I am attempting to solve part (d) of the following problem from Artin's algebra book. I have already solved parts (a) through (c).  (a) Let $p$ be an odd prime. Prove that exactly half of the elements of $\mathbb{F}_p^\times$ are squares and that if $α$ and $β$ are nonsquares, then $αβ$ is a square. (b) Prove the same assertion for any finite field of odd order. (c) Prove that in a finite field of even order, every element is a square. (d) Prove that the irreducible polynomial for $γ=\sqrt2+\sqrt3$ over $\mathbb{Q}$ is redubile modulo $p$ for every prime $p$.  Now, I have shown that $g(x)=x^4-10x^2+1\in \mathbb{Q}[x]$ is the irreducible polynomial for $γ$. Let $p$ be a prime. Let $f(x)=x^4-10x^2+1 \in \mathbb{F}_p[x]$. I want to show $f$ is reducible. I see that either 2, 4, or 6 is equal to $s^2$ for some $s\in\mathbb{F}_p$. I think it will be useful to proceed by cases here. It someone could show me a complete, elementary, simple proof for the $s^2=2$ case, I should be able to finish the rest on my own.
120574
So I have seen the similar question and answers on here for $x^4 +1$, but I am having trouble extending anything there to this polynomial... I understand it is fairly trivial with Galois theory, but my class has just barely covered Field Extensions, so suffice it to say we have no Galois theory to play with.  I managed to prove it for the primes such that $p \equiv 1, 7 \pmod 8$, by noting that $2$ is a square modulo those primes and thus $x^4 - 2x^2 +1 = (x^2 -1 + 2qx)(x^2 - 1 - 2qx)$ for those $\mathbb{Z}_p$... however, trying to get a similar result for $3 \pmod 8$ and $5 \pmod 8$ has been stumping me for a long time, I am having a hard time making $q^2 = -1$ and $q^2 = -2$ give me something factorable...  I guess the worst part of all of this is that I don't think this solution is even particularly enlightening, in terms of abstract algebra. It's really just some number theory trickery. I don't think my course has prepared me theoretically for this problem, does anyone have an elementary approach to it?
107461
I have to prove the following theorem :     Let $p$ be a prime number and let $n \ge 1$,be any integer, then there exists a field of order $p^n$.   My attempt  I started off by considering the polynomial $f(x)$=$x^{p^n}-x \in \Bbb Z_p[x]$.  I took $F$ to be the splitting field of $f(x)$ over $\Bbb Z_p$.  Since $F$ is the splitting field of $f(x)$ over $\Bbb Z_p$ therefore $f(x)$ has exactly $p^n$ zeros in $F$ counting multiplicity.  Also since $f'(x)$ = $p^nx^{p^n-1}-1$ , therefore $f(x)$ will not have any multiple zeros and hence it will have $p^n$ distinct zeros in $F$.  Then I took $K =\{k \in F | k^{p^n} = k\}$.  I was trying to prove that $K$ is a sub-field of $F$.  $1 \in K$ thus $K$ is non-empty.  For $a,b \in K$ $a^{p^n}=a$ and  $b^{p^n}=b$  So $(a+b)^{p^n}$ (should be) = $a^{p^n}$+$b^{p^n}$, this is where I am currently stuck!  I know that this is true when  $p$ is the characteristic of a field, but why would this expansion be true here?
10320
If $a$ and $b$ are two relatively prime positive integers such that $ab$ is a square, then $a$ and $b$ are squares.   I need to prove this statement, so I would like someone to critique my proof. Thanks  Since $ab$ is a square, the exponent of every prime in the prime factorization of $ab$ must be even.   Since $a$ and $b$ are coprime, they share no prime factors.   Therefore, the exponent of every prime in the factorization of $a$ (and $b$) are even, which means $a$ and $b$ are squares.
36969
So I am trying to show that $f(x) = x^4-3$ will factor over $\Bbb{F}_p$ when $p\equiv 3\mod 4$, with $p\neq 3$. Specifically, I am trying to show that $f$ will factor into irreducibles such that there is at least one factor of degree 2.  My Work So Far  So if I assume that it can be factored, we would have $$ x^4-3 = (x^2+ax+b)(x^2-ax+d) = x^4 + (b+d-a^2)x^2 + a(d-b)x + bd.$$ Because we want the linear term to be 0, we need either $a=0$ or $d-b = 0$. Now if $a=0$ we get that $b=-d$, so $-b^2 = -3$. That is, $3$ is a square. If $d-b = 0$, then we would have $b^2 = -3$, so $-3$ is a square. Because $-1$ is not a square when $p\equiv 3\mod{4}$, we know that one of $-3$ or $3$ must be a square, but not both, so $a=0$ or $d=b$, but not both.  If $a=0$, that is $3$ is a square, then we have $f(x) = (x^2 +b)(x^2-b)$. Then one of $b$ or $-b$ is a square, but not the other, so $f$ will factor into two linear terms and a quadratic. So this case works.  The trouble I am having is with the second case. If $d=b$, then we have  $$ f(x) = (x^2 + ax + b)(x^2 -ax +b)$$ with $2b = a^2$. This is where I get stuck. How do I know that such an $a$ can exist?  Edit: If there is a better approach to solving this, that would be great to know as well. I haven't been taught any of the relevant number theory to solve this, so I'm kind of out of my element here.
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?
74153
Suppose $K = \mathbb{Q}(\alpha)$ with $\alpha = a + b\sqrt{D_1}+c\sqrt{D_2}+d\sqrt{D_1D_2}$ with $D_1,D_2 \in \mathbb{Z}$. Prove that the minimal polynomial $m_\alpha(x)$ for $\alpha$ over $\mathbb{Q}$ is irreducible of degree 4 over $\mathbb{Q}$ but is reducible modulo every prime $p$. In particular show that the polynomial $x^4 - 10x^2 +1$ is irreducible in $\mathbb{Z}[x]$ but is reducible modulo every prime. [Use the fact that there are no biquadratic extensions over finite fields.]  So far I have established the following:  $[\mathbb{Q}(\alpha):\mathbb{Q}]=\deg(m_\alpha(x))=4$  $Gal(\mathbb{F}_{p^n}/\mathbb{F}_p)$ is cyclic, hence no biquadratic extension (which is iso to $V_4$) exists over finite fields.  I'm having a problem proving the reducibility mod every prime though. Any hints?
91654
Let $K$ be a field and $f \in K[x]$ a polynomial of degree $n&gt;0$. Assume that $E$ is an extension of $K$ such that $f$ can be expressed as a product of linear factors over $E$:  $$f(x)=\gamma (x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_n).$$  I want to show that $\lvert K(\alpha_1,\dots,\alpha_2) : K \rvert$ is a divisor of $n!$. I also want to show that if $\lvert K(\alpha_1,\dots,\alpha_n): K \geq (n-1)!$ then $f$ is irreducible over $K$ or $f$ already has a root in $K$.  The problem has a hint. It says: "Use induction on $n=\deg f$. If $f=gh$ in $K[x]$ with $\deg g =r$ and $\deg h=n-r$, then $r!(n-r)!$ is a divisor of $n!$.  Any idea for solving this problem would be appreciated.
16690
Let $F$ be a finite field. How do we prove that for each $n \in \mathbb{N}$ there is an irreducible polynomial of degree $n$?  One can assume that $F = \mathbb{F}_{p^m}$ where $p$ is prime. If $n \ge |F|$ then I can construct an irreducible polynomial, namely  $ p(x) = 1 + \prod_{j=1}^{|F|} ( x - a_j )$  where $a_j$ are all the field elements. It is clear that $p(x)$ has no roots in $F$.  This trick doesn't work for $n &lt; |F|$. A counter-example: Let $F = \mathbb{F}_3$ and $p(x) = 1 + (x-1)(x-2)$, then $p(1) = 1$, $p(2) = 1$, $p(0) = 1 + (-1)(-2) = 1 + 2 = 3 = 0 \pmod 3$.  I know there is a way to count them using the Möbius function $\mu(n)$ but I want a proof without it that just shows existence.
65602
Prove that any polynomial in $F[x]$ can be written in a unique manner as a product of irreducible polynomials in $F[x]$.  $F$ is a field of fractions of a UFD.  Can someone tell me how I can solve this problem. It is one of my lecture note result, but my lecturer did not provide a proof.
55256
A polynomial $p$ over a field $k$ is called irreducible if $p=fg$ for polynomials $f,g$ implies $f$ or $g$ are constant. One can consider the determinant of an $n\times n$ matrix to be a polynomial in $n^2$ variables. Does anyone know of a slick way to prove this polynomial is irreducible?  It feels like this should follow quite easily from basic properties of the determinant or an induction argument, but I cannot think of a nice proof. One consequence of this fact is that $GL_n$ is the complement of a hypersurface in $M_{n}$. Thanks.
78866
I have proven that in $\mathbb F_{p^2}^*$ exists an element $\alpha$ with $\alpha^8 = 1$.      Let $f(X) := X^4+1 \in \mathbb F_p[X]$. How can I prove that $f$ is reducible over $\mathbb F_p$?    Has $f$ a zero in $\mathbb F_p$ ?
57055
Let $K$ be an algebraically closed field ($\operatorname{char}K=p$). Denote $${\mathbb F}_{p^n}=\{x\in K\mid x^{p^n}-x=0\}.$$ It's easy to prove that ${\mathbb F}_{p^n}$ consists of exactly $p^n$ elements.  But if $|K|&lt;p^n$, we have collision with previous statement (because ${\mathbb F}_{p^n}$ is subfield of $K$).  So, are there any finite algebraically closed fields? And if they exist, where have I made a mistake?  Thanks.
61503
Problem: Is there an irreducible $f\in \mathbb{Z}[x]$, whose image in every $(\mathbb{Z}/p\mathbb{Z})[x]$ has a root for $p$ prime? If there is, what is the minimal degree possible?   I can only prove that $x^2-c$ is impossible, by quadratic reciprocity and Chinese remainder theorem. Even the case of $a x^2 - c$ is unknown to me.  Meanwhile, if $f$ is not required  to be irreducible, but only have no root in $\mathbb{Z}$, then $(x^2-a)(x^2-b)(x^2-ab)$ is a solution for non-sqaure $a,b,ab$, since if both $a,b$ are not squares in $\mathbb{Z}/p\mathbb{Z}$, then $ab$ is a square. Yet I still don't know if this is of minimal degree.  Also, it is natural to pose generalizations   When $p$ is not necessarily prime (equivalently for all prime powers). When $p$ is odd prime. When $p$ represents sufficiently large primes (equivalently all but a finite number left).
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!
121085
Let $p,q$ be primes, $p≠q$, then I have to show that $\mathbb{Q}(\sqrt{p},\sqrt[3]{q})= \mathbb{Q}(\sqrt{p}\cdot \sqrt[3]{q})$  So far I've tried a lot of things with minimal polynomials and bases, but I'm far from any argument that I find sufficient.  This is an example from my algebra course, first chapter about field extension. So far we know about basic definitions, minimal polynomials and their properties, algebraic elements and algebraic fields.  Could someone maybe give me a hint about how to proceed? Thanks a lot!!!
85214
A polynomial is called a Fermat polynomial if it can be written as the sum of the squares of two polynomials with integer coefficients. Suppose that $f(x)$ is a Fermat polynomial such that $f(0) = 1000$. Prove that $f(x) + 2x$ is not a Fermat polynomial.  What I have done:  Let $p(x)$ be a Fermat polynomial such that $p(0)$ is divisible by $4$. Suppose that $p(x) = g(x)^2+h(x)^2$ where $g(x)$ and $h(x)$ are polynomials with integer coefficients. Therefore $g(0)^2 + h(0)^2$ is divisible by $4$. Since $g(0)$ and $h(0)$ are integers, their squares are either $1 (mod 4)$ or $0 (mod 4)$. It therefore follows that $g(0)$ and $h(0)$ are even. From here, I can not proceed any further. Please help
79858
Prove that $P_n(X) = X^n - X^{n-1} - X^{n-2} - ... - X - 1$ is irreducible over $\mathbb{Z}$ for all $n$.  I was able to prove the result for $n=2^k-1$ by applying  to $P_n(X+1)$. But for other values of $n$, I'm stuck. Has anyone an idea on this ?