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As I'm new into p-adic Field Theory, I learned Hensel's Lemma by which I can show if a polynomial has roots over the p-adic numbers (i.e. in $\mathbb{Q}_2$ or $\mathbb{Q}_3$) . For polynomials of degree 2 or 3 over those fields, with Hensels Lemma I can show they are irreducible over those fields if there aren't any roots in $\mathbb{Q}_2$ or $\mathbb{Q}_3$.

But how can I show a polynomial of degree 4 is irreducible over such a field? Since for degree 4, its not sufficient to show there are no roots in $\mathbb{Q}_2$ or $\mathbb{Q}_3$. Is there an easy way to show it? Or is there a polynomial of degree 4 which obviously is irreducible in general over such a p-adic Field?

DoggyRoot
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    There is a polynomial version of Hensel's Lemma (see the wikipedia article for a statement) that relates the splitting of $f\in \mathbb Z_p[X]$ to the splitting of its reduction mod $p$. – Mathmo123 Jul 29 '22 at 16:02
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    For your last question, you can take something like $X^4 - p$, which is irreducible in $\mathbb Q_p$ by Eisenstein's criterion – Mathmo123 Jul 29 '22 at 16:05
  • First comment of yours: I've read that version of Hensels' Lemma but doesn't it only tell me there is a factorization of f in $\mathbb{Z}_2$ if it has one mod p? As far as I understood it doesn't tell me that if there's none mod p, there's none $\mathbb{Z}_2$ – DoggyRoot Jul 29 '22 at 18:37
  • Second comment of yours: So I can use it by saying $\mathbb{Z}_2$ is the p-adic ring of integers and $\mathbb{Q}_2$ is it's Field of Fractions? Thank you !! – DoggyRoot Jul 29 '22 at 18:40
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    @ultraDlion: The direction "if there is no factorization mod $p$, then there is none over $\mathbb Z_p$" goes without Hensel's Lemma: Any reasonable (i.e. monic) factorization in $\mathbb Z_p [X]$ induces one in $\mathbb F_p[X]$. – Torsten Schoeneberg Jul 29 '22 at 18:50
  • Mr Schoeneberg, could you give a short explanation where that comes from? Is it a Lemma? – DoggyRoot Jul 29 '22 at 18:57
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    The mod $p$ reduction map $\mathbb Z_p[x] \rightarrow \mathbb F_p[x]$ is a ring homomorphism, i.e. the image of a product $\overline{g \cdot h}$ is the product of the images $\bar g \cdot \bar h$. If $g$ and $h$ are monic (and even under lesser conditions), their images $\bar g$ and $\bar h$ will be as well. – Torsten Schoeneberg Jul 29 '22 at 19:42
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    One useful trick that can show a p-adic polynomial is irreducible is by using the Newton polygon, which is a more general method than Eisenstein's criteria. If the Newton polygon has a single slope and that slope doesn't pass through other lattice points (otherwise that may mean it is a product of two polynomials), $\gcd(v_p(a_n)-v_p(a_0),n)=1$, then it is irreducible. Here's a MSE answer that goes into a bit more depth. – Merosity Jul 30 '22 at 01:17

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