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Question:

Check if: $$ f(x) = x^4 + 4x^3 + 6x^2 + 2x + 1 $$ is reducible or irreducible over $\mathbb{Q}[x]$

My Answer [Edited]:

Suppose that $f(x)$ is irreducible over $Z_p[x]$ for a prime $p$, therefore it's going to be irreducible over $\mathbb{Q}[x]$. Let $p=2$. Therefore: $$ f_2(x) = f(x) \mod{2} = x^4 + 1 = a_2(x)\cdot b_2(x) \in \mathbb{Z}_2[x] $$ Since $x^4+1$ have roots in $\mathbb{Z}_2[x]$, there are two cases to check... First let's suppose that $a_2(x),b_2(x)$ have degree equal to two, therefore: \begin{align*} x^4+1&=(a_2x^2+a_1x+a_0)\cdot(b_2x^2+b_1x+b_0)=\\ &=(a_2b_2)x^4+(a_2b_1+a_1b_2)x^3+(a_2b_0+a_1b_1+a_0b_2)x^2+(a_1b_0+a_0b_1)x^+(a_0b_0) \end{align*} Since $a_i,b_i \in \mathbb{Z}_2 \rightarrow a_i,b_i\in \{0,1\}$, it follows that: $$ a_0b_0=1 \longrightarrow a_0=b_0=1\\ a_1+b_1=0 \longrightarrow a_1=b_1=0\\ a_2b_2=1 \longrightarrow a_2=b_2=1 $$ Hence: \begin{align*} a_2(x)&=x^2+1\\ b_2(x)&=x^2+1\\ a_2(x)\cdot b_2(x) &= (x^2+1)^2 = x^4 + 2x + 1 \overbrace{\longrightarrow}^{\in \mathbb{Z}_2} x^4 + 1 \end{align*}

Now let's check WLOG the case for $a_2(x)$ having degree equal to $3$ and $b_2(x)$ having degree equal to $1$:

\begin{align*} x^4+1 &= a_2(x)\cdot b_2(x) = (a_3x^3+a_2x^2+a_1x+a_0)\cdot(b_1x+b_0)\\ &= (b_1a_3)x^4 + (b_1a_2+b_0a_3)x^3 + (b_1a_1+b_0a_2)x^2 + (b_1a_0+b_0a_1)x + (b_0a_0) \end{align*}

By the same reasoning as before: \begin{align*} b_0a_0=1 \longrightarrow b_0=a_0=1\\ b_1a_3=1 \longrightarrow b_1=a_3=1\\ a_2+b_0 = 0 \longrightarrow a_2=b_0=0\\ a_1+a_2 = 0 \longrightarrow a_1=a_2=0\\ \end{align*}

Conclusion, it is reducible over $\mathbb{Z}_2[x]$ and therefore I can't conclude anything.

Bill Dubuque
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Bruno Reis
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  • Seems to me you have shown that $f(x)$ is reducible over $\Bbb Z_2[x]$, not whether or not it's reducible over $\Bbb Q[x]$. – Rolf Hoyer Dec 06 '18 at 03:34
  • @RolfHoyer Yeah, I was checking what I've done... I just know that if a monic polynomial is reducible over $\mathbb{Q}[x]$, then it is going to be reducible over $\mathbb{Z}_2[x]$... But if that polynomial is irreducible, I can't assume the same... How you would approach that question? – Bruno Reis Dec 06 '18 at 03:37
  • Why doesn't $x^4+1$ have roots mod $2$? – Michael Burr Dec 06 '18 at 03:38
  • @MichaelBurr in mod 2, $x \in {0,1}$ and from that we have $f(0) = 1$ and $f(1)=2=0$... Yeah, you are correct... We have roots... – Bruno Reis Dec 06 '18 at 03:39
  • I'd suggest Eisenstein's criterion after shifting $x$ to $x+a$ for some $a$. Perhaps $x+1$ in this case. – Michael Burr Dec 06 '18 at 03:40
  • Is shifting allowed in this kind of process ?? I also thought about Eisenstein, but couldn't find any such prime. – Anik Bhowmick Dec 06 '18 at 03:41
  • @Michael Burr So If I continue what I've done, suposing WLOG we have a degree $3$ for $a_2(x)$ and a degree $1$ for $b_2(x)$, if the monic polynomial is irreducible over $\mathbb{Q}[x]$, then we'll get that it is going to be irreducible over $\mathbb{Z}_2[x]$ right? – Bruno Reis Dec 06 '18 at 03:42
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    @AnikBhowmick Shifting is an invertable operation, so irreduciblity must be preserved. In this case, factoring commutes with shifting. – Michael Burr Dec 06 '18 at 03:45
  • The direction you've chosen is unlikely to work, at least mod $2$. Irreducible mod $p$ implies irreducible in $\mathbb{Q}$, but the converse is false. – Michael Burr Dec 06 '18 at 03:46
  • Yeah, but check the edit in 5 minutes... – Bruno Reis Dec 06 '18 at 03:47
  • related https://math.stackexchange.com/q/1147706/173147, https://math.stackexchange.com/q/674908/173147, https://math.stackexchange.com/q/2556519/173147, https://math.stackexchange.com/q/687402/173147 – glS Feb 28 '23 at 08:10
  • As described in the dupe, shifted Eisenstein applies since $2$ divides the discriminant of $f$ and $f$ a prime power $\bmod 2,,$ viz. $(x+1)^4\ \ $ – Bill Dubuque Feb 28 '23 at 08:24

2 Answers2

2

The expression for $f(x)$ looks pretty much like a binomial expansion: to be precise, $$f(x)=(x+1)^4-2x\ .$$ Hence $$f(x-1)=x^4-2x+2\ ;$$ this is irreducible by Eisenstein, therefore $f(x)$ is also irreducible.

David
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2

Your polynomial is very close to $(x+1)^4$. Making the substitution, $x \rightarrow y-1$, we get $ y^4 - 2y + 2$, which is irreducible by Eisenstein's criterion.

Eric Towers
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