Master Polynomial Concepts with Comprehensive Resources

Is it meaningful to calculate $(x+1{)}^{x+2}$ in $GF(3^2)$, e.g. using discrete logs?
I constructed GF$3^2$ using the polynomials degree $<2$ and arithmetic modulo the irreducible polynomial $(x^2+1)$. Then I built a table of logs to the base of the primitive element $(2x+2)$.
I can see that the log function maps only the non-zero elements of the field, so it seems sensible to use modulo 8 arithmetic between log values. This seems to work - for example:
$\begin{array}{rl}{\log }_{2x+2}<!--\; \; -->({(2x+1{)}^2}_{mod{x}^2+1})& =(2\times <!--\; \times \; -->{\log }_{2x+2}<!--\; \; -->(2x+1){)}_{mod8}\\ {\log }_{2x+2}<!--\; \; -->(x)& =(2\times <!--\; \times \; -->(7){)}_{mod8}\\ 6& =6\end{array}$
But how should I calculate, for example, $(x+1{)}^{(x+2)}$? Clearly modulo 8 arithmetic isn't going to do the trick. I guess that I need to use an irreducible polynomial but I am uncertain as to which one. Can I just use $(x^2+1)$ again?

What is the remainder when $1^n+2^n+3^n+\dots <!--\; \dots \; -->+{99}^n$ is divided by $1+2+3+\dots <!--\; \dots \; -->+99$? My first idea on how to approach was to create a polynomial expansion for the first few terms and then try to find a pattern for the rest but this became cumbersome and I don't think that this is a right approach as this would quickly become a problem that involves factorials and I have not covered modular arithmetic, is there a way to approach this problem in such a way that does not involve using factorials with mods.

If a polynomial is zero on a field F, is it zero on every extension of F?
Let p be a univariate polynomial over a field F, and let K be an extension of F.
If $p(x)=0$ for all $x\in <!--\; \in \; -->F$, does this imply that $p(x)=0$ for all $x\in <!--\; \in \; -->K$? How about if p is multivariate?
For context, I'm trying to understand if doing Schwartz-Zippel-style arithmetic circuit identity-testing over a large enough extension field gives the right answer when the degree of the expression may be high.

Can the concept of congruence be applied to the remainder of a polynomial division?
Can the concept of (modular arithmetic) congruence be applied to the remainder of a polynomial division? (e.g. when the divisor is a polynomial of at least degree 1?
For instance, $\frac{x^2-<!--\; -\; -->1}{x+1}=x-<!--\; -\; -->1$ with remainder = 0. Then:
Does it means that $x^2-<!--\; -\; -->1\equiv <!--\; \equiv \; -->0(\mathrm{mod}x+1)?$
If so, what happens if the remainder is not an integer? If I am not wrong the remainder of the polynomial division can be for instance a fraction as well so I am not sure what happens (what is the meaning) when the remainder is not exactly an integer.
I have seen that there is a tag in MSE about polynomial congruences and did some searching but the questions are very few and different, and some of them not related with a question regarding a division in which the divisor is also a polynomial.
If somebody could provide me a basic explanation as a starting point (or an online resource to learn a minimum base) would be very appreciated.

$\mathbb{Q}(\sqrt{1-<!--\; -\; -->\sqrt{2}})$ is Galois over Q
I am trying to show that $E=\mathbb{Q}\left(\sqrt{1-<!--\; -\; -->\sqrt{2}}\right)$ is galois over Q. The extension has the minimal polynomial
$(x-<!--\; -\; -->\sqrt{1-<!--\; -\; -->\sqrt{2}})(x+\sqrt{1-<!--\; -\; -->\sqrt{2}})(x-<!--\; -\; -->\sqrt{1+\sqrt{2}})(x+\sqrt{1+\sqrt{2}})$
but I can't manage to show using elementary arithmetic that $\sqrt{1+\sqrt{2}}$ is in E (i.e. multiplying, adding etc.) so that its the splitting field.
The trick would be to use that $\sqrt{1+\sqrt{2}}=\frac{i}{\sqrt{1-<!--\; -\; -->\sqrt{2}}}$ but we only know that $i\sqrt{\sqrt{2}-<!--\; -\; -->1}$ is in the field, we dont know if we have i.
Can someone give a hint on how to proceed, maybe there is a theorem I could use that I'm not thinking of.

Value Sets miss an AP
For any polynomial $p\in <!--\; \in \; -->\mathbb{Z}[X]$, the set $\{p(n):n\in <!--\; \in \; -->Z\}$ is called the value set of p.
Problem:- Let $p\in <!--\; \in \; -->Z[X]$ be such that deg $p>1.$ Then there exists an infinite arithmetic sequence none of who terms can be expressed as p(x) for some $x\in <!--\; \in \; -->Z.$
Progress:-
For this problem, I first took example which was $P(x)=x^2-<!--\; -\; -->1$ then we get the value set as {0,3,4,15,24,…}. Since, $x^2-<!--\; -\; -->1\equiv <!--\; \equiv \; -->-<!--\; -\; -->1,0\mathrm{mod}4.$. So clearly the AP 2,6,10,… never come.
I assumed the contrary. Hence for any n,d there will exist a x such that $P(x)\equiv <!--\; \equiv \; -->n\mathrm{mod}d.$. If there doesn't then we can select the AP $n,n+d,n+2d,\dots <!--\; \dots \; -->$
Now we know that when $p\in <!--\; \in \; -->\mathbb{[}X]$, and $m\equiv <!--\; \equiv \; -->n\mathrm{mod}d⟹<!--\; ⟹\; -->P(m)\equiv <!--\; \equiv \; -->P(n)\mathrm{mod}d.$. (*)
So, we can say that $p(n),p(n+1),\dots <!--\; \dots \; -->,p(n+d-<!--\; -\; -->1)$ forms a complete residue set, because if it doesn't then by PHP, there will be an $y\in <!--\; \in \; -->1,2,\dots <!--\; \dots \; -->d$ not being a residue in p(x), when $n\le <!--\; \le \; -->x\le <!--\; \le \; -->n+d-<!--\; -\; -->1.$
Now, take any d consecutive numbers,say $l,l+1,\dots <!--\; \dots \; -->,l+d-<!--\; -\; -->1,$ by (*), residues mod d of $\{P(l),\dots <!--\; \dots \; -->,P(l+d+1)\}=\{p(n),p(n+1),\dots <!--\; \dots \; -->,p(n+d-<!--\; -\; -->1)\}.$
This is what I have got till now.

Arithmetic progression and polynomials
Suppose the quadratic polynomial
$p(x)=a{x}^2+bx+c$
has positive coefficients a,b,c such that these are in AP in the given order. If m and n are the integer zeros of the polynomial then what is the value of $m+n+mn$? I have tried quadratic formula but ain't getting the answer.

Algorithm to find the irreducible polynomial
What algorithm can be used find an irreducible polynomial of degree n over the field GF(p) for prime p. The reason I ask is I want to make a program for finite field arithmetic, but creating a field $GF(p^k)$ requires finding a irreducible polynomial. (I'm mostly interested in $GF(2^k)$.)

Smallest prime in arithmetic progressions: upper bounds?
Given $N>2$, find a prime p such that
$\begin{array}{}\text{(1)}& p\equiv <!--\; \equiv \; -->1(\mathrm{mod}N).\end{array}$
(Of course, Dirichlet's theorem assures us that infinitely many such p exist.)
Now, I do not know of a clever way to find such a p, so I suggested simply iterating through the arithmetic progression $1+kN$ for $k=0,1,2,\dots <!--\; \dots \; -->$, and terminating when we hit the first prime number. Clearly, this procedure is efficient if and only if the smallest prime p satisfying (1) is at most $N^{O(1)}$. (Edit: This claim is incorrect; see below.) I want to know if such a bound holds or not.
More generally,
Given an integer N, what upper bound do we know on the smallest prime satisfying (1)?
I did a quick check on Wikipedia and Mathworld, but cannot find any relevant results.

Showing that $\mathbb{R}[X,Y]/(X^2+1,Y^2+1)\cong <!--\; \cong \; -->\mathbb{C}\times <!--\; \times \; -->\mathbb{C}$
A problem in my textbook asks me to show that $\mathbb{R}[X,Y]/(X^2+1,Y^2+1)\cong <!--\; \cong \; -->\mathbb{C}\times <!--\; \times \; -->\mathbb{C}$. After I've tried solving this for a long time, I looked at the textbook's hint and it said that I should consider the homomorphism $\mathrm{\varphi <!--\; \varphi \; -->}:\mathbb{R}[X,Y]\to <!--\; \to \; -->\mathbb{C}\times <!--\; \times \; -->\mathbb{C}$, $\mathrm{\varphi <!--\; \varphi \; -->}(f)=(f(i,i),f(i,-<!--\; -\; -->i))$ and this would be a surjective ring homomorphism whose kernel is the ideal $(X^2+1,Y^2+1)$, so I would be done.
The fact that this is a ring homomorphism is pretty obvious to me. However, I am really struggling to show that this function is surjective and to find the kernel.
I think that this all stems from the fact that I am not used to the arithmetic properties of polynomial rings in two indeterminates. Anyway, back to the problem. For the surjectivity I tried playing arround with a few polynomials and I hoped to kind of guess the right one, but no succes. Showing that $(X^2+1,Y^2+1)\subset <!--\; \subset \; -->\mathrm{Ker}<!--\; \; -->\mathrm{\varphi <!--\; \varphi \; -->}$ is really straightforward, but the other inclusion is really killing me. So, how should I go about doing this?

I am reading Fried-Jarden's book Field Arithmetic and I have a question on Pseudo Algebraically closed (PAC) fields.
We say that a field K is PAC if every variety absolutely irreducible over K has a K-rational point. Recall that a variety is absolutely irreducible if it is irreducible over $\stackrel{¯<!--\; ¯\; -->}{K}$, and a K-rational point is an element on the variety with all its coordinates in K.
My question is the following: Is Q a PAC field? And more generally, is any number field PAC?
My attempt: I would like to find a polynomial in two variables with coefficients in Q which is irreducible over C and that does not have any root in ${\mathbb{Q}}^2$. This would show that Q is not PAC, which is my guess, but so far i've been unable to find this example.
I am also guessing that, more generally, every number field is not PAC, although this question is much more difficult...

Definition and decidability of bounded quantifiers
Consider quantifier-free formulas $P(x,y)=Q(x,y)$ of Peano arithmetic.
Consider P(x,y),Q(x,y) to be terms composed of variables $x,y,\mathrm{succ},+,\times <!--\; \times \; -->$.
Note that these are equivalent with polynomials $P(x,y),Q(x,y)\in <!--\; \in \; -->{\mathbb{Z}}_{+}[x,y]$.
Let R(x) be correspondingly a term, resp. a polynomial in ${\mathbb{Z}}_{+}[x]$.
The quantifier of a formula $(\mathrm{\exists <!--\; \exists \; -->}y)P(x,y)=Q(x,y)$ is bounded if there is a polynomial R(y) such that
$$(\mathrm{\forall <!--\; \forall \; -->}x)(((\mathrm{\exists <!--\; \exists \; -->}y)P(x,y)=Q(x,y))\to <!--\; \to \; -->(y\le <!--\; \le \; -->R(x)))$$
which is equivalent with
$$(\mathrm{\forall <!--\; \forall \; -->}x)(((\mathrm{\exists <!--\; \exists \; -->}y)P(x,y)=Q(x,y))\to <!--\; \to \; -->((\mathrm{\exists <!--\; \exists \; -->}z)y+z=R(x)))$$
which is a pure PA sentence again.
(Note that the righthand side of $\to <!--\; \to \; -->$ is a only a shorthand!)
I wonder:
Is this the correct definition of a bounded quantifier?
Has the defining sentence to be true or provable?
Is the property of a sentence to have a bounded quantifier decidable?

Reading the mind of Prof. John Coates (motive behind his statement)
To start with the issue, I have been thinking from many days that Birch-Swinnerton-dyer conjectures should have some association with the Galois theory, but one day I got the Article of Tate called as
" J. Tate, On the conjectures of Birch and Swinnerton-Dyer and a geometric analogue, Sem. Bonrbaki 18 (1966)",
(please don't ask me to mention the reference of the above article, I have missed the link, I have only soft-copy with me) after a deep internet search, but the article was very technical and very sophisticated, rather after going through it dozens of times, I understood that Prof. J. Tate was trying to convey that one must relate the L-function to some Galois-groups, that was the rough picture in my mind! .
But later luckily to my surprise it was a coincidence that in an work of John Coates(John Coates, The Arithmetic of Elliptic Curves with Complex Multiplication, Proceedings of the International Congress of Mathematicians Helsinki, 1978), I encountered the same words where Coates describe that
".....They also gave a conjectural formulathe coefficient of $(s-<!--\; -\; -->1{)}^{g_Q}$ ($g_Q$ is rank) in the expansion of L(E,s) about $s=1$ but we shall not discuss this here.
Now Tate's work on the geometric analogue suggests that, in order to attack this conjecture, one must relate L(E,s) to the characteristic polynomial of some canonical element in a representation of a certain Galois group.
Now I request anyone who is currently working in that area/ know that area, answer me what does the above statement mean.
i.e. what was the intuition behind linking the L(E,s) to the characteristic polynomial of some canonical element in a representation of a certain Galois group ? .
And why does one need to look at characteristic polynomial of some element in Galois group in order to proceed with the Birch and Swinnerton Dyer conjecture ? .
Can anyone give a detailed summary of what was the Tate's idea and in which sense Coates refer to that sentence ? .
Please frame your answer not in a high-technical manner, but in the way a beginner can understand, but please answer me in a detailed manner.
I hope this question doesn't go unanswered, and I get the answer in a detailed form(describing to the maximum extent). Please help me.

Approximating polynomial roots without complex arithmetics
Suppose that we have polynomial of degree greater than two
$P(x)=a_n{x}^n+a_{n-<!--\; -\; -->1}{x}^{n-<!--\; -\; -->1}{+}_{\cdots <!--\; \cdots \; -->}+a_{1}x+a_0$
with real coefficients and we want to approximate all roots of this polynomial without complex arithmetic
It seems to be good idea to get quadratic factors of P(x)
$P(x)=Q(x)(x^2-<!--\; -\; -->px+q)+R(p,q)x+S(p,q)$
Now we want that after each iteration R(p,q) and S(p,q) be closer to zero until given tolerance $\mathrm{\epsilon <!--\; \epsilon \; -->}$
Now we need to solve system of nonliear equations
$$\begin{gathered} R(p,q)=0 \\ S(p,q)=0 \end{gathered}$$
and how to solve this system and present solution in such way it is easy to write code for it

How to calculate multiplicative inverses in $GF(2^3)$ without the Euclidean algorithm
The problem I have concerns finite field arithmetic in $GF(p^k)$
I know how to find multiplicative inverses using the extended Euclidean algorithm, but for my exams I need to calculate multiplicative inverses in $GF(2^3)$ without it.
What's the best way to do this? Is there even a convenient way?
The irreducible polynomial I have is $x^3+x+1$.

Some search on the internet and this site didn't result in any topic about this question of Silverman's The Arithmetic of Elliptic Curves:
Let $W\subset <!--\; \subset \; -->{\mathbb{P}}^{\mathbb{n}}$ be a smooth algebraic set, each of whose component varieties has dimension $n-<!--\; -\; -->1$. Prove that W is a variety.
Hint: use Krull's Hauptidealsatz to show W is the zeroset of a single homogenous polynomial.
my ideas: So if I(V) contains at least two linear independent polynomials, we have to prove the height of I(V) is at least 2.
After that, it is enough to find one singular point, where we assume $I(V)=fg$ with f,g nonconstant polynomials.

Confusion about genus-degree formula
My question regards a perplexity I have on how to apply the genus-degree formula for irreducible, projective, complex plane curves. Consider first the affine complex plane curve given by the equation
$$C:(x-<!--\; -\; -->2)(x-<!--\; -\; -->1)(x+1)(x+2)-<!--\; -\; -->y^2=x^4-<!--\; -\; -->5x^2+4-<!--\; -\; -->y^2=0.$$
The Jacobian is given by $(x(2x+\sqrt{10})(2x-<!--\; -\; -->\sqrt{10}),-<!--\; -\; -->2y)$, so it never vanishes on C. Let us now look at the compactification $\stackrel{^<!--\; ^\; -->}{C}$, which has the same points as C plus a point at infinity with coordinates $[x:y:z]=[0:1:0]$. This point lies in the affine chart with $y=1$, and the affine equation for $\stackrel{^<!--\; ^\; -->}{C}$ in terms of x and z in that chart is
$$x^4-<!--\; -\; -->5z^2{x}^2+4z^4-<!--\; -\; -->z^2=0.$$
The differential $(4x^3-<!--\; -\; -->10x{z}^2,-<!--\; -\; -->10x^{2}z+16z^3-<!--\; -\; -->2z)$ vanishes at $(x,z)=(0,0)$, hence $\stackrel{^<!--\; ^\; -->}{C}$ has one singularity: the point at infinity. If we pretend for a moment that it doesn't (i.e., that it is smooth), one can do the "usual construction" to see that it is topologically a torus: one can draw two cuts along [-2,-1] and [1,2] on two copies of the Riemann sphere and glue them together with the right orientation. So if $\stackrel{^<!--\; ^\; -->}{C}$ were regular, it would be an elliptic curve, and in particular have genus 1. However, the genus-degree formula for projective plane curves predicts genus 3, since the equation of $\stackrel{^<!--\; ^\; -->}{C}$ has degree 4. But the Wikipedia article on the genus-degree formula also mentions that the formula actually gives the arithmetic genus and that for every ordinary singularity of multiplicity r, the geometric genus is smaller than the arithmetic genus by $\frac{1}{2}r(r-<!--\; -\; -->1)$. Now, I am not really sure about how to measure the multiplicity of a singularity, but in this case it seems that for any value of $r\ge <!--\; \ge \; -->0$, we never have that $3-<!--\; -\; -->\frac{1}{2}r(r-<!--\; -\; -->1)=1$. So the geometric intuition and the formula seem to disagree. The only other thing that comes to my mind is that I have not checked yet that $\stackrel{^<!--\; ^\; -->}{C}$ is irreducible, but this can be checked on C using Eisentein's criterion applied to the polynomial ring $(\mathbb{C}[x])[y]$ using the prime ideal $\mathfrak{p}=(x+1)$.
Reassuming, my question is: what is the genus of $\stackrel{^<!--\; ^\; -->}{C}$? If it is 1, why is the genus-degree formula wrong? If it is 3, why is the geometric intuition wrong? After all, also the article of Wikipedia on elliptic curves seems to confirm that $\stackrel{^<!--\; ^\; -->}{C}$ should have genus 1.

What is the difference between the largest and smallest possible positive roots?
I am faced with the following question:
What is the difference between the largest and the smallest possible positive roots of 4x5+3x3−5x2+7x−12?
Now, my first attempt was to try substituting arbirtrary values to find one root and then long division to find the others. However, no integer (or fractional) value seemed to satisfy this.
Is the another way to approach this problem, or am I just making a simple arithmetic mistake?
Any help will be appreciated.

Is the Polynomial $X^{p^n}-<!--\; -\; -->X$ the zero polynomial in characteristic p?
Suppose that p is a prime number and that ${\mathbb{F}}_{p^n}[X]$ is the ring of polynomials with one variable, with coefficients in the field ${\mathbb{F}}_{p^n}$ for some $n\in <!--\; \in \; -->\mathbb{N}$. Then
$p(X):=X^{p^n}-<!--\; -\; -->X\in <!--\; \in \; -->{\mathbb{F}}_{p^n}[X]$
is a polynomial, which appears at many places in number theory, for example in the definition of the Carlitz Exponential. (In books about function field arithmetics this polynomial is often called [1], where in general $[k]:=X^{p^{n\cdot <!--\; \cdot \; -->k}}-<!--\; -\; -->X$)
Now the question is: Since $p(X)=0$ for any $X\in <!--\; \in \; -->{\mathbb{F}}_{p^n}$, isn't f just the zero polynomial? The same should be true for all [k], but as this is not mentioned in any book on function field arithmetic, I think I might overlook something.
Maybe someone has something enlightening to say here.

Let P and Q be polynomials with integer coefficients such that P(a) divides Q(a) in Z for every integer a. Does it mean that P divides Q in Z[X]?
P(a) divides Q(a) means that there exists an integer k(a) such that $Q(a)=k(a)P(a)$, so $Q=kP$. This means that P divides Q if k is polynomial with integer coefficients. But I can't find a proof or a counter-example. Any idea?
For the context, I was looking at a proof of the fact that $X^2-<!--\; -\; -->X+1$ divides $X^{2n+1}+(X=1{)}^{n+2}$ for every $n\ge <!--\; \ge \; -->1$ (in a video of Michael Penn I think but I am not sure). The method used was to consider congruences modulo $P(X)=X^2-<!--\; -\; -->X+1$. This is fine and clever method, but my students don't know arithmetic in Z[X], so I tried to circumvent the problem by looking at congruence in Z modulo P(a) for every $a\in <!--\; \in \; -->\mathbb{Z}$. But I hit the problem stated above to conclude the reasoning.