Let, $n\in \mathbb{N}$ and $p$ be a prime. If $f(x)=\sum_{i=1}^{r} a_{i} x^{i}$ and $g(x)=\sum_{i=1}^{s} b_{i} x^{i}$ be polynomial in $\mathbb{Z}_{p^{n}}$ such that $f(x)g(x)=0$ , (1) $a_{i} b_{j} =0$,for  all $i,j$ (2) $a_{i} b_{j} \neq 0$,for  some  $i,j$ My try, $a_{1} b_{1}=0 $ Either $a_1=0$ or $b_1=0$ or they are divisor of zero. Then $a_{1} b_{2}+a_{2} b_{1}=0 $ Now, if $a_1=0$, then $a_{2} b_{1}=0 $ After, $a_{1} b_{3}+a_{2} b_{2}+a_{2} b_{3} = 0 $ If, $a_1=0$ then $a_{2} b_{2}+a_{2} b_{3} = 0 $ My thought here, if $\mathbb{Z_{p^n}}$ is $\mathbb{Z_2} $ here, then, I think option (2) is true if we take all of $a_2,b_2,b_3$ non zero. But is option (1) the answer? I am really in doubt here. Please anyone help!
123135
Let $n \in \mathbb{N}$ and $p$ be a prime number. Let $f(x),g(x) \in \mathbb{Z}_{p^n}[x]$ such that, in $\mathbb{Z}_{p^n}[x]$, $f(x)g(x)=0$. Prove that $a_ib_j=0$ for all $i,j$, where $a_i$ and $b_j$ are the coefficients of $f(x)$ and $g(x)$.  I have the set of equations (for degree $r$ and degree $s$): $$a_rb_s=0, a_rb_{s-1}+a_{r-1}b_s=0,...,a_0b_0=0.$$ From these, how do I conclude $a_ib_j=0$ for all $i,j$? Also, I found that for $n=1$, the polynomial ring is an integral domain, so the conclusion holds. I had an idea to use induction on $n$, but do not know how to proceed.
2317
In one of my assignment, I was told that $(\mathbb Z[x], +)$ and $(\mathbb Q_{&gt;0}, .)$ are isomorphic, with $$\phi (\sum_{k=0}^n a_k x^k) = \prod_{k=0}^n p_k^{a_k}$$ is a one-to-one surjective group homomorphism, where $p_1 = 2$; $p_2 = 3$; $p_3 =5$, and so on.  I know that this proof is based on the decomposition of numbers into a product of primes. But I don't think that it is correct. For instance, for $\frac12 \in \mathbb Q$, I cannot find a corresponding polynomial $f \in \mathbb Z[x]$ such that $\phi(f) = \frac12$.  If this proof is wrong, then are $(\mathbb Z[x], +)$ and $(\mathbb Q_{&gt;0}, \cdot)$ really isomorphic?
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).
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.
87297
Let $f(x) = x^2 + ax + b$ be a quadratic polynomial in which $a$ and $b$ are integers. If for a given $n$ , $f(n)f(n+1) = f(m)$ for some integer $m$ then the value of $m$ is ?  Do we have to substitute $x$ by $n$ and $n+1$ and then multiply it? or is there another method to get the solution because the mentioned process will give an extremely large equation.  Answer is $n(n+1)+an+b$
81764
Let $f(x) \in \mathbb{Z}[x]$. If we reduce the coefficents of $f(x)$ modulo $p$, where $p$ is prime, we get a polynomial $f^*(x) \in \mathbb{F}_p[x]$. Then if $f^*(x)$ is irreducible and has the same degree as $f(x)$, the polynomial $f(x)$ is irreducible.   This is one way to show that a polynomial in $\mathbb{Z}[x]$ is irreducible, but it does not always work. There are polynomials which are irreducible in $\mathbb{Z}[x]$ yet factor in $\mathbb{F}_p[x]$ for every prime $p$. The only examples I know are $x^4 + 1$ and $x^4 - 10x^2 + 1$.  I'd like to see more examples, in particular an infinite family of polynomials like this would be interesting. How does one go about finding them? Has anyone ever attempted classifying all polynomials in $\mathbb{Z}[x]$ with this property?
10635
Let $P \in \mathbb{Z}[x], P(x) = \displaystyle\sum\limits_{j=0}^n a_j x^j, a_n \neq 0$ and $a_0 \neq 0$; if $p/q$ is a root of P (with p and q coprimes) then $p|a_0$ and $q|a_n$  I've managed to prove the first part ($p|a_0$) and I suppose I'm not far from proving the second, though I'd really like some feedback since I'm just starting with making proofs of my own.  Proof:  $P(x) = a_n(x-p/q)\displaystyle\prod\limits_{j=2}^n (x-r_j)$, with $r_j$ being the other n-1 roots of P(x).  It follows that $a_0 = a_n(-p/q)\displaystyle\prod\limits_{j=2}^n (-r_j)$  Then, $-p/q|a_0$ and obviously $p/q|a_0$. Rephrasing, $a_0 = l\frac{p}{q} = \frac{l}{q} p$ with $l \in \mathbb{Z}$.  This implies $p|a_0$ if $l/q \in \mathbb{Z}$, but this is trivial since $q|lp$ and q and p are coprimes, so $q|l$.  Therefore, $p|a_0$.  As for the second part, we want to see that $q_i|a_n \forall i \leq n$.   We define $d$ as the least common multiple of $\{q_1, q_2,...,q_n\}$. Then, $q_i|a_n \forall i \leq n \iff d|a_n$.  Also, it follows that $d|\displaystyle\prod\limits_{j=1}^n q_j$, so we want to see that $a_n = l \displaystyle\prod\limits_{j=1}^n q_j$ with $l \in \mathbb{Z}$. Here's where I have my doubts with the proof as I have no way to show that l is indeed an integer.  Rearranging the previously given equation for $a_0$:  $a_n = a_0 \displaystyle\prod\limits_{j=1}^n \frac{-q_j}{p_j}$  Using the previous reasoning, as $p_i|a_0 \forall i \leq n$, then $a_0 = k \displaystyle\prod\limits_{j=1}^n p_j$.  Replacing $a_0$:  $a_n = k \displaystyle\prod\limits_{j=1}^n -q_j$, which is equivalent to $q|a_n$ as shown earlier.
105683
We know that if $F$ is a field and $f(X)$ a non-zero polynomial in $F[X]$, then for every polynomial $g(X)$ we can find $q(X),r(X)$ such that $$g(X)=f(X)\cdot q(X)+r(X)$$ with $r(X)$ the zero polynomial or $\deg r(X)&lt;\deg f(X)$. My question is: the same algorithm holds for many variables? So can i write $$g(X_1,\ldots,X_n)=f(X_1,\ldots,X_n)\cdot q(X_1,\ldots,X_n)+r(X_1,\ldots,X_n)$$
66110
Let $R = \mathbb{Z} + x\mathbb{Q}[x] \subset \mathbb{Q}[x]$. Find the irreducibles of $R$. Show that the irreducible elements in $R$ are $\pm p$ for prime integers $p$ and the irreducible polynomials $p(x) \in \mathbb{Q}[x]$ whose constant coefficient is $\pm 1$. Prove these irreducibles are prime in $R$.  The first part is pretty clear. ($\pm p$ for prime integers $p$), if $\alpha \in R$ s.t. $\deg(\alpha)=0$ then $\alpha$ is a constant polynomial thus in $\mathbb{Z}$. If $\alpha$ is a prime then it is irreducible. Having trouble showing that for $p(x) \in R$ s.t. $\deg(p(x)) &gt; 0$ with constant term $\pm 1$ are irreducible.
84192
Let $f$ be a rational function mapping to $\mathbb{C}$, $$f(z)=\frac{P(z)}{Q(z)}$$ with $\deg P\leq \deg Q -2$. I want to show that the sum over all residues is zero.  What am I asked to show? I think: The only singularities are in the zeros of $Q$. So if $N_Q$ is the set of zeros of $Q$ I have to show $$\sum_{q\in N_Q} Res(f,q)=0,$$haven't I? Since there are max. $\deg Q$ many zeros of $Q$, the sum is finite.  In order to show the statement - how can I use the $\deg$ condition?
89088
Despite the fact that $\forall n, n^3 + 2n \equiv 0 \pmod 3$, I understand that $n^3 + 2n$ (considered as a polynomial with coefficients in $\mathbb Z/3\mathbb Z$) is not equal to the zero polynomial.  What is the value of defining polynomials in this (strange) way? What situations does it make things simpler?  I ask this because it seemed natural to me to define polynomials as a subset of functions, so I was surprised by this.
24677
Let $a\in \mathbb{Z}$ be such that $a=b^2+c^2$ where, $b,c \in \mathbb{Z}-\{0\}$. Then $a$ cannot be written as    $pd^2$ where $d \in \mathbb{Z}$ and $p$ is prime with $p \equiv 1~(mod~4)$ $pd^2$ where $d \in \mathbb{Z}$ and $p$ is prime with $p \equiv 3~(mod~4)$ $pqd^2$ where $d \in \mathbb{Z}$ and $p,q$ are primes with $p \equiv 1~(mod~4)$ and $q \equiv 3~(mod~4)$ $pqd^2$ where $d \in \mathbb{Z}$ and $p,q$ are primes with $p,q \equiv 3~(mod~4)$   The similar question with some part missing is asked , but I want to discuss additional options too.  Here, I got 1) is false Since $13=3^2+2^2$ For other options What should I do
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
3445
I know that for every $n\in\mathbb{N}$, $n\ge 1$, there exists $p(x)\in\mathbb{F}_p[x]$ s.t. $\deg p(x)=n$ and $p(x)$ is irreducible over $\mathbb{F}_p$.     I am interested in counting how many such $p(x)$ there exist (that is, given $n\in\mathbb{N}$, $n\ge 1$, how many irreducible polynomials of degree $n$ exist over $\mathbb{F}_p$).   I don't have a counting strategy and I don't expect a closed formula, but maybe we can find something like "there exist $X$ irreducible polynomials of degree $n$ where $X$ is the number of...".  What are your thoughts ?
66378
Suppose $p$ and $q$ primes and $p$ is odd. Then, are there nice and clever ways to factorize polynomials of the form $$f(x)=1+x+\cdots +x^{p-1}$$ in the ring $\mathbb{F}_q[x]$ ? What about the case when $q=2$ ?   I know that there are factorization algorithms but they are too general. I want to know if there are clever ways to do this for these special type of polynomials. Like for example, in $\mathbb{F}_2$ one might add terms $x^r+x^r$ and rearrange them so that $f(x)$ gets factorized.  In case there are no good methods to factorize, are there nice ways to check whether $f(x)$ is irreducible in $\mathbb{F}_q[x]$ ?
28992
Let $F$ be a finite field and $i$ an integer. Calculate the sum of all the elements of $F$, each raised to the $i$th power.                                                            My approach so far:  Let $F=(0,1,\alpha,\alpha^2,...,\alpha^{p^n-1})$, where $p$ is a prime number and let $\sigma=1+\sum_{k=1}^{p^n-1}(α^k)^i$ be the desired sum.  Since $F$ is a ring, each element has an additive inverse. Thus, the trivial case of $i=1$ results in $\sigma=1+[\alpha+(-\alpha)+\alpha^2+(-\alpha^2)+\cdots+\alpha^{p^n-1}+(-\alpha^{p^n-1})]=1$.   We can also calculate the trivial case of $i=0$, where $\sigma=1+p^n-1=p^n$   In the general case, let $i=t (\text{mod } p^n)$. Then: $i=t+mp^n$, where $t,m$ are integers. We take into account the fact that the order of the multiplicative group of the finite field is $p^n$, so each element raised to the $p^n$ equals $1$. Thus $σ=1+α^t+α^{2t}+\cdots+α^{t(p^n-1)}$. This is the sum of a geometric progression, plus $1$. Thus, $σ=1+α(α^{p^n-1})/α-1$ = $ 1$+($α^{p^n}-α$)/${\alpha-1}$. But according to Lagrange's theorem, for every $\alpha\in F$, the polynomial $x^{p^n}-x$ is zero. It follows that $\sigma=1$.  An important question is raised: Is it true that the characteristic of the finite field is $p$ if its order is $p^n$? If so, the case for $i=0$ leads to $σ=0≠1$ and all cases have been considered?
53386
I have to find $P(0)$ from the polynomial with minimum degree given that  $$(x-1)^3|(P(x)+1)$$  $$(x+1)^3|(P(x)-1)$$  Plugging in $x=\pm 1$ gets something nice, also division by a polynomial of third order gives successively:  $$P(1)+1 =0; \ P'(1)=0; \ P''(1)=0$$ $$P(-1)-1 =0; \ P'(-1)=0; \ P''(-1)=0$$  Furthermore since $P(1)=-1$ and $P(-1)=1$, also $P(1)=-P(-1)$  and $$P'(1)=P''(1)=P'(-1)=P''(-1)$$ I don't see how to use this stuff.