MHB How Does Eisenstein's Criterion Confirm the Irreducibility of a Polynomial?

[mathmari]

Hey!

Let $f(x)=x^3-3x-1\in \mathbb{Q}[x]$.

Then $f(x+1)=(x+1)^3-3(x+1)-1=x^3+3x^2+3x+1-3x-3-1=x^3+3x^2-3$.

The prime $p=3$ divides all the coefficients except of the one of the term of the highest degree and $p^2$ doesn't divide the constant term.
So, from Eisenstein's criterion we have that $f(x+1)$ is irreducible in $\mathbb{Q}[x]$, and so is $f(x)$.

Let $a\in \mathbb{C}$ be a root of $f$, $f(a)=0 \Rightarrow a^3=3a+1$.

We have that $f(2-a^2)=(2-a^2)^3-3(2-a^2)-1=8-12 a^2+6 a^4-a^6-6+3a^2-1 \\ =-a^6+6 a^4-9 a^2+1=-(a^3)^2+6 a^3a-9 a^2+1 \\ =-(3a+1)^2+6 (3a+1)a-9 a^2+1=-9a^2-6a-1+18a^2+6a-9a^2+1=0$
so $2-a^2$ is also a root of $f$.

We have that $f$ is of degree $3$ and $a$ and $2-a^2$ are two roots, so we have to find the third root of $f$ by applying the Euclidean algorithm. $f(x)=(x-a)(x-(2-a^2))(x-(a^2-a-2))$

Since $a, 2-a^2, a^2-a-2\in \mathbb{Q}(a)$, i.e., all the roots belong to $\mathbb{Q}(a)$, we have that the extension $\mathbb{Q}(a)/\mathbb{Q}$ is normal.

Is this correct? (Wondering) Let $n$ a positive integer and $c_0, c_1, c_2\in \mathbb{Z}$ such that $(3+a-a^2)^n=c_0+c_1a+c_2a^2$.
Why do for every $n$ exist such $c_0, c_1, c_2$ ? (Wondering)

How could we show the following relation:
$(1-a)^n=(c_0+2c_1+4c_2)+c_2a-(c_1+c_2)a^2$
? (Wondering)

 

[caffeinemachine]

Hey!

Let $f(x)=x^3-3x-1\in \mathbb{Q}[x]$.

Then $f(x+1)=(x+1)^3-3(x+1)-1=x^3+3x^2+3x+1-3x-3-1=x^3+3x^2-3$.

The prime $p=3$ divides all the coefficients except of the one of the term of the highest degree and $p^2$ doesn't divide the constant term.
So, from Eisenstein's criterion we have that $f(x+1)$ is irreducible in $\mathbb{Q}[x]$, and so is $f(x)$.

Let $a\in \mathbb{C}$ be a root of $f$, $f(a)=0 \Rightarrow a^3=3a+1$.

We have that $f(2-a^2)=(2-a^2)^3-3(2-a^2)-1=8-12 a^2+6 a^4-a^6-6+3a^2-1 \\ =-a^6+6 a^4-9 a^2+1=-(a^3)^2+6 a^3a-9 a^2+1 \\ =-(3a+1)^2+6 (3a+1)a-9 a^2+1=-9a^2-6a-1+18a^2+6a-9a^2+1=0$
so $2-a^2$ is also a root of $f$.

We have that $f$ is of degree $3$ and $a$ and $2-a^2$ are two roots, so we have to find the third root of $f$ by applying the Euclidean algorithm. $f(x)=(x-a)(x-(2-a^2))(x-(a^2-a-2))$

Since $a, 2-a^2, a^2-a-2\in \mathbb{Q}(a)$, i.e., all the roots belong to $\mathbb{Q}(a)$, we have that the extension $\mathbb{Q}(a)/\mathbb{Q}$ is normal.

Is this correct? (Wondering)

Yes. This much is fine.

Let $n$ a positive integer and $c_0, c_1, c_2\in \mathbb{Z}$ such that $(3+a-a^2)^n=c_0+c_1a+c_2a^2$.
Why do for every $n$ exist such $c_0, c_1, c_2$ ? (Wondering)

You can expand out $(3+a-a^2)^n$ using the binomial theorem in the form $\sum b_ja^j$ for some integers $b_j$. Now since $a^3=3a+1$, an inductive argument shows that $\sum_jb_ja^j = c_0+c_1a+c_2a^2$.

How could we show the following relation:
$(1-a)^n=(c_0+2c_1+4c_2)+c_2a-(c_1+c_2)a^2$
? (Wondering)

For this give me some time.

 

[mathmari]

You can expand out $(3+a-a^2)^n$ using the binomial theorem in the form $\sum b_ja^j$ for some integers $b_j$. Now since $a^3=3a+1$, an inductive argument shows that $\sum_jb_ja^j = c_0+c_1a+c_2a^2$.

The formula of the binomial theorem when we have $3$ terms is the following:

$$(3+a-a^2)^n=\sum_{k_1+k_2+k_3=n}\binom{n}{k_1+k_2+k_3}3^{k_1}a^{k_2}(a^2)^{k_3}$$

So, to show that $(3+a-a^2)^n=c_0+c_1a+c_2a^2$, do we use induction in respect to $n$ ?

Base case: For $n=1$ we have that $(3+a-a^2)^1=1+1\cdot a+(-1)\cdot a^2$.

Inductive hypothesis: We assume that $(3+a-a^2)^n=c_0+c_1a+c_2a^2, \forall n\in \mathbb{N}$.

Inductive step: We want to show that it holds for $n+1$:
$$(3+a-a^2)^{n+1}=(3+a-a^2)^n(3+a-a^2)\overset{\text{Ind. Hyp}}{=}(c_0+c_1a+c_2a^2)(3+a-a^2) \\ =3c_0+c_0a-c_0a^2+3c_1a+c_1a^2-c_1a^3+3c_2a^2+c_2a^3-c_2a^4 \\ =3c_0+c_0a-c_0a^2+3c_1a+c_1a^2-c_1(3a+1)+3c_2a^2+c_2(3a+1)-c_2a(3a+1) \\ =(3c_0-c_1+c_2)+(c_0+2c_2)a+(-c_0+c_1+3c_2)a^2$$
So, there are $\tilde{c_0}=(3c_0-c_1+c_2), \tilde{c_1}=(c_0+2c_2), \tilde{c_2}=(-c_0+c_1+3c_2)$ such that $(3+a-a^2)^{n+1}=\tilde{c_0}+\tilde{c_1}a+\tilde{c_2}a^2$.

Therefore, we have that for each $n$ there are $c_0, c_1, c_2\in \mathbb{Z}$ such that $(3+a-a^2)^n=c_0+c_1a+c_2a^2$.
Is this correct? (Wondering)

 

[caffeinemachine]

The formula of the binomial theorem when we have $3$ terms is the following:

$$(3+a-a^2)^n=\sum_{k_1+k_2+k_3=n}\binom{n}{k_1+k_2+k_3}3^{k_1}a^{k_2}(a^2)^{k_3}$$

So, to show that $(3+a-a^2)^n=c_0+c_1a+c_2a^2$, do we use induction in respect to $n$ ?

Base case: For $n=1$ we have that $(3+a-a^2)^1=1+1\cdot a+(-1)\cdot a^2$.

Inductive hypothesis: We assume that $(3+a-a^2)^n=c_0+c_1a+c_2a^2, \forall n\in \mathbb{N}$.

Inductive step: We want to show that it holds for $n+1$:
$$(3+a-a^2)^{n+1}=(3+a-a^2)^n(3+a-a^2)\overset{\text{Ind. Hyp}}{=}(c_0+c_1a+c_2a^2)(3+a-a^2) \\ =3c_0+c_0a-c_0a^2+3c_1a+c_1a^2-c_1a^3+3c_2a^2+c_2a^3-c_2a^4 \\ =3c_0+c_0a-c_0a^2+3c_1a+c_1a^2-c_1(3a+1)+3c_2a^2+c_2(3a+1)-c_2a(3a+1) \\ =(3c_0-c_1+c_2)+(c_0+2c_2)a+(-c_0+c_1+3c_2)a^2$$
So, there are $\tilde{c_0}=(3c_0-c_1+c_2), \tilde{c_1}=(c_0+2c_2), \tilde{c_2}=(-c_0+c_1+3c_2)$ such that $(3+a-a^2)^{n+1}=\tilde{c_0}+\tilde{c_1}a+\tilde{c_2}a^2$.

Therefore, we have that for each $n$ there are $c_0, c_1, c_2\in \mathbb{Z}$ such that $(3+a-a^2)^n=c_0+c_1a+c_2a^2$.
Is this correct? (Wondering)

Yes this is good. In fact there is no need to invoke the binomial theorem as your solution makes clear.

 

[mathmari]

How could we show the following relation:
$(1-a)^n=(c_0+2c_1+4c_2)+c_2a-(c_1+c_2)a^2$
? (Wondering)

To show this relation we have to consider the automorphism $\tau : \mathbb{Q}[a]\rightarrow \mathbb{Q}[a]$ that is defined as follows:
$\tau (1) = 1 \\ \tau (a) = a^2 − a − 2$

How do we know that such an automorphism exists? (Wondering)

 

[caffeinemachine]

To show this relation we have to consider the automorphism $\tau : \mathbb{Q}[a]\rightarrow \mathbb{Q}[a]$ that is defined as follows:
$\tau (1) = 1 \\ \tau (a) = a^2 − a − 2$

How do we know that such an automorphism exists? (Wondering)

Right. I forgot about this problem.

In general, if $f(x)$ is an irreducible polynomial over a field $F$ and $K$ is an extension of $F$, and if $a$ and $b$ are two roots of $f$ in $K$, then there is an isomorphism $F(a)\to F(b)$ which takes $a$ to $b$ and fixes $F$ pointwise.

Now we have $a$ and $b:=a^2-a-2$ are both roots (in some extension of $\mathbf Q$, namely $\mathbf Q(a)$) of an irreducible polynomial $f$ over $\mathbf Q$, we have an isomorphism $\mathbf Q(a)\to \mathbf Q(b)$ which takes $a$ to $b$. Since $\mathbf Q(b)$ is contained in $\mathbf Q(a)$, it follows from this isomorphism that $\mathbf Q(b)=\mathbf Q(a)$ and we are done.

 

[mathmari]

In general, if $f(x)$ is an irreducible polynomial over a field $F$ and $K$ is an extension of $F$, and if $a$ and $b$ are two roots of $f$ in $K$, then there is an isomorphism $F(a)\to F(b)$ which takes $a$ to $b$ and fixes $F$ pointwise.

We have that $K=F[x]/\langle f(x)\rangle$ is a field that contains $F$ and inside of which $f(x)$ has a root.
From a theorem we have that there is a field $K$, an extenion of $F$ such that $a\in K$ wih $f(a)=0$ and $K=F[a]$.
Therefore, we have that $F[x]/\langle f(x)\rangle=F[a]$.

Applying the same for $b$ instead of $a$, we get $F[x]/\langle f(x)\rangle=F$. Is everything correct so far? (Wondering)

Do we get from that that there is an isomorphism $F(a)\to F(b)$ which takes $a$ to $b$ and fixes $F$ pointwise? Or do we prove it in an other way? (Wondering)

 

[caffeinemachine]

We have that $K=F[x]/\langle f(x)\rangle$ is a field that contains $F$ and inside of which $f(x)$ has a root.
From a theorem we have that there is a field $K$, an extenion of $F$ such that $a\in K$ wih $f(a)=0$ and $K=F[a]$.
Therefore, we have that $F[x]/\langle f(x)\rangle=F[a]$.

Applying the same for $b$ instead of $a$, we get $F[x]/\langle f(x)\rangle=F$.

It seems that you do not understand the theorem I stated. Writing $F[x]/\langle f(x)\rangle = F(a)$ is not even valid if $a$ is not contained in $F[x]/\langle f(x)\rangle$. What you probably meant is that $F[x]/\langle f(x)\rangle \cong F(a)$. You also have $F[x]/\langle f(x)\rangle \cong F(b)$. But this does not tell us that there is an isomorphism $F(a)\to F(b)$ which takes $a$ to $b$.

I suggest you think about this theorem more carefully and try to come up with a proof. If you are unable to supply a proof then tell me and I will help you.

 

[mathmari]

It seems that you do not understand the theorem I stated. Writing $F[x]/\langle f(x)\rangle = F(a)$ is not even valid if $a$ is not contained in $F[x]/\langle f(x)\rangle$. What you probably meant is that $F[x]/\langle f(x)\rangle \cong F(a)$. You also have $F[x]/\langle f(x)\rangle \cong F(b)$. But this does not tell us that there is an isomorphism $F(a)\to F(b)$ which takes $a$ to $b$.

I thought the following:
When $F[x]/\langle f(x)\rangle \cong F[a]$ and $F[x]/\langle f(x)\rangle \cong F$ we get that $F[a]\cong F $ and so there is an isomorphism $F[a]\to F$.
Why does it not hold? (Wondering)

 

[caffeinemachine]

I thought the following:
When $F[x]/\langle f(x)\rangle \cong F[a]$ and $F[x]/\langle f(x)\rangle \cong F$ we get that $F[a]\cong F $ and so there is an isomorphism $F[a]\to F$.
Why does it not hold? (Wondering)

It is true that $F[x]/\langle f(x)\rangle F(a), F(b)$ and thus $F(a)\cong F(b)$. But this does not mean that there is an isomorphism $F(a)\to F(b)$ which takes $a$ to $b$ and fixes $F$ pointwise.

 

[mathmari]

I suggest you think about this theorem more carefully and try to come up with a proof. If you are unable to supply a proof then tell me and I will help you.

Could you give me a hint for the proof? (Wondering)

 

[mathmari]

$f(x)=(x-a)(x-(2-a^2))(x-(a^2-a-2))$

Since $a, 2-a^2, a^2-a-2\in \mathbb{Q}(a)$, i.e., all the roots belong to $\mathbb{Q}(a)$, we have that the extension $\mathbb{Q}(a)/\mathbb{Q}$ is normal.

So we have that one irreducible polynomial of $\mathbb{Q}$ that has a root in $\mathbb{Q}(a)$ has all the roots there.
To conclude that the extension is normal do we not have to show that all the irreducible polynomials of $\mathbb{Q}$ that have a root in $\mathbb{Q}(a)$ have all the roots there? (Wondering)

 

[caffeinemachine]

So we have that one irreducible polynomial of $\mathbb{Q}$ that has a root in $\mathbb{Q}(a)$ has all the roots there.
To conclude that the extension is normal do we not have to show that all the irreducible polynomials of $\mathbb{Q}$ that have a root in $\mathbb{Q}(a)$ have all the roots there? (Wondering)

This is a separate question. It is true that to show normality of an extension $K:F$, one has to show that each irreducible polynomial in $f$ that has a root in $K$ splits in $K$. But there is a theorem which states that if $K:F$ is the splitting field for a polynomial $f$ over $F$ then $K:F$ is normal. Using this theorem allows us to say that $\mathbf Q(a):\mathbf Q$ is a normal extension.

 

[caffeinemachine]

Could you give me a hint for the proof? (Wondering)

Okay. Here is the thing.

Theorem. Let $F$ be a field and $f(x)$ be an irreducible polynomial over $F$. Let $L:F$ and $K:F$ be extensions of $F$. Suppose $K$ has a root $a$ of $f$ and $L$ has a root $b$ of $F$. Then there is an isomorphism $F(a)\to F(b)$ which takes $a$ to $b$ and fixes $F$ pointwise.

Proof. (Sketch). Let $:\varphi: F[x]\to F(a)\subseteq K$ be the map which sends $p(x)$ to $p(a)$. Then $\varphi$ is a ring homomorphism with kernel $\langle f(x)\rangle$ (this uses the irreducibility of $f$ over $F$). Thus we have an isomorphism $\bar \varphi: F[x]\langle f(x)\rangle \to F(a)$. Similarly we have a a ring homomorphism $\psi: F[x]\to F(b)$ inducing an isomorphism $\bar \psi:F[x]/\langle f(x)\rangle \to F(b)$. Can you see that $\bar \psi\circ \bar \varphi^{-1}:F(a)\to F(b)$ has the required properties.