# Field Theory - Dummit and Foote Ch 13 - Exercise 1, page 519

#### Peter

##### Well-known member
MHB Site Helper
I am studying Dummit and Foote Chapter 13: Field Theory.

Exercise 1 on page 519 reads as follows:

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"Show that [TEX] p(x) = x^3 + 9x + 6 [/TEX] is irreducible in [TEX] \mathbb{Q}[x] [/TEX]. Let [TEX] \theta [/TEX] be a root of p(x). Find the inverse of [TEX] 1 + \theta [/TEX] in [TEX] \mathbb{Q} ( \theta ) [/TEX]."

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Now to show that [TEX] p(x) = x^3 + 9x + 6 [/TEX] is irreducible in [TEX] \mathbb{Q}[x] [/TEX] use Eisenstein's Criterion

[TEX] p(x) = x^3 + 9x + 6 = x^3 + a_1 x + a_0 [/TEX]

Now (3) is a prime ideal in the integral domain [TEX] \mathbb{Q} [/TEX]

and [TEX] a_1 = 9 \in (3) [/TEX]

and [TEX] a_0 = 6 \in (3) [/TEX] and [TEX] a_0 \notin (9) ([/TEX]

Thus by Eisenstein, p(x) is irreducible in [TEX] \mathbb{Q}[x] [/TEX]

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However, I am not sure how to go about part two of the problem, namely:

"Let [TEX] \theta [/TEX] be a root of p(x). Find the inverse of [TEX] 1 + \theta [/TEX] in [TEX] \mathbb{Q} ( \theta ) [/TEX]."

I would be grateful for some help with this problem.

Peter

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#### johng

##### Well-known member
MHB Math Helper
Hi Peter,
You can't say (3) is a prime ideal in Q -- Q is a field, whose only ideals are (0) and Q.
If your polynomial were reducible over Q, it would have to have a linear factor and hence a root in Q. The Eisenstein criterion says the only possible roots of p are $$\displaystyle \{\pm1,\pm2,\pm3,\pm6\}$$. None of these are roots.
Now any element of $$\displaystyle Q(\theta)$$ is of the form:
$$\displaystyle a_0+a_1\theta+a_2\theta^2$$
So multiply $$\displaystyle 1+\theta$$ by such an element and use the fact that $$\displaystyle \theta^3+9\theta+6=0$$.

#### Peter

##### Well-known member
MHB Site Helper
Hi Peter,
You can't say (3) is a prime ideal in Q -- Q is a field, whose only ideals are (0) and Q.
If your polynomial were reducible over Q, it would have to have a linear factor and hence a root in Q. The Eisenstein criterion says the only possible roots of p are $$\displaystyle \{\pm1,\pm2,\pm3,\pm6\}$$. None of these are roots.
Now any element of $$\displaystyle Q(\theta)$$ is of the form:
$$\displaystyle a_0+a_1\theta+a_2\theta^2$$
So multiply $$\displaystyle 1+\theta$$ by such an element and use the fact that $$\displaystyle \theta^3+9\theta+6=0$$.
Thanks for the help Johng

You write:

"You can't say (3) is a prime ideal in Q -- Q is a field, whose only ideals are (0) and Q."

Yes, of course you are right ... careless of me!

You also write:

"The Eisenstein criterion says the only possible roots of p are $$\displaystyle \{\pm1,\pm2,\pm3,\pm6\}$$. None of these are roots."

Is this really using the Eisenstein Criterion? I think it is actually using Propositions 10 and 11 of Dummit and Foote Chapter 9: Polynomial Rings:

They are given on page 308 D&F as follows:

Proposition 10: A polynomial of degree two or three over a field F is reducible if and only if it has a root in F.

Proposition 11. Let $$\displaystyle p(x) = a_nx^n + a_{n-1} x^{n-1} + ... ... + a_0$$ be a polynomial of degree n with integer coefficients. If $$\displaystyle r/s \in \mathbb{Q}$$ is in lowest terms (i.e. r and s are relatively prime integers) and r/s is a root of p(x), then r divides the constant term and s divides the leading coefficient of p(x): $$\displaystyle r|a_0$$ and $$\displaystyle s|a_n$$. In particular, if p(x) is a monic polynomial with integer coefficients and $$\displaystyle p(d) \ne 0$$ for all integers d dividing the constant term of p(x), then p(x) has no roots in $$\displaystyle \mathbb{Q}$$.

Mind you, this is a rather trivial and pedantic point ... also I could be wrong if either of the above propositions are attributed to Eisenstein ...

Essentially you have solved the problem ... so thanks again ...

Please let me know if you think that you are not using the above propositions ...

Peter

#### Peter

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MHB Site Helper
Thanks for the help Johng

You write:

"You can't say (3) is a prime ideal in Q -- Q is a field, whose only ideals are (0) and Q."

Yes, of course you are right ... careless of me!

You also write:

"The Eisenstein criterion says the only possible roots of p are $$\displaystyle \{\pm1,\pm2,\pm3,\pm6\}$$. None of these are roots."

Is this really using the Eisenstein Criterion? I think it is actually using Propositions 10 and 11 of Dummit and Foote Chapter 9: Polynomial Rings:

They are given on page 308 D&F as follows:

Proposition 10: A polynomial of degree two or three over a field F is reducible if and only if it has a root in F.

Proposition 11. Let $$\displaystyle p(x) = a_nx^n + a_{n-1} x^{n-1} + ... ... + a_0$$ be a polynomial of degree n with integer coefficients. If $$\displaystyle r/s \in \mathbb{Q}$$ is in lowest terms (i.e. r and s are relatively prime integers) and r/s is a root of p(x), then r divides the constant term and s divides the leading coefficient of p(x): $$\displaystyle r|a_0$$ and $$\displaystyle s|a_n$$. In particular, if p(x) is a monic polynomial with integer coefficients and $$\displaystyle p(d) \ne 0$$ for all integers d dividing the constant term of p(x), then p(x) has no roots in $$\displaystyle \mathbb{Q}$$.

Mind you, this is a rather trivial and pedantic point ... also I could be wrong if either of the above propositions are attributed to Eisenstein ...

Essentially you have solved the problem ... so thanks again ...

Please let me know if you think that you are not using the above propositions ...

Peter
Hi johng,

Now working on the second part of the problem and using your guidance - specifically:

"Now any element of [FONT=MathJax_Math]Q[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]θ[/FONT][FONT=MathJax_Main])[/FONT] is of the form:
$$\displaystyle a_0 + a_1 \theta + a_2 {\theta}^2$$
So multiply [FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]θ[/FONT] by such an element and use the fact that $$\displaystyle {\theta}^3 + 9 \theta = 6 = 0$$ "

OK so we have

$$\displaystyle (a_0 + a_1 \theta + a_2 {\theta}^2) (1+ \theta) = 1$$

So $$\displaystyle a_0 + (a_0 +a_1) \theta + (a_1 + a_2) {\theta}^2 + a_2{\theta}^3 = 1$$ ... ... (1)

But where to from here?

You mention that we should use $$\displaystyle {\theta}^3 + 9 \theta + 6 = 0$$ ... ...

What to do ....

We could write (1) above as follows:

$$\displaystyle [a_2 {\theta}^3 + (a_0 + a_1) \theta + a_0] + (a_1 + a_2){\theta}^2 = 1$$

and then assign values to a_2, a_1 and a_0 to make the term in the sqare brackets equal to zero ... but, well, I think I am wandering around here

Can anyone help?

Peter

#### Opalg

##### MHB Oldtimer
Staff member
I think what johng is suggesting is that you should divide $\theta^3 + 9\theta + 6$ by $\theta+1$: $$0 = \theta^3 + 9\theta + 6 = (\theta + 1)(\text{quadratic quotient}) + \text{remainder}.$$ It will then follow that $$(\theta+1)^{-1} = -\frac{\text{quotient}}{\text{remainder}}.$$

#### Peter

##### Well-known member
MHB Site Helper
I think what johng is suggesting is that you should divide $\theta^3 + 9\theta + 6$ by $\theta+1$: $$0 = \theta^3 + 9\theta + 6 = (\theta + 1)(\text{quadratic quotient}) + \text{remainder}.$$ It will then follow that $$(\theta+1)^{-1} = -\frac{\text{quotient}}{\text{remainder}}.$$
Thanks Opalg.

So then after dividing we have the following:

$$\displaystyle 0 = {\theta}^3 + 9 \theta + 6 = ( \theta + 1 ) ( {\theta}^2 - \theta + 8) -2$$

$$\displaystyle \Longrightarrow (\theta + 1)( {\theta}^2 - \theta +8) = 2$$

$$\displaystyle \Longrightarrow (\theta + 1)( {\theta}^2 - \theta +8)/2 = 1$$

$$\displaystyle \Longrightarrow (\theta + 1)^{-1} = ( {\theta}^2 - \theta +8)/2$$

Peter

#### Opalg

##### MHB Oldtimer
Staff member
Thanks Opalg.

So then after dividing we have the following:

$$\displaystyle 0 = {\theta}^3 + 9 \theta + 6 = ( \theta + 1 ) ( {\theta}^2 - \theta + 8) -2$$ ${}\qquad\color{red}{( \theta + 1 ) ( {\theta}^2 - \theta + 10) -4\ ?}$
...

#### Deveno

##### Well-known member
MHB Math Scholar
1. When applying Eisenstein to rational polynomials, we typically turn them into integral polynomials by multiplying by the lcm of the denominators of the coefficients. This turns our polynomial into one over an integral domain, the integers. In the integers, we HAVE prime ideals, namely the ideals generated by a prime integer.

2. The method Opalg suggests only works because $\theta + 1$ is a linear polynomial in $\theta$. For finding inverses of higher degree elements of $\Bbb Q(\theta)$ the gcd method is preferred.

3. I believe johng is referred to the Rational Root Theorem, not the Eisenstein criterion. Since the degree of our original irreducible polynomial is 3, it follows that if it factors at all, one of the factors must be linear: that is, it must have a rational root, which it does not.

EDIT: this is the "hard way" to find the inverse:

Assuming we already KNOW that $\Bbb Q(\theta)$ is a field (by dint of the irreducibility of $x^3 + 9x + 6$), we seek $a_1,a_2,a_3 \in \Bbb Q$ such that:

$(\theta + 1)(a_1\theta^2 + a_2\theta + a_3) = 1$.

Multiplying this out we get:

$a_1\theta^3 + (a_1+a_2)\theta^2 + (a_2+a_3)\theta + a_3 = 1$.

Since $\theta$ is a root of $x^3 + 9x + 6$, we have: $\theta^3 = -9\theta - 6$.

Substituting this in the above, gives us:

$(a_1+a_2)\theta^2 + (a_2+a_3-9a_1)\theta + (a_3-6a_1) = 1$.

Since $\{\theta^2,\theta,1\}$ is a basis for $\Bbb Q(\theta)$ (considered as a vector space over $\Bbb Q$, again, this follows from the irreducibility of the cubic above), we must have:

$a_1+a_2 = 0$
$a_2+a_3-9a_1 = 0$
$a_3-6a_1 = 1$

The last equation tells us $a_3 = 1+6a_1$ so substituting into the second equation we obtain:

$a_2 = 3a_1 - 1$.

Substituting THIS into the first equation, we get:

$a_1 = \frac{1}{4}$, and thus: $a_2 = -\frac{1}{4}, a_3 = \frac{10}{4}$.

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#### Peter

##### Well-known member
MHB Site Helper
1. When applying Eisenstein to rational polynomials, we typically turn them into integral polynomials by multiplying by the lcm of the denominators of the coefficients. This turns our polynomial into one over an integral domain, the integers. In the integers, we HAVE prime ideals, namely the ideals generated by a prime integer.

2. The method Opalg suggests only works because $\theta + 1$ is a linear polynomial in $\theta$. For finding inverses of higher degree elements of $\Bbb Q(\theta)$ the gcd method is preferred.

3. I believe johng is referred to the Rational Root Theorem, not the Eisenstein criterion. Since the degree of our original irreducible polynomial is 3, it follows that if it factors at all, one of the factors must be linear: that is, it must have a rational root, which it does not.

EDIT: this is the "hard way" to find the inverse:

Assuming we already KNOW that $\Bbb Q(\theta)$ is a field (by dint of the irreducibility of $x^3 + 9x + 6$), we seek $a_1,a_2,a_3 \in \Bbb Q$ such that:

$(\theta + 1)(a_1\theta^2 + a_2\theta + a_3) = 1$.

Multiplying this out we get:

$a_1\theta^3 + (a_1+a_2)\theta^2 + (a_2+a_3)\theta + a_3 = 1$.

Since $\theta$ is a root of $x^3 + 9x + 6$, we have: $\theta^3 = -9\theta - 6$.

Substituting this in the above, gives us:

$(a_1+a_2)\theta^2 + (a_2+a_3-9a_1)\theta + (a_3-6a_1) = 1$.

Since $\{\theta^2,\theta,1\}$ is a basis for $\Bbb Q(\theta)$ (considered as a vector space over $\Bbb Q$, again, this follows from the irreducibility of the cubic above), we must have:

$a_1+a_2 = 0$
$a_2+a_3-9a_1 = 0$
$a_3-6a_1 = 1$

The last equation tells us $a_3 = 1+6a_1$ so substituting into the second equation we obtain:

$a_2 = 3a_1 - 1$.

Substituting THIS into the first equation, we get:

$a_1 = \frac{1}{4}$, and thus: $a_2 = -\frac{1}{4}, a_3 = \frac{10}{4}$.

Thank you Deveno, this post is invaluable to me re understanding the application of Eisenstein's Criterion (just to mention one issue). It was so helpful when you wrote:

"1. When applying Eisenstein to rational polynomials, we typically turn them into integral polynomials by multiplying by the lcm of the denominators of the coefficients. This turns our polynomial into one over an integral domain, the integers. In the integers, we HAVE prime ideals, namely the ideals generated by a prime integer."

Thus when I wrote:

"Now to show that [TEX] p(x) = x^3 + 9x + 6 [/TEX] is irreducible in [TEX] \mathbb{Q}[x] [/TEX] use Eisenstein's Criterion

[TEX] p(x) = x^3 + 9x + 6 = x^3 + a_1 x + a_0 [/TEX]

Now (3) is a prime ideal in the integral domain [TEX] \mathbb{Q} [/TEX]

and [TEX] a_1 = 9 \in (3) [/TEX]

and [TEX] a_0 = 6 \in (3) [/TEX] and [TEX] a_0 \notin (9) ([/TEX]

Thus by Eisenstein, p(x) is irreducible in [TEX] \mathbb{Q}[x] [/TEX]"

I was incorrect when I wrote:

"Now (3) is a prime ideal in the integral domain [TEX] \mathbb{Q} [/TEX]"

as I should have written

"Now (3) is a prime ideal in the integral domain [TEX] \mathbb{Z} [/TEX]"

Then my use of Eisenstein's criterion would have been OK. [Can someone please confirm this?]

BUT, then I would have shown that p(x) was irreducible in Z[x]. How can I be sure it has no root in Q[X]? Can someone please help? [Thought ... can I appeal somehow to the Rational Roots Theorem?]

If we show using Eisenstein that p(x) is irreducible in Z[x], how can we be sure it is irreducible in Q[x}?

Peter

Last edited:

#### Peter

##### Well-known member
MHB Site Helper
Thank you Deveno, this post is invaluable to me re understanding the application of Eisenstein's Criterion (just to mention one issue). It was so helpful when you wrote:

"1. When applying Eisenstein to rational polynomials, we typically turn them into integral polynomials by multiplying by the lcm of the denominators of the coefficients. This turns our polynomial into one over an integral domain, the integers. In the integers, we HAVE prime ideals, namely the ideals generated by a prime integer."

Thus when I wrote:

"Now to show that [TEX] p(x) = x^3 + 9x + 6 [/TEX] is irreducible in [TEX] \mathbb{Q}[x] [/TEX] use Eisenstein's Criterion

[TEX] p(x) = x^3 + 9x + 6 = x^3 + a_1 x + a_0 [/TEX]

Now (3) is a prime ideal in the integral domain [TEX] \mathbb{Q} [/TEX]

and [TEX] a_1 = 9 \in (3) [/TEX]

and [TEX] a_0 = 6 \in (3) [/TEX] and [TEX] a_0 \notin (9) ([/TEX]

Thus by Eisenstein, p(x) is irreducible in [TEX] \mathbb{Q}[x] [/TEX]"

I was incorrect when I wrote:

"Now (3) is a prime ideal in the integral domain [TEX] \mathbb{Q} [/TEX]"

as I should have written

"Now (3) is a prime ideal in the integral domain [TEX] \mathbb{Z} [/TEX]"

Then I take it my analysis would have been OK. Can someone confirm this for me?

Peter
Deveno,

In your post above you write:

"Substituting this in the above, gives us:

$$\displaystyle (a_1 + a_2) {\theta}^2 + (a_2 + a_3 - 9a_1) \theta + (a_3 - 6a_1) = 1$$

Since $$\displaystyle \{ {\theta}^2, \theta, 1 \}$$ is a basis for [FONT=MathJax_AMS]Q[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]θ[/FONT][FONT=MathJax_Main])[/FONT] (considered as a vector space over [FONT=MathJax_AMS]Q[/FONT], again, this follows from the irreducibility of the cubic above), we must have:

[FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]
[FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]9[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]
[FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT]"

Can you explain why it is necessary that $$\displaystyle \{ {\theta}^2, \theta, 1 \}$$[FONT=MathJax_Main][/FONT] is a basis for [FONT=MathJax_AMS]Q[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]θ[/FONT][FONT=MathJax_Main])[/FONT] - can you explain how and why $$\displaystyle \{ {\theta}^2, \theta, 1 \}$$ being a basis for [FONT=MathJax_AMS]Q[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]θ[/FONT][FONT=MathJax_Main])[/FONT] comes into your reasoning - why exactly do we need this condition?

Peter

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#### Peter

##### Well-known member
MHB Site Helper
Thank you Deveno, this post is invaluable to me re understanding the application of Eisenstein's Criterion (just to mention one issue). It was so helpful when you wrote:

"1. When applying Eisenstein to rational polynomials, we typically turn them into integral polynomials by multiplying by the lcm of the denominators of the coefficients. This turns our polynomial into one over an integral domain, the integers. In the integers, we HAVE prime ideals, namely the ideals generated by a prime integer."

Thus when I wrote:

"Now to show that [TEX] p(x) = x^3 + 9x + 6 [/TEX] is irreducible in [TEX] \mathbb{Q}[x] [/TEX] use Eisenstein's Criterion

[TEX] p(x) = x^3 + 9x + 6 = x^3 + a_1 x + a_0 [/TEX]

Now (3) is a prime ideal in the integral domain [TEX] \mathbb{Q} [/TEX]

and [TEX] a_1 = 9 \in (3) [/TEX]

and [TEX] a_0 = 6 \in (3) [/TEX] and [TEX] a_0 \notin (9) ([/TEX]

Thus by Eisenstein, p(x) is irreducible in [TEX] \mathbb{Q}[x] [/TEX]"

I was incorrect when I wrote:

"Now (3) is a prime ideal in the integral domain [TEX] \mathbb{Q} [/TEX]"

as I should have written

"Now (3) is a prime ideal in the integral domain [TEX] \mathbb{Z} [/TEX]"

Then my use of Eisenstein's criterion would have been OK. [Can someone please confirm this?]

BUT, then I would have shown that p(x) was irreducible in Z[x]. How can I be sure it has no root in Q[X]? Can someone please help? [Thought ... can I appeal somehow to the Rational Roots Theorem?]

If we show using Eisenstein that p(x) is irreducible in Z[x], how can we be sure it is irreducible in Q[x}?

Peter
My apologies to members of the MHBs ... Regarding my post above, I think I have been careless in not reading D&F carefully enough on Eisenstein's Criterion.

On page 310 of Dummit and Foote, we find the following:

Corollary 14: (Eisenstein's Criterion for $$\displaystyle \mathbb{Z}[x]$$). Let p be a prime in $$\displaystyle \mathbb{Z}$$ and let $$\displaystyle f(x) = x^n + a_{n-1}x^{n-1} + ... ... a_1x + a_0 \in \mathbb{Z}[x], n \ge 1$$. Suppose p divides $$\displaystyle a_i$$ for all $$\displaystyle i \in \{ 0, 1, 2, ... ... n-1 \}$$ but that $$\displaystyle p^2$$ does not divide $$\displaystyle a_0$$. Then f(x) is irreducible in $$\displaystyle \mathbb{Z}[x]$$ and $$\displaystyle \mathbb{Q}[x]$$

I obviously should have noted: :-(

Then f(x) is irreducible in $$\displaystyle \mathbb{Z}[x]$$ and $$\displaystyle \mathbb{Q}[x]$$

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
It is a theorem (Gauss' lemma) that if a polynomial with integer coefficients factors over the rationals, it factors over the integers. A proof would take too long to state here, but rests on the notion of primitive polynomials (one for which the gcd of the coefficients is 1). In fact, a similar result holds for any UFD R, and its field of fractions Q(R).

As to the use of the basis $\{\theta^2,\theta,1\}$...we employ this to show that the coefficients $a_1,a_2,a_3$ of $a_1\theta^2 + a_2\theta + a_3 \in \Bbb Q(\theta)$ are UNIQUE. Thus if two "polynomials" (which are linear combinations of powers of $\theta$ over $\Bbb Q$, in the same way that polynomials are linear combinations of powers of $x$ (an "indeterminate") over $\Bbb Q$) in $\theta$ of degree < 3 are equal, they have the same coefficients. One can also use the division algorithm to establish this. But if you are going to study field theory, you really need to get comfortable with linear algebra.

It is a theorem of linear algebra that the coefficients of a vector $v$ in any basis are uniquely determined. This is NOT true in the quotient ring $\Bbb Q[x]/(p(x))$, the coefficients of the polynomial $f(x)$ are not uniquely determined by $\tilde{f}(x) = f(x) + (p(x))$ UNLESS $\text{deg}(f) < \text{deg}(p)$ (many, many polynomials lie in the coset).