This a new chapter that we are learning and the teacher is flying through it and this are also new concept that i have just learn and i was wondering if i can have some guidance in this problem.     Show that $x^3 + x + 1$ is irreducible over $F_{2}$ and let $\theta$ be a root. Compute the powers of $\theta$ in $F_{2} (\theta)$.    Proof:  $F_2$ has only two element $0$ and $1$ thus to show that the polynomial is irreducible i did $$0^3+0+1 = 1 \not=0$$ and $$ 1^3+1+1 = 3 = 1 \not=0$$ since there is no root therefore the polynomial is irreducible.   Now my question is how to you go about computing the powers of $\theta$
120946
Problem statement:     Show that $x^3+x+1$ is irreducible over $\mathbb{F}_2$ and let $\theta$ be a root. Compute the powers of $\theta$ in $\mathbb{F}_2$.   I am having trouble computing the powers of $\theta$. $x^3+x+1$ is irreducible, since any cubic polynomial over a field is reducible if and only if it has a root in the field. Since $\mathbb{F}_2=\{0,1\}$, we see that none of these is a root of $x^3+x+1$.  To compute the powers of $\theta$, note that $\theta^3+\theta+1=0$. Hence $\theta^3=-(\theta+1)$. Is this a good enough answer for $\theta^3$?  To compute $\theta^2$, note that $\theta = -(\theta^3+1)$. Therefore, $$\theta^2=(\theta^3+1)^2 = \theta^6 + 2\theta^3 + 1 = (-\theta-1)^2-2\theta-2+1,$$ which just reduces to $\theta^2$. Any ideas for computing these powers?  Thanks!
86347
I'm hoping someone can give me a nudge in the right direction...     Let $F$ be a finite field, and let $f(x)$ be a nonconstant polynomial whose derivative is the zero polynomial.  Prove that $f$ cannot be irreducible over $F$.   I've got that every root of $f$ is a multiple root and that for $F=\mathbb{F}_{p^r}$, the exponent of every term of $f$ is a multiple of $p$.
21584
My Question is to check Irreducibility for polynomials not satisfying Eisenstein Criterion.  As an Illustration, to check whether $x^{p-1}+.....+x+1$ for p a prime is irreducible or not, we replaced $x$ by $x+1$ and by using Eisenstein's Criterion for the resulting polynomial we conclude that resulting polynomial is irreducible and so is the original polynomial.  In general, for a given irreducible polynomial $f(x)$ with coefficients in a known U.F.D, is there some element $a$ such that we can apply Eisenstein's criterion to $f(x+a)$?  I am sure there would be no general structure for this but I expect there to be at least some special cases.   Any Reference/suggestion would be appreciated.  Thank You.
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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 $F$?
48957
Let $F$ be a field and $f\in F[X]$ a separable polynomial. Let $K_f$ be the splitting field of $f$ and $G_f=\text{Gal}(K_f/F)$ its Galois group. Show that $f$ is irreducible $\iff$ $G_f$ acts transitively on the roots on $f$. My work $\implies :$ Let $f\in F[x]$ irreducible, $\alpha \in K_f$ a root and $\alpha _1,...,\alpha _n$ the other roots of $f$ (so $K_f=F(\alpha ,\alpha _1,...,\alpha _n)$). What I have to show (I think) is that for all $i$ there is a $\sigma \in G_f$ s.t. $\sigma \alpha =\alpha _i.$ Q1) Is it right ? Q2) Is $f$ the minimal polynomial of $\alpha $ ? (To me it is, but my teacher said that it can be not the case, but I don't understand why). Since $\alpha $ and $\alpha _1$ have the same minimal polynomial, there is an isomorphism $\sigma :\alpha \longmapsto \alpha _1$. We can observe that $\sigma : F(\alpha )\longrightarrow K_f$. Now $F(\alpha ,\alpha _1)/F(\alpha )$ is algebraic, then we can extend $\sigma $ to a morphism $F(\alpha ,\alpha _1)\longrightarrow K_f$. Now, we can continue like that and extend $\sigma $ as $\psi: K_f\longrightarrow K_f$ s.t. $\psi(\alpha )=\alpha _1$. Q3) The fact that $\psi(\alpha )=\alpha _1$ comes from the fact that $\psi|_{F(\alpha )}=\sigma $, right ? The bijectivity is obvious since it's a field homomorphism and that $\sigma (K_f)=K_f$. Q4) How to conclude that $G_f$ act transitively on the roots ? I don't have particular problem for the converse.
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Let $h$ the irreducible polynomial over $\Bbb F_2 = \{0, 1\}$  $$h  =  x^4 +x +1$$  Let $E=\Bbb F_2(\alpha)$, i.e. the field $\Bbb F_2$ extended with $\alpha$, where $\alpha$ is a root of $h$.   How do we determine the numbers of elements in the extension $E$? Can every non-zero element in the extension be expressed in the form $(\alpha)^n$ for some $n\in\Bbb N$ ? How do we find all the roots of $h$ over the extension $E$?
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This is a question from Putnam and Beyond, and I have a question about the proof. The question is:  Show $x^{p-1} + 2x^{p-2} + 3x^{p-3} + \dots + (p-1)x + p$ is irreducible over $\mathbb{Z}[X]$.  The proof first shows that the polynomial $P(x) = x^{p-1} + 2x^{p-2} + 3x^{p-3} + \dots + (p-1)x + p$ only has zeros of absolute value $&gt; 1$. Then, it assumes it can be factored into two polynomials, namely, $P(X) = Q(X)R(X)$, and since $p = P(0) = Q(0)R(0)$, one of $Q(0), R(0) = \pm 1$, so WLOG $Q(0)$ has a zero of absolute value $&gt; 1$ which contradicts the fact that $P(X)$ only has zeros of absolute value $&gt; 1$. I don't see why $Q(0) = \pm 1$ means that it has a zero with absolute value greater than one. (The following is the proof in question.) &quot;We will show that all the zeros of $P (x)$ have absolute value greater than $1$. Let $y$ be a complex zero of $P(x)$. Then $0 = (y − 1)P(y) = y^p + y^{p−1} + y^{p−2} +···+ y − p.$ Assuming $|y| ≤ 1$, we obtain $p = |y^p + y^{p−1}+ y^{p−2}+···+ y| ≤ \sum_{i=1}^p |y|^ i ≤  \sum_{i=1}^p 1 = p.$ This can happen only if the two inequalities are, in fact, equalities, in which case $y = 1$. But $P (1) &gt; 0$, a contradiction that proves our claim. Next, let us assume that $P (x) = Q(x)R(x)$ with $Q(x)$ and $R(x)$ polynomials with integer coefficients of degree at least $1$. Then $p = P (0) = Q(0)R(0)$. Since both $Q(0)$ and $R(0)$ are integers, either $Q(0) = ±1$ or $R(0) = ±1$. Without loss of generality, we may assume $Q(0) = ±1$. This, however, is impossible, since all zeros of $Q(x)$, which are also zeros of $P (x)$, have absolute value greater than $1$. We conclude that $P(x)$ is irreducible.&quot;
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.
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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.
4763
I want to know if there is a way to decide if a cyclotomic polynomial is irreducible over a field $\mathbb{F}_q$?
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How do I show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$? Someone says I should use the rational root test, but I don't exactly know how that applies. Thanks for any input.
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Let $p$ be a prime. How do I prove that $x^p-x+a$ is irreducible in a field with $p$ elements when $a\neq 0$?   Right now I'm able to prove that it has no roots and that it is separable, but I have not a clue as to how to prove it is irreducible. Any ideas?
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This  shows that if every polynomial of degree 2 and odd degree has a root in $K[X]$, then $K$ is algebraically closed.   I wonder what examples can I find to see that the condition "odd degree" is necessary so that I cannot prove the statement:     Let $K$ be a field, if every polynomial of degree two is reducible   then $K$ is algebraically closed.   This  seems to give a procedure.  If I start with $\mathbb{F}_p$ then I know that adding a root of an irreducible of degree two over $\mathbb{F}_p$ will give me:   $\mathbb{F}_{p^2}$ by Moore's theorem there is only one field of cardinality $p^2$.  Any other root of an irreducible of degree 2 over $\mathbb{F}_p$ must be here .    The problem is that I do not get the roots of irreducible over $\mathbb{F}_{p^2}$ and therefore this process has to be iterated obtaining at the end the union $\cup_{i \ge 1} \mathbb{F}_{p^{2^i}}$. This does not need to be a fieldso that I shoud take the minimal field over it $\langle \cup_{i \ge 1} \mathbb{F}_{p^{2^i}} \rangle$.   In $\cup_{i \ge 1} \mathbb{F}_{p^{2^i}}$, it is clear that there cannot be a root of an irreducible of degree $3$. However, in principle I could find such a root in $\langle \cup_{i \ge 1} \mathbb{F}_{p^{2^i}} \rangle$. Why is it not the case?
20598
Determine the number of irreducible monic polynomials of degree 3 in $\mathbb F_p[x]$ where $p$ is a prime.  I'd like to start off by acknowledging that I know there are many posts relating to similar problems both on here and mathoverflow, however none address my problem.  In class, my professor went over the monic polynomials of degree 2 by reasoning that there are $p^2$ distinct polynomials, the reducible ones look like $(x - a)^2$ and $(x - a)(x - b)$ for the first one there are $p$ of them and for the second one there are $p(p-1)/2$ of them, subtracting the reducible ones from the total ones gives $$p^2 - p - \frac{p(p - 1)}{2} = \frac{2p^2 - 2p - p(p-1)}{2} = \frac{2p^2 - 2p - p^2 + p}{2} = \frac{p^2 - p}{2}.$$  I'm trying to use similar reasoning for polynomials of degree three using the fact that there are $p$ irreducible monic linear polynomials and $p(p-1)/2$ irreducible monic quadratic polynomials. But I can never get the result shown  and .  My reasoning is as follows: Observe that reducible polynomials would look like $(x - a)(x - b)(x - c)$, $(x - a)^2(x - b)$, $(x - c)^3$ and $(x^2 + ax + b)(x - c)$ formed from irreducible linear and quadratic monic polynomials.    For the first type where all three are distinct we can surmise that there are $p$ choices for the first, $p-1$ choices for the second and $p - 2$ choices for the third. So we have $p(p-1)(p-2)$. (is it $p$ choose $3$? Do I have to account for order not mattering?) For the second type there are $p$ choices for the square, and $p-1$ choices for the linear polynomial, so we have $p(p-1)/2$. For the third type, we have shown that there are $p(p-1)/2$ choices for the quadratic polynomials and there are $p$ choices for the linear polynomial. So we have $p^2(p-1)/2$.   Subtracting these values from $p^3$ should give me what I'm looking for, which is $p(p^2 - 1)/3$. But I'm not getting that.  Can someone tell me where my reasoning is wrong? Or tell me how to go about counting the number of irreducibles? (I know there are other methods of getting the answer, even more general ones, but I'm looking for a simple counting method that can help me better understand where my reasoning is wrong and help me correct it.)
21199
I want to prove that     $p(x):=x^n-x-1 \in \mathbb Q[x]$ for $n\ge 2$ is irreducible.   My attempt.   GCD of coefficients is $1$, $\mathbb Q$ is the field of fractions of $\mathbb Z$, and $\mathbb Z$ is UFD. Hence, $p(x)$ is irreducible over $\mathbb Q$ iff it's irreducible over $\mathbb Z$ (Gauss's lemma). Let $m\in \mathbb Z$ such that $\varphi(m)=n$ (Euler's totient). Make reduction of $p(x)$ by modulo $m$. Because of $\overline{x^n}=\overline{x^{\varphi(m)}}=\overline{1}$, we get $\overline{p(x)}=\overline{1-x-1}=\overline{-x}$, which is irreducible. Hence, $p(x)$ is irreducible.   Does this proof is correct?  UPDATE. Thanks to Calvin Lin. My mistake is: not for all $n$ we can find such $m$. OK, but as for the rest, does my proof is correct for such $n$, that $n=\varphi(m)$ for an integer $m$? And can it be some changed for all $n$, i.e. can we find such modulo that $\overline{p(x)}$ is irreducible for every $n\ge 2$?
19236
Show that the polynomial $X^5 + X^3 + \bar{1}$ in $(\mathbb{Z}/2\mathbb{Z})[X]$ is irreducible. (Hint: if it were reducible, it would either have a root or be of the form $g(X) \cdot h(x)$, where deg$g(X) = 2$ and deg$h(X) = 3$) Recall that $\bar{a}$ is shorthand for the coset $a + 2\mathbb{Z} \space $in$ \space (\mathbb{Z}/2\mathbb{Z})[X]$ (for any $a \in \mathbb{Z}$)  This question is worth $7$ marks out of a possible $75$, this is my solution..  $f(X) = X^5 + X^3 + \bar{1}$ is irreducible in $\mathbb{Z}/2\mathbb{Z}[X]$ iff it has no roots in $\mathbb{Z}/2\mathbb{Z}$. Since $f(\bar{0}) = \bar{1}$ and $f(\bar{1}) = \bar{1}$, it follows that $f(x)$ is irreducible in $\mathbb{Z}/2\mathbb{Z}[X]$.  Would this be a sufficient answer? am I missing something?
47441
Is there a general or elegant way to approach this problem? One can show that $\sqrt 5+\sqrt[3] 2$ is a root of the hexic $x^6-6x^4-10x^3+12x^2-60x+17$, which should then be its minimal polynomial to get the result. I see no straightforward way to show this is irreducible, though. It has a root in $\mathbf{F}_4$, and to show it's irreducible over $\mathbf{F}_3$ already seems to require a lot of grunt work.  One thing I can do is rule out the possibility that $[\mathbf{Q}[\sqrt 5+\sqrt[3] 2]:\mathbf{Q}]=2$, because $[\mathbf{Q}(\sqrt 5,\sqrt[3] 2):\mathbf{Q}]=6$ and $\mathbf{Q}(\sqrt 5,\sqrt[3] 2)$ is the compositum of $\mathbf{Q}[\sqrt 5+\sqrt[3] 2]$ and $\mathbf{Q}[\sqrt 5]$, where the latter has degree two. But I don't seem to be able to play an analogous trick to rule out $3$, because I know of no general lower bound on the degree of a compositum.   Could there be some way to bring Galois theory into the picture? These extension aren't Galois, and so it's by no means clear to me how to reason from their normal closures, but perhaps that's the direction to go in. Suggestions are very much appreciated.