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This is a past exam question.

Decompose each of the following elements as a product of irreducible:

(a) $X^4+2 \in \mathbb{Z}_5[X]$

(b) $X^5+X \in \mathbb{Z}_2[X]$

(c) $X^5+4X^4-3X^3+X^2+7X+11 \in \mathbb{Q}[X]$


(a) I run over the value $0$ to $4$, and show this polynomial has no linear factors, but how do I show this polynomial also has no quadratic factor? Do I write down $$X^4+2 = (aX^2+bX+c)(a^\prime X^2+b^\prime X+c^\prime)$$ and show there are no such $a,b,c,a^\prime,b^\prime,c^\prime \in \mathbb{Z}_5$ Which is something that I don't want to do in the exam, will take quite a long time.

(b) $X=0,1$ is a root for the polynomial. and after some long divisions. $X^5+X=X(X+1)^4$

(c) $X^5+4X^4-3X^3+X^2+7X+11$ has no linear factors in $\mathbb{Z}_2[X]$. And I know all the irreducible polynomials in $\mathbb{Z}_2[X]$ with degree $\le$ 3. So I can use long division to see can they divide $X^5+4X^4-3X^3+X^2+7X+11$.


And for a more general question, how do I test the irreducibility for a polynomial with degree higher than 3. Since they might have quadratic factor or factor of higher power.

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  • $\begingroup$ Concerning (c), it is said that this polynomial is in Q[X], why are you looking for solution in Z2[X]? $\endgroup$ – lisyarus Jun 17 '14 at 8:26
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    $\begingroup$ @lisyarus If a polynomial over the integers is reducible, then surely it is reducible over $\mathbb Z_p$. So by the contrapositive, if it is irreducible over $\mathbb Z_p$, it also is irreducible over $\mathbb Z$. I'm guessing the OP is thinking about somthing along these lines. $\endgroup$ – Git Gud Jun 17 '14 at 8:30
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    $\begingroup$ @lisyarus No, just need to find one prime that work. $\endgroup$ – SamC Jun 17 '14 at 8:36
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    $\begingroup$ @lisyarus Sorry for the imprecision, the statement should go as follows. For all polynomials $f(x)\in \mathbb Z[x]$ and for all primes $p\in \mathbb P$, denote the $\mathbb Z_p$-polynomial whose coefficients are those of $f(x)$ reduced modulo $p$ by $f_p(x)$. The following conditional is true: if $f(x)$ is reducible over $\mathbb Z$, then for all $p\in P$, $(f_p(x))$ is reducible over $\mathbb Z_p$. Badly written proof: If $f=gh$, then $f_p=g_ph_p\square$. So the useful contrapositive becomes existential. $\endgroup$ – Git Gud Jun 17 '14 at 8:42
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    $\begingroup$ @SamC Note that in $\left(aX^2+bX+c\right)\left(a^\prime X^2+b^\prime X+c^\prime\right)$ you need to require $a'=a^{-1}$, thus yielding $\left(aX^2+bX+c\right)\left(a^{-1} X^2+b^\prime X+c^\prime\right)$. Now multiply this $1\color{grey}{=aa^{-1}}$ in the following manner: $\left[a^{-1}\left(aX^2+bX+c\right)\right]\left[a\left(a^{-1} X^2+b^\prime X+c^\prime\right)\right]$. $\endgroup$ – Git Gud Jun 17 '14 at 14:06
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Hope this answer can help people who are doing algebra.

(a) $f(X)=X^4+2 \in \mathbb{Z}_5[X]$

$f(0)=2,f(1)=3,f(2)=3,f(3)=3,f(4)=2 \Rightarrow f(X)$ has no linear factors in $\mathbb{Z}_3[X]$.

Hence if $f(X)$ is reducible, then it will have 2 quadratic factors for which they are monic.

$$X^4+2 = (X^2+bX+c)(X^2+b^\prime X+c^\prime)$$

Gives,

$$b = -b^\prime \space\space\space\space\space\space\space\space -(1)\\ c+c^\prime+bb^\prime=0 \space\space\space\space\space\space\space\space -(2)\\ b^\prime c+bc^\prime = 0 \space\space\space\space\space\space\space\space -(3)\\ cc^\prime=2 \space\space\space\space\space\space\space\space -(4)$$

Sub. $(1)$ into $(3)$, we get $-bc+bc^\prime=0 \Rightarrow b=0$ or $c=c^\prime$

If $b=0$, $(2) \Rightarrow c=-c^\prime$, $(4) \Rightarrow c^2 =2$ (contradiction)

If $c=c^\prime$, $(4) \Rightarrow c^2 =2$ (contradiction)

Therefore, $X^4+2$ is irreducible in $\mathbb{Z}_5[X]$

(b) $X=0,1$ is a root for the polynomial. and after some long divisions. $X^5+X=X(X+1)^4$ in $\mathbb{Z}_2[X]$

(c) $f(X)=X^5 + 4 X^4 - 3 X^3 + X^2 + 7 X + 11 \in \mathbb{Q}[X]$

$$\bar{f}(X) = X^5+X^3+X^2+X+1 \in \mathbb{Z}_2[X]$$

$\bar{f}(0)=\bar{f}(1)=1 \Rightarrow \bar{f}(X)$has no linear factors in $\mathbb{Z}_2[X]$.

So, $\bar{f}(X)$ must be decompose into $$\text{(polynomial of degree 2) * (polynomial of degree 3)}$$

There are only one irreducible polynomial of degree 2 in $\mathbb{Z}_2[X]$ which is $X^2+X+1$. By long division, we can show that $X^2+X+1 \nmid \bar{f}(X)$. Hence $\bar{f}(X)$ is irreducible in $\mathbb{Z}_2[X]$ also $deg(f(X))=deg(\bar{f}(X))$ (Note: this is same as saying the prime number 2 does not divide the highest order coefficient of $f(X)$), then $f(X)$ is irreducible in $\mathbb{Q}[X]$.

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