# Integral polynomial irreducible over $\mathbb Q$ but reducible mod $2$, $3$, and $5$ [duplicate]

Anyone know an $f\in \mathbb{Z}[x]$ that is irreducible over $\mathbb{Q}$ but whose reduction mod $p$ is reducible over the first three positive primes?

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## marked as duplicate by Arturo Magidin, Eugene, BenjaLim, William, Rudy the ReindeerSep 3 '12 at 10:27

This was discussed recently, e.g., here and here give you examples that are reducible over all primes, not just the first three. – Arturo Magidin Jun 25 '12 at 21:49
@Arturo The question is not an exact duplicate of the linked problems. Rather, it is a much simpler case. – Bill Dubuque Jun 25 '12 at 22:18

Hint $\rm\ \ x^2 +\, 2\cdot3\cdot 5$

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"Most" polynomials are irreducible. So if we can find any polynomial reducible modulo 2, 3, and 5, we should have no trouble finding one that is also irreducible over the integers.

And we can even consider the three primes independently, and use the Chinese Remainder Theorem (CRT) to combine our results.

The simplest reducible polynomial modulo 2 is $x^2$. The same modulo 3 and 5. So applying the CRT is trivial in this case, since $x^2$ reduces to $x^2$ modulo 2 3 and 5.

Alas, $x^2$ is reducible over the integers. But "most" polynomials are irreducible, so just pick another polynomial that reduces to $x^2$ modulo 2, 3, and 5. As $30$ is equivalent to $0$ modulo 2, 3, and 5, we can add it to any coefficient without changing its residues modulo 2, 3, and 5.

We were, in some sense, incredibly unlucky with our first guess of $x^2$; any other guess is extremely likely to find a good example by choosing any other one. $x^2 + 30$ turns out to be irreducible, so there you go!

Let's do a more involved example. Suppose we choose

• $f(x) \equiv x(x-1) = x^2 + x\pmod 2$
• $f(x) \equiv (x-1)(x-2) = x^2 + x + 2 \pmod 3$
• $f(x) \equiv x^3 \pmod 5$

Applying the CRT to each coefficient gives

• $f(x) \equiv 6 x^3 + 25 x^2 + 25x + 20 \pmod {30}$

Happily, $6 x^3 + 25 x^2 + 25x + 20$ is irreducible over the integers.

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Yes. The polynomial $x^4+1$ does the trick.

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KReiser as taste that this polynomial works? – Andres Jun 26 '12 at 8:12
I'm sorry, what do you mean? – KReiser Jun 26 '12 at 22:26