Let $f:=X^5-9 \in \mathbb{Q}[X]$. Show: $f$ is irreducible.   So I can't use the criterion, that f has no roots in $\mathbb{Q}[X]$, beacuse $\deg(f)&gt;3$. Eisenstein is also not possible, because the only divider of $9$ is $3$. So I tried the reduction criterion, which doesn't really help me because $\deg(f)$ is still $5$. And in $\mathbb{Z}/7$ for example there are no prim elements, So I can't use Eisenstein. Is there another way I don't see?
32287
Let $F$ be a field, let $\omega$ be a primitive $n$th root of unity in an algebraic closure of $F$. If $a$ in $F$ is not an $m$th power in $F(\omega)$ for any $m\gt 1$ that divides $n$, how to show that $x^n -a$ is irreducible over $F$?
106784
Determine whether $X^4-16X^2+4$ is irreducible in $\mathbb{Q}[X]$.  To solve this problem, I reasoned that since $X^4-16X^2+4$ has no rational roots hence irreducible.  But there is a hint to this question that uses different approach: Try supposing it is reducible, then it must factor into a product of two monic quadratic polynomials with integer coefficients. Then show that it is impossible, then conclude the original polynomial is irreducible.  My questions regarding the hints: 1. Why can't we just show that since $X^4-16X^2+4$ has no rational roots hence irreducible? 2. Why do we have to factorise into a product of two monic quadratic polynomials with integer coefficients? Why monic? And why can't we factorise into polynomial with degree 1 and 3? Also, lastly why the coefficients have to be integers?  Thank you for any explanations!    Edit: Thanks for all the answers.  I saw that Second Gauss Lemma is used. But I only learned the first one in class. Which is: Let $R$ be a Unique Facorisation Domain. If $f,g\in\mathbb{R}[X]$ are primitive, then so too is their product $fg$.  Is it inevitable to use Second Gauss Lemma? Is there any other way around?
68768
Show that $2,3, 1-\sqrt{-5}, 1+\sqrt{-5}$ are irreducible over $\mathbb{Z}[\sqrt{-5}]$, but not prime and that 1 and -1 are the only units.   Let $N$ be the norm map into $\mathbb{Z}$ and let u denote the unit, then because it is a homomorphism it follows that :   $N(xy)=N(x)N(y)$ and thus with $N(u)N(u^{-1})=N(uu^{-1})=N(1)$ it follows that the norm of u is a divider of 1 and thus also a unit of $\mathbb{Z}$. Therefore we have in $\mathbb{Z}[\sqrt{-5}]$: $\forall a,b \in \mathbb{Z}: N(a+b\sqrt{-5})=(a+b\sqrt{-5})(a-b\sqrt{-5})=a^{2}+5b^{2}=1 \ \ \Rightarrow u= \pm 1$  Now it is to show that $2,3,1-\sqrt{-5}, 1+\sqrt{-5}$ are irreducible over $\mathbb{Z}[\sqrt{-5}]: $  Assume $1-\sqrt{-5}$ is reducible, then there must exist $a,b \in \mathbb{Z}[\sqrt{-5}]$ so that $N(1-\sqrt{-5})=N(a)N(b) \Rightarrow N(a)=N(b)= \pm (1-\sqrt{-5})$ But since $1-\sqrt{-5}$ is not a quadratic remainder of $5$, there doesn't exist a solution for the equations $a^{2}+5b^{2}= \pm (1-\sqrt{-5})$ And the exactly same argument also works for $2,3$ and $1+\sqrt{-5}$.  Thus we have shown that $2,3,1\pm \sqrt{-5}$ are not reducible over $\mathbb{Z}[\sqrt{-5}]$.   Now we show that they are not prime:   Assume that 2 is a prime in $\mathbb{Z}[\sqrt{-5}]$, then because of $2\cdot 3 = (1-\sqrt{-5})(1+\sqrt{-5})=6$ it must hold that $2|(1-\sqrt{-5})$ or $2|(1+\sqrt{-5})$. But with $a,b \in \mathbb{Z}[\sqrt{-5}]$ it immediately follows that for :   $(1\pm\sqrt{-5})= 2(a+b\sqrt{-5})$ $2b = \pm 1$. So 2 is not a prime in $\mathbb{Z}[\sqrt{-5}]$.   Assume that 3 is a prime in $\mathbb{Z}[\sqrt{-5}$, then: $(1\pm \sqrt{-5}) = 3(a+b\sqrt{-5}) \Rightarrow 3b= \pm 1$ it follows that 3 is not a prime in $\mathbb{Z}[\sqrt{-5}]$.  Assume that $1\pm \sqrt{-5}$ is a prime in $\mathbb{Z}[\sqrt{-5}]$, then: $3 = (1\pm \sqrt{-5})(a+b\sqrt{-5}) \Rightarrow (1\pm \sqrt{-5} )a = 3$ which is not solvable in $\mathbb{Z}$. And also $2=(1\pm \sqrt{-5})(a+b\sqrt{-5})=(1\pm \sqrt{-5})a$ which is also not solvable in $\mathbb{Z}$ and thus $(1\pm \sqrt{-5})$ can not be a prime in $\mathbb{Z}[\sqrt{-5}]$.   Tell me if this proof is correct. Please.
75159
Question:  If $p(x) \in F[x]$ is of degree $3$, and $p(x)=a_0+a_1x+a_2x^2+a_3x^3$, show that $p(x)$ is irreducible over $F$ if there is no element $r\in F$ such that $a_0+a_1r+a_2r^2+a_3r^3 =0$.  My understanding: If $p(x)$ is reducible, then there exists $ax + b$ such that $a, b \in F$ and $a\ne 0$. And $p(x) = (ax + b)(cx^2 + dx + e)$. Then an $r$ exists such that $p(r) = 0$. Is that how you're supposed to tackle it? I understand that I am supposed to first explain the argument for a polynomial of degree 1, but I do not understand how to go about that, and why that helps.
124213
Let $K$ be a field and $f,g$ irreducible polynomials in $K[X]$, is there a nice iff condition for $K[X]/(f)\cong K[X]/(g)$?  ($\cong$ denotes an isomorphism that is the identity on restriction to $K$).  Thoughts: It is sufficient that they are $K^\times$ multiples of each other. I'd hoped this was necessary but it isn't as $\mathbb{Q}[X]/(X^2-2)\cong\mathbb{Q}(\sqrt2)=\mathbb{Q}(\sqrt2+1)\cong\mathbb{Q}[X]/(X^2-2X-1)$ with the middle two fields viewed as subfields of $\mathbb{C}$. It is necessary that they have the same degree. Please let me know if there are any other simple necessary conditions. Thanks!
16855
Let $F=\Bbb{Q}(i)$. Prove that $x^3-2$ and $x^3-3$ are irreducible over $F$.   How do I go about this? Should I just say that the roots of $x^3-2$ are $2^{1/3},2^{1/3}e^{i2\pi/3},2^{1/3}e^{i4\pi/3}$, which do not belong to $\Bbb{Q}(i)$?
81607
This question is generalized based on my previous question:   Problem:  Consider an irreducible polynomial $f = x^4 + x^3 + 1$ in $\mathbb{F}_2 [x]$. I am interested in its solvability and reducibility in a "chain" of fields. Namely, does $f$ have root in the following fields? Is $f$ reducible in the following fields?  $$ \color{red}{ \mathbb{F}_2},\mathbb{F}_4,\mathbb{F}_8,\color{red}{\mathbb{F}_{16}},\mathbb{F}_{32},\cdots,\color{red}{\mathbb{F}_{256}}$$  My progress:  $\mathbb{F}_2,\mathbb{F}_{16},\mathbb{F}_{256}$:  It is easy to verify that $f$ has no root in $\mathbb{F}_2$ and it is irreducible as well. As a result, $f$ has a root in $\mathbb{F}_{16} \simeq \mathbb{F}_2/f(x)$ and thus is reducible in $\mathbb{F}_{16}$. Since $\mathbb{F}_{256} = \mathbb{F}_{2^{4\times 2}}$, $\mathbb{F}_{16}$ is a subfield of $\mathbb{F}_{256}$, $f$ has root and is reducible in $\mathbb{F}_{256}$ as well.    $\mathbb{F}_{32},\mathbb{F}_{64},\cdots$:  For the fields "between" $\mathbb{F}_{16}$ and $\mathbb{F}_{256}$, I verified that $f$ cannot have root in them. Take $\mathbb{F}_{64}$ for example. Suppose $f$ actually has root in it, call it $\zeta$. Then apparently $\mathbb{F}_2 (\zeta)$ is a subfield of $\mathbb{F}_{64}$. But since $\mathbb{F}_2 (\zeta) \simeq \mathbb{F}_2 [x]/f(x)$, this would imply that $\mathbb{F}_{16}$ is a subfield of $\mathbb{F}_{64}$, which is impossible. But I haven't figured out how to deal with the reducibility part?    $\mathbb{F}_4,\mathbb{F}_8$:  For the fields "between" $\mathbb{F}_2$ and $\mathbb{F}_{16}$, I am not sure what's the best strategy here. For the existence of root, I am attempted to say that $f$ has no root in both of them since "the smallest" field extension that has a root is $\mathbb{F}_{16}$. I forgot the proof but I think it is correct.   With the operation tables of $\mathbb{F}_4$ from WolframAlpha, I was able to show that $f$ has no root in $\mathbb{F}_4$, and it is irreducible in $\mathbb{F}_4$ as well. But is there a better way?  For $\mathbb{F}_8$, I mimicked the answer in my previous question and managed to show that $f$ is irreducible in $\mathbb{F}_8$. But I don't know how I should deal with the existence of root?  I understand that this is a long post and I appreciate your help. Thanks.
21335
Let $f(x)$ and $g(x)$ be irreducible polynomials over a field $F$ and let $a,b \in E$ where $E$ is some extension of $F$. If $a$ is a zero of $f(x)$ and $b$ is a zero of $g(x)$, show that $f(x)$ is irreducible over $F(b)$ if and only if $g(x)$ is irreducible over $F(a)$.   Attempt:  Since $f(x),g(x)$ are irreducible over $F \implies a,b \notin F$.  $f(x)$ is irreducible over $F(b) $ and $f(x)$ is irreducible over $F \implies a \neq b$ (As, $a$ is the zero of $f(x)$)  Which means $b \notin F(a)$ either $\implies g(x)$ is irreducible over $F(a)$.  Similarly, the other half can be proved in a similar way.  Is my solution attempt correct?  Thank you for your help..
38390
Prove that $f=x^4-4x^2+16\in\mathbb{Q}[x]$ is irreducible.  I am trying to prove it with Eisenstein's criterion but without success: for p=2, it divides -4 and the constant coefficient 16, don't divide the leading coeficient 1, but its square 4 divides the constant coefficient 16, so doesn't work. Therefore I tried to find $f(x\pm c)$ which is irreducible:  $f(x+1)=x^4+4x^3+2x^2-4x+13$, but 13 has the divisors: 1 and 13, so don't exist a prime number p such that to apply the first condition: $p|a_i, i\ne n$; the same problem for $f(x-1)=x^4+...+13$ For $f(x+2)=x^4+8x^3+20x^2+16x+16$ is the same problem from where we go, if we set p=2, that means $2|8, 2|20, 2|16$, not divide the leading coefficient 1, but its square 4 divide the constant coefficient 16; again, doesn't work.. is same problem for x-2  Now I'll verify for $f(x\pm3)$, but I think it will be fall... I think if I verify all constant $f(x\pm c)$ it doesn't work with this method...  so have any idea how we can prove that $f$ is irreducible?
1406
Let $K$ be an arbitrary field, $p$ a prime and $a\in K$. Show $f=x^p-a$ is either irreducible in   $K[x]$ or has a root in $K$.   My strategy was to split this up into a case for each characteristic.   The characteristic $p$ case is easy. Assume $f$ is reducible $=gh$ both smaller degree $g$ irreducible, with $\alpha$ a root of $g$. Then in $K(\alpha)[x]$, $f=(x-\alpha)^p$. So $g=(x-\alpha)^k$, and all the other factors of $f$ are of the same form. Hence they are all equal to the $minpoly(\alpha)$ and the degree of $minpoly(\alpha)|p$ so it is $1$ or $p$ and we are done.  The characteristic $0$ case  and characteristic $q\ne p$ I get stuck on. If $K$ contains a primitive $p$th root of unity $\zeta$, then over $K(\alpha)$, $f$ splits into linear factors. So the degree of the field extension for any root of $f$ is the same, so the degree of such a root is either $1$ or $p$. If $K$ does not have a primitive $p$th root of unity, then simply adjoin one. We know $[K(\zeta):K]|p-1$ and by the previous step either $f$ is irreducible over $K(\zeta)$ or splits into linear factors. If it is irreducible we are done, if not we have it splits into linear factors of the form $x-\zeta^ia^{1/p}$, but I am unsure what to do from here. The characteristic $q$ case seems similar to the characteristic $0$ case but again I am stuck trying to finish the proof.   Some googling has lead me to believe that this problem is generally approached using the field norm and if an accessible online reference exists I would love to see it, but I feel that in the context I have seen it should be entirely possible that this is done without needing the norm. As I have not seen the field norm before, I would like to know if there is a possible way to prove this without using it and hopefully that follows in the same vein my attempt at a proof did.
48542
Is there any criterion (theorem or algorithm) to find out a given polynomial $f(x)$ is irreducible over the field $\mathbb{F}_2$?
115644
I know that irreducible polynomials over fields of zero characteristic have distinct roots in its splitting field.  seems to show that irreducible polynomials over $\Bbb F_p$ have distinct roots in its splitting field (and all the roots are powers of one root). Is the proof correct? I have never seen this result anywhere else. The proof is very convincing to me. Does the result hold for $\Bbb F_q$ where $q$ is a power of prime? I don't think it holds because I've heard there are irreducible polynomials with repeated roots? Please help.
4760
I want to argue that argue that $\pi(\alpha)=\alpha^3+3\alpha+3$ is an irreducible polynomial over the finite field with 5 elements $\mathbb{F}_5$. My approach was just to check that $\pi$ has no roots in $\mathbb{F}_5$. Is this right?
97413
I was “playing” with Eisenstein's criterion and the polynomials $X^n-p$ ($p$ being a prime number).  Over $\mathbf Z$, the classical Eisenstein's criterion directly applies and shows that $X^n - p$ is an irreducible element of $\mathbf Z[X]$. The question was “what happens over $\mathbf Z[i]$?”  We know that in this new ring, a prime number $p$ has one of three different fates:   if $p \equiv -1\ (\mathrm{mod}\, 4)$, $p$ still is an irreducible element of $\mathbf Z[i]$, or equivalently $(p) \subset \mathbf Z[i]$ still is prime. (Unless I'm mistaken, number theorists say that $p$ is inert). Eisenstein's criterion still applies and the polynomial is irreducible over $\mathbf Z[i]$. if $p \equiv 1\ (\mathrm{mod}\, 4)$, $p$ can be written as a sum of two squares: $p = n^2 + m^2$ and is therefore reducible on $\mathbf Z[i]$: $p = (n+im)(n-im)$. These two factors aren't associated (the units in $\mathbf Z[i]$ being $\pm 1$ and $\pm i$) so $(p)$ splits: $(p) = (n+im)(n-im)$ is the product of two different prime ideals. One can still use Eisenstein's criterion with either of these prime ideals, and $X^n - p$ is irreducible over $\mathbf Z[i]$. if $p = 2$, the oddest of all primes, one still has a factorisation $p = (1+i)(1-i)$ but this time, $1+i$ and $1-i$ are associated. In NT slang, $(2)$ is ramified: $(2) = (1+i)(1-i) = (1+i)^2$ and one cannot directly use Eisenstein's criterion.   So here's my question: how do we prove (or disprove) the irreducibility of $X^n - 2$ over $\mathbf Z[i]$?
93989
Let $K\subset \mathbb{C}$ be a subfield and $f\in K[t]$ an irreducible polynomial. Show that $f$ has no multiple roots in $\mathbb{C}$.   If I understand this question correctly, I must show that there is no $a \in \mathbb{C}$ such that $(t-a)^n|f$ in $F[t]$ with $n&gt;1$. So suppose $(t-a)^2|f$ and $f=(t-a)^2h$. Then we have $f'=2(t-a)h+(t-a)^2h' \Rightarrow (t-a)|f'$ so $\gcd(f,f'$) is not constant. Therefore $f$ is divisible by some square of non-constant polynomial in $F[t]$, which is a contradiction. Is my argument correct? Thank you.
107699
Let $\mathbb F_9$ be a finite field of size $9$ obtained via the irreducible polynomial $x^2 + 1$ over the base field $\mathbb F_3$.    How can you find a primitive element?  Make a list of the elements of $\mathbb F_9$ together with a primitive element and all the powers of the primitive element.
48811
Given an arbitrary finite field $K$ (not necessarily $\mathbb{F}_p$ with $p \in \mathbb{P}$) with $|K| = q$ and an irreducible polynomial $f$ with $\alpha$ as root and degree of $n$. Is $|K(\alpha)| = q^n$ and why? Its clear to me for $K$ isomorphic to $\mathbb{Z}_p$ with $p$ is a prime number.