🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

To find a polynomial with the same remainder we first do things via long division:

Therefore the polynomial $-2x^3-35136x+3$ is congruent modulo $x^4-16$, because they both have the same remainder upon division by $x^4-16$ .

Recall the definition of a zero divisor, note that $0$ in this quotient ring is simply the polynomial with all coefficients set to $0\in \mathbb{Z}$, therefore we can clearly see that $x-2,x+2$ have at least one non-zero coefficient therefore they cannot be zero elements. Now at the same time we can factor $x^4-16=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4)$. We get two equations from this : $(x-2)(x^3+2x^2+4x+8)$ and $(x+2)(x^3-2x^2+4x-8)$, since we can verify that $x^3-2x^2+4x-8,x^3+2x^2+4x+8$ are both non-zero polynomials, then so far we can conclude that $x-2,x+2$

Now notice that $x^4-16+{(x^4-16)}_{\diamond }=0+{(x^4-16)}_{\diamond }$. For convience set $I={(x^4-16)}_{\diamond }$ So then we have that $[x-2+I][x^3+2x^2+4x+8+I]=x^4-16+I=0+I$ and also that $[x+2+I][x^3-2x^2+4x-8+I]=0+I$

We start by finding the gcd in $\mathbb{Q}\left[x\right]$ by using the euclidean algorithm:

Therefore we conclude that $x^5-3x^4+2x^2+1=(x^2-x-3)(x^3-x^2+x-3)-8$, therefore we know that their gcd is one, we can express the linear combination as $${\displaystyle \left(1\right)f\left(x\right)+g\left(x\right)(x^3-x^2+x-3)=-8}$$

Now we do the same in ${\mathbb{Z}}_5^{\%}\left[x\right]$, but since in the above long division we never used any inverses, or elements that cannot be represented in ${\mathbb{Z}}_5^{\%}$ (we can see 6 as 1 here), then we get the same answer, but we can re-write $\overline{-8}$ as $2$, so we $${\displaystyle \left(1\right)f\left(x\right)+g\left(x\right)(x^3-x^2+x-3)=2}$$

Consider the map $\varphi :{\mathbb{Z}}_3^{\%}\left[x\right]\to {\mathbb{Z}}_3^{\%}\left(i\right)$ defined by $\varphi \left(p\left(x\right)\right)=p\left(i\right)$, note that the domain is correct because for any $n\in {\mathbb{N}}_0$, $i^n\in \{1,-1,i,-i\}$ so therefore $p\left(i\right)=c+di$ for some $c,d\in {\mathbb{Z}}_3^{\%}$. Moreover see that it is a homomorphism quite easily by verifying multiplication, addition and that clearly it maps $1$ to $1$.

Additionally this map is surjective because given any $a+bi\in {\mathbb{Z}}_3^{\%}\left(i\right)$ we see that the polynomial $a+bx$ will map to it under $\varphi$, this means that $im\left(\varphi \right):=\varphi \left({\mathbb{Z}}_3^{\%}\left[x\right]\right)={\mathbb{Z}}_3^{\%}\left(i\right)$, with that knowledge therefore we have that $${\displaystyle {\mathbb{Z}}_3^{\%}\left[x\right]/ker\left(\varphi \right)\cong {\mathbb{Z}}_3^{\%}\left(i\right)}$$

We'll now prove that $ker\left(\varphi \right)={(x^2+1)}_{\diamond }$. Starting with $\supseteq$ recall that ${(x^2+1)}_{\diamond }$ are all the multiples of $x^2+1$, moreover $\varphi (x^2+1)=1+{\left(i\right)}^2=0$ therefore $x^2+1\in ker\left(\varphi \right)$, since any element of ${(x^2+1)}_{\diamond }$ is a multiple of $x^2+1$ then by taking an arbitrary element from it, it must be of the form $m\left(x\right)·(x^2+1)$ where $m\left(x\right)\in {\mathbb{Z}}_3^{\%}\left[x\right]$ but since $\varphi$ is a homomorphism, then we know that $\varphi \left(m\left(x\right)(x^2+1)\right)=\varphi \left(m\left(x\right)\right)\varphi (x^2+1)=0$, thus ${(x^2+1)}_{\diamond }\subseteq ker\left(\varphi \right)$.

Now approaching the other inclusion we suppose we have some $p\left(x\right)\in ker\left(\varphi \right)$, clearly if $p\left(x\right)=0$ tne $\varphi \left(phi\left(x\right)\right)=0$ on the other hand if $p\left(x\right)$ is a constant other than zero, then also $\varphi \left(p\left(x\right)\right)\ne 0$ therefore we may assume that $p\left(x\right)$ is non-constant, in such a case we know that either $gcd(p\left(x\right),x^2+1)=1$ or that $(x^2+1)\mid p\left(x\right)$.

Suppose that $gcd(p\left(x\right),x^2+1)=1$ therefore we get polynomials $a\left(x\right),b\left(x\right)$ such that $1=a\left(x\right)g\left(x\right)+b\left(x\right)(x^2+1)$ but then $1=\varphi \left(1\right)=\varphi \left(a\left(x\right)\right)·0+\varphi \left(b\left(x\right)\right)·0=0$ which is a contradiction, and therefore we must have that $(x^2+1)\mid p\left(x\right)$ so that $p\left(x\right)$ is a multiple of $x^2+1$, in other-words $p\left(x\right)\in {(x^2+1)}_{\diamond }$

To show that $I$ is a maximal ideal, reacall that we've proven that the ideal generated by a polynomial is maximal iff the polynomial is irreducible

Note that $p\left(x\right)=x^2+x+1$ doesn't have any roots because of the following facts

• $0+0+1\ne 0$
• $1+1+1\ne 0$
• $4+2+1=2\ne 0$
• $9+3+1=3\ne 0$
• $16+4+1=1\ne 0$

Therefore since it's degree is 2, then we can conclude that $p\left(x\right)$ is irreducible, which implies that $I$ is maximal.

As we look to find the gcd, we just observe the following simple calculation $${\displaystyle (3x+2)2x=x^2+4x\text{ and }(3x+2)(-1)=-3x-2}$$ Therefore we have that $${\displaystyle x^2+x+1=(3x+2)(2x-1)+3}$$ So that the gcd is also one. This allows us to conclude that $2x-1+I$ is the multiplicative inverse of $3x+2+I$ in ${\mathbb{Z}}_5^{\%}\left[x\right]/I$

We'll note that since $p\left(x\right)$ is of degree 3 and has no root in ${\mathbb{Z}}_2^{\%}$ because $0+0+1\ne 0$, $1+1+1=1\ne 0$, therefore $p\left(x\right)$ is irreducible. Therefore ${\mathbb{Z}}_2^{\%}\left[x\right]/I$ is a field.

Note that is possible that the equation we wrote down doesn't specific that $\psi$ is a function, so let's determine that first. That is, we must verify that our function doesn't every map one thing to two different things. So suppose that we have $a+4\mathbb{Z}=b+4\mathbb{Z}$ where $a,b\in \mathbb{Z}$, and note that this is only true when $a\%4=b\%4$, since $2\mid 4$, then also we conclude that $a\%2=b\%2$, meaning that $a+2\mathbb{Z}=b+2\mathbb{Z}$, therefore $\psi (a+4\mathbb{Z})=\psi (b+4\mathbb{Z})$, so we know that indeed $\psi$ is well defined.

Now we continue by showing that it's a crone homomorphism, so we see the following $${\displaystyle \psi ((a+4\mathbb{Z})+(b+4\mathbb{Z}))=\psi ((a+b)+4\mathbb{Z})=(a+b)+2\mathbb{Z}=a+2\mathbb{Z}+b+2\mathbb{Z}=\psi (a+4\mathbb{Z})+\psi (b+4\mathbb{Z})}$$ So that $\psi$ respects addition, and also that $${\displaystyle \psi \left((a+4\mathbb{Z})·(b+4\mathbb{Z})\right)=\psi (a·b+4\mathbb{Z})=a·b+2\mathbb{Z}=\psi (a+4\mathbb{Z})·\psi (b+4\mathbb{Z})}$$ so that $\psi$ respects multiplication, and then we note $${\displaystyle \psi \left(1\right)=\psi (1+4\mathbb{Z})=1+2\mathbb{Z}=1}$$ Thus we conclude that $\psi$ is a crone homomorphism.

since $p\left(x\right)$ by definition irreducible over $F$ we have that $E$ is a field containing $F$ and a root of $p\left(x\right)$, at the same time we know that $E$ is the collection of polynomials of degree less than one, in $F\left[x\right]$ which is in a clear isomorphism to $F$ by the identity map

Let's first look at what it means to be an irreducible polynomial in $\mathbb{C}\left[x\right]$, for any polynomial with degree greater or equal to $2$ we know that by the fundamental theorem of algebra it has exactly two roots, and that can be used to re-write that polynomial as a product of two polynomials in $C\left[x\right]$ which means by definition none of these polynomials are irreducible. This only leaves polynomials of degree one in $\mathbb{C}\left[x\right]$ each of which are irreducible over $\mathbb{C}$.

Therefore we can now conclude that $p\left(x\right)$ is a polynomial of degree $1$, so that it is a linear polynomial, therefore by the prvious exercise we conclude that $E\cong \mathbb{C}$

Recall the correspondance theorem for rings which states that if $I$ is a proper ideal in a ring $R$, then there is a bijection from all ideals $J$ such that $I\subseteq J\subseteq R$ t the family of ideals in $R/I$ given by $${\displaystyle \pi \left(J\right)=J/I}$$

In our given question $\mathbb{R}\left[x\right]$ plays the role of $R$ and ${(x^2+1)}_{\diamond }$ that of $I$. If we consider the family of ideals in $R\left[x\right]/{(x^2+1)}_{\diamond }$ since $x^2+1$ is irreducible in $\mathbb{R}\left[x\right]$, then we know that this quotient is a field, and thus only has two trivial ideals, (everything and just zero) which correspond to $R\left[x\right]$ and ${(x^2+1)}_{\diamond }$, thus those are the only ideals which contain $I$.

For the first part, we'll notice that if $\theta$ is a root then we know that ${\theta }^2+\theta +1=0$, but on that same thought we also note $${\displaystyle {(\theta +1)}^2+(\theta +1)+1={\theta }^2+2\theta +1+\theta +1+1={\theta }^2+\theta +1=0}$$ So that $E$ also contains the other root of $x^2+x+1$

For the second part we set $s=2^{\frac{1}{3}}$ for convience. Now note the following factorization $${\displaystyle p\left(x\right)=x^3-2=(x-s)(x^3+sx+s^2)}$$ Therefore if $\mathbb{Q}\left(s\right)$ has some root of $p\left(x\right)$ other than $s$, then it must be a root of $x^3+sx+s^2$, but we can compute the roots of this polynomial via the quadratic equation, which yields $${\displaystyle -\frac{s}{2}\pm \sqrt{s^2-4s^2}}$$ and therefore we see these roots are in $\mathbb{C}⧵\mathbb{R}$, therefore there are no other real roots other than $s$ in $\mathbb{Q}\left(s\right)$.

Now recall that a quadratic polynomial over any field $F$ has no roots in $F$ iff it is irreducible in $F\left[x\right]$, and that is preserved over isomorphisms of $F$ with another field. Now observe that through $\varphi$ we can verify that this bijection forms a homomorphism, and thus is an isomorphism, so that $\mathbb{Q}\left(s\right)$ and $E$ are isomorphic, and therefore $E$ doesn't contain any other root of $p\left(x\right)$

Let's first observe that there is no element in $\mathbb{Q}\left(2\right)$ that is a root of $x^2-3$, to see why suppose we have some $a+b\sqrt{2}\in \mathbb{Q}\left(\sqrt{2}\right)$, if it's a root, then we know $${\displaystyle {(a+b\sqrt{2})}^2=3}$$ and the left hand side is ${(a+b\sqrt{2})}^2=a^2+2b^2+2ab\sqrt{2}$, since the product and sum of a non-zero rational with an irrational is irrational, then since $3$ is not irrational, we conclude that $ab=0$,

If $a=0$, then we know that $b^2=\frac{3}{2}$ which is not solvable in $\mathbb{Q}$ , however given $b=0$ then we have that $3=a^2$ which is also not solvable in $\mathbb{Q}$, so that there is no element in $\mathbb{Q}\left(2\right)$ that is a root of $(x^2-3)$

We show that they are not isomorphic by contradiction, so we suppose that they are isomorphic and see what happens, if they are isomorphic, then we have

$${\displaystyle \varphi {\left(\sqrt{3}\right)}^2=\varphi \left(3\right)=\varphi \left(1\right)+\varphi \left(1\right)+\varphi \left(1\right)=3}$$ but then $\varphi \left(\sqrt{3}\right)$ is an element of $\mathbb{Q}\left(\sqrt{2}\right)$ which is a root of $x^2-3$ which is clearly impossible, so no such isomorphism exists.

A transcendental number is a real or complex number that is not algebraic, that is to say it is not a root of a non-zero polynomial of finite degree with rational coefficients. Suppose that $\alpha$ is transcendental over $\mathbb{Q}$.

Consider the map $f:\mathbb{Q}\left[x\right]\to \mathbb{Q}\left(\alpha \right)$ so that for any $p\in \mathbb{Q}\left[x\right]$ we have $${\displaystyle f\left(p\right)=p\left(\alpha \right)}$$

Let $p,q\in \mathbb{Q}\left[x\right]$, then $f(p+q)=(p+q)\left(\alpha \right)=p\left(\alpha \right)+q\left(\alpha \right)=f\left(p\right)+f\left(q\right)$ which follows from basic facts about finite summations. Additionally we have that $f\left(p·q\right)=\left(p·q\right)\left(\alpha \right)=p\left(\alpha \right)·q\left(\alpha \right)=f\left(p\right)f\left(q\right)$ which follows from the definition of polynomial multiplication (which holds when $x$ holds a real number). Finally $f\left(1\right)=1$ since there is no varaible to evaluate, so that $f$ is a homomorphism.

Note that the kernel of $f$ is empty, for if an element were to be in the kernel then $\alpha$ would be a root of that polynomial, which is impossible because $\alpha$ is transcendental. Therefore by the first isomorphism theorem we have an isomorphism between $\mathbb{Q}\left[x\right]/ker\left(f\right)=\mathbb{Q}\left[x\right]$ and $im\left(f\right)=\mathbb{Q}\left(\alpha \right)$. That is to say that $\mathbb{Q}\left(\alpha \right)$'s elements are fully described by $\mathbb{Q}\left[x\right]$ elements, which are polynomials with rational coefficients. We also know $\mathbb{Q}\left(\alpha \right)=\{\frac{f\left(\alpha \right)}{g\left(\alpha \right)}:f,g\in \mathbb{Q}\left(x\right),g\left(\alpha \right)\ne 0\}$, which is another way of describing the elements of $\mathbb{Q}\left(\alpha \right)$.

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Moving onto the next question, let's observe that $F\left(\pi \right)/F$ is a field extension, and that $\pi \in F\left(\pi \right)$, note that $x^3-{\pi }^3$ is a monic polynomial in $F\left[x\right]$ such that $\pi$ is a root, therefore $\pi$ is algebraic over $F$.

We can also observe that $x^3-{\pi }^3$ is irreducible in $\mathbb{Q}\left({\pi }^3\right)$ this follows from what we showed earlier and then we have that $${\displaystyle \mathbb{Q}\left({\pi }^3\right)\left[x\right]/{(x^3-{\pi }^3)}_{\diamond }\cong \mathbb{Q}\left({\pi }^3\right)\left(\pi \right)}$$ As seen earlier $\mathbb{Q}\left({\pi }^3\right)\left[x\right]/{(x^3-{\pi }^3)}_{\diamond }$ can be viewed as a vector space over $\mathbb{Q}\left({\pi }^3\right)$ with a basis given by $(1,x\mathbf{+}I,{\left(x\mathbf{+}I\right)}^2)$. Now the isomorphism between these two structures is actually defined pointwise as $\Phi (x+{(x^3-{\pi }^3)}_{\diamond })=\pi$, in other words we obtain that a basis for $F\left(\pi \right)$ is $(1,\pi ,{\pi }^2)$ through this isomorphism.

Let $E$ be the splitting field of $f$ over $E$, note that $E$ is a field extension of $F$. Since $f$ splits here, then also $E$ contains all the roots $r_1,\dots ,r_n$ of $f$.

Observe that $r_1-a,\dots ,r_n-a$ are roots of $f(x+a)$, and that they reside within $E$ as well, this is seen because since $E$ is a field extension of $F$ then $a\in E$ and therefore $r_i-a\in E$ as it's a field.

At this point we want to claim that $E$ is a splitting field for $f(x+a)$, but to do that we need to show that there is no proper subfield in which $f(x+a)$ splits. For the sake of contradiction, if there was, then we'd have a problem because we could use this proper subfield as a splitting field for $f\left(x\right)$ by a similar argument, which contradicts the fact that $E$ was a splitting field.

For the reverse direction, a symmetric proof may be obtained.

We start by observing the following factorization:

$${\displaystyle x^4-x^2-2=(x^2-2)(x^2+1)={(x^2+1)}^2}$$

By brute force, we can observe that there are no roots to the polynomial in ${\mathbb{Z}}_3^{\%}$, therefore it is irreducible, so that we obtain the field extension $E:={\mathbb{Z}}_3^{\%}\left[x\right]/{(x^2+1)}_{\diamond }$ that contains a root $a$, we can see that $a\ne 0$ and also that since it's a root we have that $a^2+1=0$, but then also that ${\left(2a\right)}^2+1=4a^2+1=a^2+1=0$ so then $2a$ is a different root, therefore $E$ contains all roots of $x^2+1$ so that it must split here, specifically we know $(x+a)(x-a)$ is the split.

Now we have to prove that there is no proper subfield of $E$ such that $x^2+1$ splits, this follows from the fact that it's a degree 2 extension and that 2 is prime in conjunction with the degree division formula. Thus we've shown that $E$ is indeed the splitting field of $f$ as needed.

We first start by looking at monic polynomials, so they are of the form $x^2+ax+b$ that of which there are $p^2$ by iterating through all possible values for $a,b\in {\mathbb{Z}}_p^{\%}$.

Now given such a polynomial, we know that it is reducible if it can be factored into linear terms of the form $(x+c)(x+d)$ where $c,d\in {\mathbb{Z}}_p^{\%}$ each of which gives rise to $x^2+(c+d)x+cd$ which is uniquely associated with this factorization. With that said, there are $\frac{p(p-1)}{2}$ such factorizations, each of which yields a unique reducible polynomial. Thus we conclude that the number of irreducible monic polynomials is given by $${\displaystyle p^2-\frac{p(p-1)}{2}=\frac{p(p-1)}{2}}$$

Now if we want to know the number of irreducible quadratic polynomials over ${\mathbb{Z}}_p^{\%}$ which aren't necessarily monic, we can realize that that such a polynomial is of the following form $${\displaystyle k(x^2+ax+b)}$$ where $k\in {\mathbb{Z}}_p^{\%}⧵\left\{0\right\}$ and $x^2+ax+b$ is irreducible in ${\mathbb{Z}}_p^{\%}$, that of which there are $\frac{p(p-1)}{2}$, thus for each such choice of $k$, which there are $p-1$ choices of we obtain a unique irreducible polynomial, and therefore there are $${\displaystyle \frac{p{(p-1)}^2}{2}}$$ many irreducible polynomials.

We'll use the cubic formula to determine the roots, let's start with $x^3+9x-1$ because it's in the correct form.

Let's recall that the roots of a polynomial of the form $x^3+qx+r$ are given by the following $${\displaystyle y+z\omega y+{\omega }^{2}z{\omega }^{2}y+\omega z}$$

Such that:

• $y={\left(\frac{1}{2}(-r+\sqrt{R})\right)}^{\frac{1}{3}}$
• $z=\frac{-q}{3y}$
• $R=r^2+\frac{4q^3}{27}$
• $\omega =e^{\frac{2\pi i}{3}}$

In our question $q=9$ and $r=-1$ so let's start by computing $R={(-1)}^2+\frac{4{\left(9\right)}^3}{27}=1+4·{\left(\frac{9}{3}\right)}^3=1+4·27=109$, let $\alpha ={\left(\frac{1}{2}(1+\sqrt{109})\right)}^{\frac{1}{3}}$ therefore $y=\alpha$ and thus $z=\frac{-9}{3\alpha }=\frac{-3}{\alpha }$, thus our roots are given by $${\displaystyle \alpha +\frac{-3}{\alpha }\omega \alpha +{\omega }^2\frac{-3}{\alpha }{\omega }^2\alpha +\omega \frac{-3}{\alpha }}$$

We now move our attention to $f\left(x\right)=x^3+x^2-1$ which is not in the correct form to apply the cubic formula, although we do have a method to deal with this, so we employ it.

In our case $n=3$ and $a_{n-1}=1$ so therefore we turn our attention to the polynomial $f(x-\frac{1}{3})$ $${\displaystyle f(x-\frac{1}{3})=(x-\frac{1}{3})3+{(x-\frac{1}{3})}^2-1}$$

Let's break down ${(x-\frac{1}{3})}^2$ first: $${\displaystyle (x-\frac{1}{3})(x-\frac{1}{3})=x^2-\frac{2}{3}x+\frac{1}{9}}$$ Now we can try ${(x-\frac{1}{3})}^3$ $$\begin{gathered} {\displaystyle {(x-\frac{1}{3})}^3\&=(x-\frac{1}{3}){(x-\frac{1}{3})}^2 \\ \&=(x-\frac{1}{3})(x^2-\frac{2}{3}x+\frac{1}{9}) \\ \&=(x^3-\frac{2}{3}{x}^2+\frac{x}{9})-(\frac{x^2}{3}-\frac{2}{9}x+\frac{1}{27}) \\ \&=x^3-(\frac{2}{3}+\frac{1}{3})x^2+\frac{3}{9}x-\frac{1}{27} \\ \&=x^3-x^2+\frac{x}{3}-\frac{1}{27}} \end{gathered}$$

Now back to computing $f(x-\frac{1}{3})$ we have $$\begin{gathered} {\displaystyle f(x-\frac{1}{3})\&={(x-\frac{1}{3})}^3+{(x-\frac{1}{3})}^2-1 \\ \&=(x^3-x^2+\frac{x}{3}-\frac{1}{27})+(x^2-\frac{2}{3}x+\frac{1}{9})-1 \\ \&=x^3-\frac{x}{3}+\frac{2}{27}-1 \\ \&=x^3-\frac{x}{3}-\frac{25}{27}} \end{gathered}$$ Now our equation has the correct form to be able to apply the cubic formula and so we can do that, suppose we obtain roots $r_1,r_2,r_3$, then the roots of $f\left(x\right)$ are given by $r_1-\frac{1}{3},r_2-\frac{1}{3},r_2-\frac{1}{3}$

We can show that $f\left(x\right)$ is irreudcible in a few ways, the first way is by using the the rational roots theorem to find the possible candidates for roots are $\pm 1,\pm 5$ none of which work, so we know that the only possible factorization is by a degree 2 polynomial and a degree 3 polynomial, we can use the gauss theorem to show that $f\left(x\right)$ is irred in $Q$ iff $f\left(x\right)$ is irred in $\mathbb{Z}$ because the gcd of all of it's coefficients are 1, which means that $f\left(x\right)$ is primitive. We proceed by then writing out a degree 3 polynomial times a degree 2 polynomial with variable coefficents and find that no such coeffients exists.

The above method works out, but I found an interesting irreducibility test that I found was interesting. It's called the Cohn's irreducibility criterion which states that:

Assume $b\ge 2$ is a natural number and $p\left(x\right)=a_k{x}^k+a_{k-1}{x}^{k-1}+\dots +a_{1}x+a_0$ is a polynomial such that $0<a_i<b-1$. If $p\left(b\right)$ is a prime number then $p\left(x\right)$ is irreducible in $\mathbb{Z}\left[x\right]$

Applying that to our polynomial, $x^5+x^2+5$ we observe that $5\le b-1$ must be true so that $b\ge 6$, by plugging in we see that $p\left(6\right)=7853$ which is a prime number, and therefore $f\left(x\right)$ is irreducible in $\mathbb{Z}\left[x\right]$ then by gauss it is irreducible in $\mathbb{Q}\left[x\right]$.

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Looking at the next question we recall that we've proven this fact after observing this proof we note that clearly by the division algorithm we have that every element in $E$ must be a coset of a polynomial of degree at most four, as a direct consequence we also note that $(1,{\theta }^1,{\theta }^2,{\theta }^3,{\theta }^4)$ is a basis for $E$ over $\mathbb{Q}$

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We divide $x^5+2x^2+5$ by $x-\theta$ to obtain $q\left(x\right)=x^4+\theta x^3+{\theta }^2{x}^2+({\theta }^3+2)x+({\theta }^4+2\theta )$