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I've given the following problem:

(a) $R$ is a factorial ring and $\mathfrak{p} \lhd R$ a prime ideal. By use of the cannonical projection $R \to R/\mathfrak{p}$ on the coefficients, we'll get a ring homomorphism $\varphi: R[x] \to (R/\mathfrak{p})[x]$. Let $p \in R[x]$ be primitive with leading coefficient not in $\mathfrak{p}$. Show, if $\varphi(p)$ is irreducible in $(R/\mathfrak{p})[x]$, then $p$ is irreducible in $R[x]$.

(b) $1 + x^3 +x^5$ is irreducible in $\mathbb{Z}[x]$

I proved (a) in the following way: Let $p \in R[x]$ with $p \neq 0$ and $\varphi(p)$ is irreducible. Then for $g,h \in R[x]$ with $p = g h$ WLOG $\varphi(g) \in (R/ \mathfrak{p})[x]^*$ (because $\varphi(p)$ is irreducible). Then we can immediatly follow that g has to be in $R^* = R[x]^*$ because $\varphi$ is an ringhomomorphism.

Is this proof correct?

I don't have any idea how to solve (b), I think I will need (a), but I don't know how I can use this.

user26857
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    For $b$, presumably you are to show that $1+x^3+x^5$ is irreducible over $\Bbb Z_2, \Bbb Z_3$ or $\Bbb Z_5$, or something similar, and use (a) to conclude that it must therefore be irreducible over $\Bbb Z$. – Arthur Jan 03 '18 at 13:51

4 Answers4

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Hint: Looking around at MSE (see also the answer of Thomas) we see that $x^5+x^3+1$ is irreducible over $\mathbb{F}_2$.

Exercise: Is it still irreducible over $\mathbb{F}_{32}$ and $\mathbb{F}_8$?

Answer: Is $x^5 + x^3 + 1$ irreducible in $\mathbb{F}_{32}$ and $\mathbb{F}_8$?

Dietrich Burde
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We will prove that it is irreducible over $\mathbb F_2$. (Why is this enough?)

We will use the more general theorem:

The irreducible factors of $x^{q^n}-x$ in $\mathbb F_q[x]$ are exactly the irreducible polynomials of degrees $d\mid n.$

Since $x^5+x^3+1$ has no roots in $\mathbb F_2$, it has no factors of degree $d=1$, so it is enough to show that $x^{2^5}-x$ is divisible by $x^5+x^3+1.$ We will compute this by computing $x^{32}\bmod{x^5+x^3+1}$ in $\mathbb F_2[x]$.

$$\begin{align}x^8&=x^3\cdot x^5\\&\equiv x^3(x^3+1)\\&\equiv x\cdot x^5+x^3\\&=x(x^3+1)+x^3\\&=x^4+x^3+x\pmod{x^5+x^3+1}\\\\ x^{16}&=(x^8)^2\\&\equiv (x^4+x^3+x)^2\\&=x^{8}+x^{6}+x^2\\&\equiv (x^4+x^3+x)+(x^4+x)+x^2\\&=x^3+x^2\pmod{x^5+x^3+1}\\\\ x^{32}&=(x^{16})^2\\&\equiv(x^3+x^2)^2\\&\equiv x^6+x^4\\&\equiv x(x^3+1)+x^4\\&=x\pmod{x^5+x^3+1}\end{align}$$

A short proof, hiding the above work, is to show that $x^{4}-x$ is relatively prime to $x^5+x^3+1$ in $\mathbb F_2[x],$ with the following equality:

$$(x^4-x)(x+1)+(x^5+x^3+1)(x^3+x+1)=1$$

This shows that $x^5+x^3+1$ has no irreducible factors of degree $1$ or $2$, and thus must be irreducible.

Aside: I'm still using $-x$ in much of the above, to be consistent with the general theorem, but, of course, $-x=x$ in $\mathbb F_2[x].$

Thomas Andrews
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    A shorter proof not hiding anything: the only irreducible polynomial of degree two over $\mathbb F_2$ is $X^2+X+1$. Now check that this doesn't divide $X^5+X^3+1$. – user26857 Jan 14 '18 at 15:44
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$x^5+x^3+1$ has no integer roots and since coefficients before $x^4$ and $x$ they are zero,

it's enough to check for integer $a$ two cases only: $$x^5+x^3+1=(x^2+ax+1)(x^3-ax^2-ax+1);$$ $$x^5+x^3+1=(x^2+ax-1)(x^3-ax^2-ax-1)$$ and easy to see that it's impossible.

Michael Rozenberg
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Here's an alternative solution:

$2^5+2^3+1=41$ is a prime number, thus $x^5+x^3+1$ is irreducible by Cohn's criterion.

Lukas Heger
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