep 17 '22 at 01:42
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Sep 17 '22 at 01:42
  • @BillDubuque, Thanks for the link for general case – kechang lee Sep 17 '22 at 01:54
  • @BillDubuque In general I do; I searched only cursorily but found no duplicate. In retrospect a search for eisenstein reciprocal polynomial brings up some relevant results. – Travis Willse Sep 17 '22 at 03:57
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    @kechanglee Yes, excluding the case where one of the polynomials has a zero constant term. It's a standard technique for preparing polynomials for an application of Eisenstein. – Travis Willse Sep 17 '22 at 03:59