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 $$ 0\in <!--\; \in \; -->\mathbb{Z}$$, 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 $$ x^4-<!--\; -\; -->16+{(x^4-<!--\; -\; -->16)}_{\diamond <!--\; \diamond \; -->}=0+{(x^4-<!--\; -\; -->16)}_{\diamond <!--\; \diamond \; -->}$$. For convience set $$ I={(x^4-<!--\; -\; -->16)}_{\diamond <!--\; \diamond \; -->}$$ 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 $\[ \left(1\right)f\left(x\right)+g\left(x\right)(x^3-<!--\; -\; -->x^2+x-<!--\; -\; -->3)=-<!--\; -\; -->8 \]$

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

Consider the map $$ \mathrm{\varphi <!--\; \varphi \; -->}:{\mathbb{Z}}_3^{\mathrm{\%<!--\; \%\; -->}}\left[x\right]\to <!--\; \to \; -->{\mathbb{Z}}_3^{\mathrm{\%<!--\; \%\; -->}}\left(i\right)$$ defined by $$ \mathrm{\varphi <!--\; \varphi \; -->}\left(p\left(x\right)\right)=p\left(i\right)$$, note that the domain is correct because for any $$ n\in <!--\; \in \; -->{\mathbb{N}}_0$$, $$ i^n\in <!--\; \in \; -->\{1,-<!--\; -\; -->1,i,-<!--\; -\; -->i\}$$ so therefore $$ p\left(i\right)=c+di$$ for some $$ c,d\in <!--\; \in \; -->{\mathbb{Z}}_3^{\mathrm{\%<!--\; \%\; -->}}$$. 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 $$ a+bi\in <!--\; \in \; -->{\mathbb{Z}}_3^{\mathrm{\%<!--\; \%\; -->}}\left(i\right)$$ we see that the polynomial $$ a+bx$$ will map to it under $$ \mathrm{\varphi <!--\; \varphi \; -->}$$, this means that $$ \mathrm{im}<!--\; \; -->\left(\mathrm{\varphi <!--\; \varphi \; -->}\right):=\mathrm{\varphi <!--\; \varphi \; -->}\left({\mathbb{Z}}_3^{\mathrm{\%<!--\; \%\; -->}}\left[x\right]\right)={\mathbb{Z}}_3^{\mathrm{\%<!--\; \%\; -->}}\left(i\right)$$, with that knowledge therefore we have that $\[ {\mathbb{Z}}_3^{\mathrm{\%<!--\; \%\; -->}}\left[x\right]/\ker <!--\; \; -->\left(\mathrm{\varphi <!--\; \varphi \; -->}\right)\cong <!--\; \cong \; -->{\mathbb{Z}}_3^{\mathrm{\%<!--\; \%\; -->}}\left(i\right) \]$

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

Now approaching the other inclusion we suppose we have some $$ p\left(x\right)\in <!--\; \in \; -->\ker <!--\; \; -->\left(\mathrm{\varphi <!--\; \varphi \; -->}\right)$$, clearly if $$ p\left(x\right)=0$$ tne $$ \mathrm{\varphi <!--\; \varphi \; -->}\left(phi\left(x\right)\right)=0$$ on the other hand if $$ p\left(x\right)$$ is a constant other than zero, then also $$ \mathrm{\varphi <!--\; \varphi \; -->}\left(p\left(x\right)\right)\ne <!--\; \ne \; -->0$$ 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 $$ (x^2+1)\mid <!--\; \mid \; -->p\left(x\right)$$.

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 $$ 1=\mathrm{\varphi <!--\; \varphi \; -->}\left(1\right)=\mathrm{\varphi <!--\; \varphi \; -->}\left(a\left(x\right)\right)\cdot <!--\; \cdot \; -->0+\mathrm{\varphi <!--\; \varphi \; -->}\left(b\left(x\right)\right)\cdot <!--\; \cdot \; -->0=0$$ which is a contradiction, and therefore we must have that $$ (x^2+1)\mid <!--\; \mid \; -->p\left(x\right)$$ so that $$ p\left(x\right)$$ is a multiple of $$ x^2+1$$, in other-words $$ p\left(x\right)\in <!--\; \in \; -->{(x^2+1)}_{\diamond <!--\; \diamond \; -->}$$

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

• $$ 0+0+1\ne <!--\; \ne \; -->0$$
• $$ 1+1+1\ne <!--\; \ne \; -->0$$
• $$ 4+2+1=2\ne <!--\; \ne \; -->0$$
• $$ 9+3+1=3\ne <!--\; \ne \; -->0$$
• $$ 16+4+1=1\ne <!--\; \ne \; -->0$$

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 $\[ (3x+2)2x=x^2+4x\text{ and }(3x+2)(-<!--\; -\; -->1)=-<!--\; -\; -->3x-<!--\; -\; -->2 \]$ Therefore we have that $\[ x^2+x+1=(3x+2)(2x-<!--\; -\; -->1)+3 \]$ 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 $$ {\mathbb{Z}}_5^{\mathrm{\%<!--\; \%\; -->}}\left[x\right]/I$$

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

Note that is possible that the equation we wrote down doesn't specific that $$ \mathrm{\psi <!--\; \psi \; -->}$$ 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 $$ a,b\in <!--\; \in \; -->\mathbb{Z}$$, and note that this is only true when $$ a\mathrm{\%<!--\; \%\; -->}4=b\mathrm{\%<!--\; \%\; -->}4$$, since $$ 2\mid <!--\; \mid \; -->4$$, then also we conclude that $$ a\mathrm{\%<!--\; \%\; -->}2=b\mathrm{\%<!--\; \%\; -->}2$$, meaning that $$ a+2\mathbb{Z}=b+2\mathbb{Z}$$, therefore $$ \mathrm{\psi <!--\; \psi \; -->}(a+4\mathbb{Z})=\mathrm{\psi <!--\; \psi \; -->}(b+4\mathbb{Z})$$, so we know that indeed $$ \mathrm{\psi <!--\; \psi \; -->}$$ is well defined.

Now we continue by showing that it's a crone homomorphism, so we see the following $\[ \mathrm{\psi <!--\; \psi \; -->}((a+4\mathbb{Z})+(b+4\mathbb{Z}))=\mathrm{\psi <!--\; \psi \; -->}((a+b)+4\mathbb{Z})=(a+b)+2\mathbb{Z}=a+2\mathbb{Z}+b+2\mathbb{Z}=\mathrm{\psi <!--\; \psi \; -->}(a+4\mathbb{Z})+\mathrm{\psi <!--\; \psi \; -->}(b+4\mathbb{Z}) \]$ So that $$ \mathrm{\psi <!--\; \psi \; -->}$$ respects addition, and also that $\[ \mathrm{\psi <!--\; \psi \; -->}((a+4\mathbb{Z})\cdot <!--\; \cdot \; -->(b+4\mathbb{Z}))=\mathrm{\psi <!--\; \psi \; -->}(a\cdot <!--\; \cdot \; -->b+4\mathbb{Z})=a\cdot <!--\; \cdot \; -->b+2\mathbb{Z}=\mathrm{\psi <!--\; \psi \; -->}(a+4\mathbb{Z})\cdot <!--\; \cdot \; -->\mathrm{\psi <!--\; \psi \; -->}(b+4\mathbb{Z}) \]$ so that $$ \mathrm{\psi <!--\; \psi \; -->}$$ respects multiplication, and then we note $\[ \mathrm{\psi <!--\; \psi \; -->}\left(1\right)=\mathrm{\psi <!--\; \psi \; -->}(1+4\mathbb{Z})=1+2\mathbb{Z}=1 \]$ Thus we conclude that $$ \mathrm{\psi <!--\; \psi \; -->}$$ 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 $$ E\cong <!--\; \cong \; -->\mathbb{C}$$

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 $$ I\subseteq <!--\; \subseteq \; -->J\subseteq <!--\; \subseteq \; -->R$$ t the family of ideals in $$ R/I$$ given by $\[ \mathrm{\pi <!--\; \pi \; -->}\left(J\right)=J/I \]$

In our given question $$ \mathbb{R}\left[x\right]$$ plays the role of $$ R$$ and $$ {(x^2+1)}_{\diamond <!--\; \diamond \; -->}$$ that of $$ I$$. If we consider the family of ideals in $$ R\left[x\right]/{(x^2+1)}_{\diamond <!--\; \diamond \; -->}$$ 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 $$ {(x^2+1)}_{\diamond <!--\; \diamond \; -->}$$, thus those are the only ideals which contain $$ I$$.

For the first part, we'll notice that if $$ \mathrm{\theta <!--\; \theta \; -->}$$ is a root then we know that $$ {\mathrm{\theta <!--\; \theta \; -->}}^2+\mathrm{\theta <!--\; \theta \; -->}+1=0$$, but on that same thought we also note $\[ {(\mathrm{\theta <!--\; \theta \; -->}+1)}^2+(\mathrm{\theta <!--\; \theta \; -->}+1)+1={\mathrm{\theta <!--\; \theta \; -->}}^2+2\mathrm{\theta <!--\; \theta \; -->}+1+\mathrm{\theta <!--\; \theta \; -->}+1+1={\mathrm{\theta <!--\; \theta \; -->}}^2+\mathrm{\theta <!--\; \theta \; -->}+1=0 \]$ 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 $\[ p\left(x\right)=x^3-<!--\; -\; -->2=(x-<!--\; -\; -->s)(x^3+sx+s^2) \]$ 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 $\[ -<!--\; -\; -->\frac{s}{2}\pm <!--\; \pm \; -->\sqrt{s^2-<!--\; -\; -->4s^2} \]$ and therefore we see these roots are in $$ \mathbb{C}\setminus <!--\; \setminus \; -->\mathbb{R}$$, 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 $$ \mathrm{\varphi <!--\; \varphi \; -->}$$ 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 $$ a+b\sqrt{2}\in <!--\; \in \; -->\mathbb{Q}\left(\sqrt{2}\right)$$, if it's a root, then we know $\[ {(a+b\sqrt{2})}^2=3 \]$ 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

$\[ \mathrm{\varphi <!--\; \varphi \; -->}{\left(\sqrt{3}\right)}^2=\mathrm{\varphi <!--\; \varphi \; -->}\left(3\right)=\mathrm{\varphi <!--\; \varphi \; -->}\left(1\right)+\mathrm{\varphi <!--\; \varphi \; -->}\left(1\right)+\mathrm{\varphi <!--\; \varphi \; -->}\left(1\right)=3 \]$ but then $$ \mathrm{\varphi <!--\; \varphi \; -->}\left(\sqrt{3}\right)$$ 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 $$ \mathrm{\alpha <!--\; \alpha \; -->}$$ is transcendental over $$ \mathbb{Q}$$. is a transcendental number.-->

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

Let $$ p,q\in <!--\; \in \; -->\mathbb{Q}\left[x\right]$$, then $$ f(p+q)=(p+q)\left(\mathrm{\alpha <!--\; \alpha \; -->}\right)=p\left(\mathrm{\alpha <!--\; \alpha \; -->}\right)+q\left(\mathrm{\alpha <!--\; \alpha \; -->}\right)=f\left(p\right)+f\left(q\right)$$ which follows from basic facts about finite summations. Additionally we have that $$ f(p\cdot <!--\; \cdot \; -->q)=(p\cdot <!--\; \cdot \; -->q)\left(\mathrm{\alpha <!--\; \alpha \; -->}\right)=p\left(\mathrm{\alpha <!--\; \alpha \; -->}\right)\cdot <!--\; \cdot \; -->q\left(\mathrm{\alpha <!--\; \alpha \; -->}\right)=f\left(p\right)f\left(q\right)$$ 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 $$ \mathrm{\alpha <!--\; \alpha \; -->}$$ would be a root of that polynomial, which is impossible because $$ \mathrm{\alpha <!--\; \alpha \; -->}$$ 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 $$ \mathrm{im}<!--\; \; -->\left(f\right)=\mathbb{Q}\left(\mathrm{\alpha <!--\; \alpha \; -->}\right)$$. That is to say that $$ \mathbb{Q}\left(\mathrm{\alpha <!--\; \alpha \; -->}\right)$$'s elements are fully described by $$ \mathbb{Q}\left[x\right]$$ elements, which are polynomials with rational coefficients. We also know $$ \mathbb{Q}\left(\mathrm{\alpha <!--\; \alpha \; -->}\right)=\{\frac{f\left(\mathrm{\alpha <!--\; \alpha \; -->}\right)}{g\left(\mathrm{\alpha <!--\; \alpha \; -->}\right)}:f,g\in <!--\; \in \; -->\mathbb{Q}\left(x\right),g\left(\mathrm{\alpha <!--\; \alpha \; -->}\right)\ne <!--\; \ne \; -->0\}$$, which is another way of describing the elements of $$ \mathbb{Q}\left(\mathrm{\alpha <!--\; \alpha \; -->}\right)$$.

---

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

We can also observe that $$ x^3-<!--\; -\; -->{\mathrm{\pi <!--\; \pi \; -->}}^3$$ is irreducible in $$ \mathbb{Q}\left({\mathrm{\pi <!--\; \pi \; -->}}^3\right)$$ this follows from what we showed earlier and then we have that $\[ \mathbb{Q}\left({\mathrm{\pi <!--\; \pi \; -->}}^3\right)\left[x\right]/{(x^3-<!--\; -\; -->{\mathrm{\pi <!--\; \pi \; -->}}^3)}_{\diamond <!--\; \diamond \; -->}\cong <!--\; \cong \; -->\mathbb{Q}\left({\mathrm{\pi <!--\; \pi \; -->}}^3\right)\left(\mathrm{\pi <!--\; \pi \; -->}\right) \]$ As seen earlier $$ \mathbb{Q}\left({\mathrm{\pi <!--\; \pi \; -->}}^3\right)\left[x\right]/{(x^3-<!--\; -\; -->{\mathrm{\pi <!--\; \pi \; -->}}^3)}_{\diamond <!--\; \diamond \; -->}$$ can be viewed as a vector space over $$ \mathbb{Q}\left({\mathrm{\pi <!--\; \pi \; -->}}^3\right)$$ with a basis given by $$ (1,x\mathbf{+}I,{(x\mathbf{+}I)}^2)$$. Now the isomorphism between these two structures is actually defined pointwise as $$ \mathrm{\Phi <!--\; \Phi \; -->}(x+{(x^3-<!--\; -\; -->{\mathrm{\pi <!--\; \pi \; -->}}^3)}_{\diamond <!--\; \diamond \; -->})=\mathrm{\pi <!--\; \pi \; -->}$$, in other words we obtain that a basis for $$ F\left(\mathrm{\pi <!--\; \pi \; -->}\right)$$ is $$ (1,\mathrm{\pi <!--\; \pi \; -->},{\mathrm{\pi <!--\; \pi \; -->}}^2)$$ 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 $$ r_1,\dots <!--\; \dots \; -->,r_n$$ of $$ f$$.

Observe that $$ r_1-<!--\; -\; -->a,\dots <!--\; \dots \; -->,r_n-<!--\; -\; -->a$$ 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 $$ a\in <!--\; \in \; -->E$$ and therefore $$ r_i-<!--\; -\; -->a\in <!--\; \in \; -->E$$ 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:

$\[ x^4-<!--\; -\; -->x^2-<!--\; -\; -->2=(x^2-<!--\; -\; -->2)(x^2+1)={(x^2+1)}^2 \]$

By brute force, we can observe that there are no roots to the polynomial in $$ {\mathbb{Z}}_3^{\mathrm{\%<!--\; \%\; -->}}$$, therefore it is irreducible, so that we obtain the field extension $$ E:={\mathbb{Z}}_3^{\mathrm{\%<!--\; \%\; -->}}\left[x\right]/{(x^2+1)}_{\diamond <!--\; \diamond \; -->}$$ that contains a root $$ a$$, we can see that $$ a\ne <!--\; \ne \; -->0$$ 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 $$ a,b\in <!--\; \in \; -->{\mathbb{Z}}_p^{\mathrm{\%<!--\; \%\; -->}}$$.

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 $$ c,d\in <!--\; \in \; -->{\mathbb{Z}}_p^{\mathrm{\%<!--\; \%\; -->}}$$ 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 $\[ p^2-<!--\; -\; -->\frac{p(p-<!--\; -\; -->1)}{2}=\frac{p(p-<!--\; -\; -->1)}{2} \]$

Now if we want to know the number of irreducible quadratic polynomials over $$ {\mathbb{Z}}_p^{\mathrm{\%<!--\; \%\; -->}}$$ which aren't necessarily monic, we can realize that that such a polynomial is of the following form $\[ k(x^2+ax+b) \]$ where $$ k\in <!--\; \in \; -->{\mathbb{Z}}_p^{\mathrm{\%<!--\; \%\; -->}}\setminus <!--\; \setminus \; -->\left\{0\right\}$$ and $$ x^2+ax+b$$ is irreducible in $$ {\mathbb{Z}}_p^{\mathrm{\%<!--\; \%\; -->}}$$, 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 $\[ \frac{p{(p-<!--\; -\; -->1)}^2}{2} \]$ 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 $\[ y+z\mathrm{\omega <!--\; \omega \; -->}y+{\mathrm{\omega <!--\; \omega \; -->}}^{2}z{\mathrm{\omega <!--\; \omega \; -->}}^{2}y+\mathrm{\omega <!--\; \omega \; -->}z \]$

Such that:

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

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

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 $$ f(x-<!--\; -\; -->\frac{1}{3})$$ $\[ f(x-<!--\; -\; -->\frac{1}{3})=(x-<!--\; -\; -->\frac{1}{3})3+{(x-<!--\; -\; -->\frac{1}{3})}^2-<!--\; -\; -->1 \]$

Let's break down $$ {(x-<!--\; -\; -->\frac{1}{3})}^2$$ first: $\[ (x-<!--\; -\; -->\frac{1}{3})(x-<!--\; -\; -->\frac{1}{3})=x^2-<!--\; -\; -->\frac{2}{3}x+\frac{1}{9} \]$ Now we can try $$ {(x-<!--\; -\; -->\frac{1}{3})}^3$$

Now back to computing $$ f(x-<!--\; -\; -->\frac{1}{3})$$ we have 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 $$ \pm <!--\; \pm \; -->1,\pm <!--\; \pm \; -->5$$ 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 $$ b\ge <!--\; \ge \; -->2$$ is a natural number and $$ p\left(x\right)=a_k{x}^k+a_{k-<!--\; -\; -->1}{x}^{k-<!--\; -\; -->1}+\dots <!--\; \dots \; -->+a_{1}x+a_0$$ 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 $$ 5\le <!--\; \le \; -->b-<!--\; -\; -->1$$ must be true so that $$ b\ge <!--\; \ge \; -->6$$, 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 $$ (1,{\mathrm{\theta <!--\; \theta \; -->}}^1,{\mathrm{\theta <!--\; \theta \; -->}}^2,{\mathrm{\theta <!--\; \theta \; -->}}^3,{\mathrm{\theta <!--\; \theta \; -->}}^4)$$ is a basis for $$ E$$ over $$ \mathbb{Q}$$

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