# Maximal Ideals - Exercise 5.1 (ii) Rotman AMA

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In summary, the conversation revolves around exercise 5.1 in Joseph J. Rotman's book, Advanced Modern Algebra (AMA), which focuses on prime ideals and maximal ideals. The exercise involves determining the maximum possible degree of an irreducible real polynomial, which is shown to be 2 with the use of the Fundamental Theorem of Algebra. The discussion also touches on the distinction between real and complex polynomials and the concept of conjugate pairs of roots. Overall, the conversation provides insight into the properties of real polynomials and their factorization.
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I am reading Joseph J. Rotman's book: Advanced Modern Algebra (AMA) and I am currently focused on Section 5.1 Prime Ideals and Maximal Ideals ...

I need some help with Exercise 5.1, Part (ii) ... ...Exercise 5.1 reads as follows:

View attachment 5942

Peter

What is the maximum possible degree of an irreducible real polynomial (Hint: it's a small integer)?

Deveno said:
What is the maximum possible degree of an irreducible real polynomial (Hint: it's a small integer)?

Sorry for the late reply, Deveno ... I have been unwell for a while ...

i suspect from memory the maximum possible degree is 2 ... but I am not sure of the reasoning that establishes this fact ...

Again ... apologies for the lateness of my reply ...

Petet

Peter said:
Sorry for the late reply, Deveno ... I have been unwell for a while ...

i suspect from memory the maximum possible degree is 2 ... but I am not sure of the reasoning that establishes this fact ...

Again ... apologies for the lateness of my reply ...

Petet

The Fundamental Theorem of Algebra states that every polynomial in $\Bbb C[x]$ splits into linear factors of the form $(x - z_i)$ for roots $z_i \in \Bbb C$. In other words, the roots of a complex polynomial are complex numbers.

Since complex-conjugation $z \mapsto \overline{z}$ is a field automorphism, it follows that for any $p(x) \in \Bbb C[x]$, that:

$\overline{p}(\overline{z}) = \overline{p(z)}$.

(Here, if $p(x) = a_0 + a_1x + a_2x^2 +\cdots + a_nx^n$ then $\overline{p}(x) = \overline{a_0} + \overline{a_1}x + \overline{a_2}x^2 + \cdots + \overline{a_n}x^n$).

In particular, if $p(x) \in \Bbb R[x]$ (so that $\overline{p} = p$), we have for any root $z$ of $p$:

$p(\overline{z}) = \overline{p}(\overline{z}) = \overline{p(z)} = \overline{0} = 0$.

This is often paraphrased as "complex roots of real polynomials come in conjugate pairs".

Furthermore, for each such conjugate-pair, we have:

$(x - z)(x - \overline{z}) = x^2 - (z + \overline{z})x + z\overline{z} = x^2 - 2\mathfrak{Re}(z)x + |z|^2 \in \Bbb R[x]$.

Deveno said:
The Fundamental Theorem of Algebra states that every polynomial in $\Bbb C[x]$ splits into linear factors of the form $(x - z_i)$ for roots $z_i \in \Bbb C$. In other words, the roots of a complex polynomial are complex numbers.

Since complex-conjugation $z \mapsto \overline{z}$ is a field automorphism, it follows that for any $p(x) \in \Bbb C[x]$, that:

$\overline{p}(\overline{z}) = \overline{p(z)}$.

(Here, if $p(x) = a_0 + a_1x + a_2x^2 +\cdots + a_nx^n$ then $\overline{p}(x) = \overline{a_0} + \overline{a_1}x + \overline{a_2}x^2 + \cdots + \overline{a_n}x^n$).

In particular, if $p(x) \in \Bbb R[x]$ (so that $\overline{p} = p$), we have for any root $z$ of $p$:

$p(\overline{z}) = \overline{p}(\overline{z}) = \overline{p(z)} = \overline{0} = 0$.

This is often paraphrased as "complex roots of real polynomials come in conjugate pairs".

Furthermore, for each such conjugate-pair, we have:

$(x - z)(x - \overline{z}) = x^2 - (z + \overline{z})x + z\overline{z} = x^2 - 2\mathfrak{Re}(z)x + |z|^2 \in \Bbb R[x]$.

Hi Deveno,

I'm not sure what question of mine you're answering ... I was asking about $$\displaystyle \mathbb{R}[x]$$ ... you seem to have provided some analysis pertaining to the case of $$\displaystyle \mathbb{C}[x]$$ ...

Please forgive me if I have completely missed the point of your post ... and please know that I very much appreciate your help ...

Peter

Peter said:
Hi Deveno,

I'm not sure what question of mine you're answering ... I was asking about $$\displaystyle \mathbb{R}[x]$$ ... you seem to have provided some analysis pertaining to the case of $$\displaystyle \mathbb{C}[x]$$ ...

Please forgive me if I have completely missed the point of your post ... and please know that I very much appreciate your help ...

Peter

Real polynomials are precisely those complex polynomials that equal their own conjugate-polynomial.

Here is an example:

$x^2 + x + 1$ is a REAL polynomial. ONE complex root is the complex number:

$\omega = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}$, since:$\omega^2 + \omega + 1 = \left(-\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}\right)^2 + -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} + 1$

$= \left(-\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}\right)\left(-\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}\right) + -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} + 1$

$=\dfrac{1}{4} -2i\dfrac{\sqrt{3}}{4} - \dfrac{3}{4} - \dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} + 1$

$= 0$

By the discussion above, $\overline{\omega} = -\dfrac{1}{2} -i\dfrac{\sqrt{3}}{2}$ is *also* a root of $x^2 + x + 1$.

These two roots of this REAL polynomial $x^2 + x + 1$ are both non-real, and any possible factorization of a quadratic would be into the two linear factors $(x - r_1)(x - r_2)$, so since both roots are non-real, we see that $x^2 + x + 1$ is an irreducible quadratic in $\Bbb R[x]$ (there AREN'T any irreducible complex polynomials over $\Bbb C$ except those of degree 0 or 1).

Deveno said:
Real polynomials are precisely those complex polynomials that equal their own conjugate-polynomial.

Here is an example:

$x^2 + x + 1$ is a REAL polynomial. ONE complex root is the complex number:

$\omega = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}$, since:$\omega^2 + \omega + 1 = \left(-\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}\right)^2 + -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} + 1$

$= \left(-\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}\right)\left(-\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}\right) + -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} + 1$

$=\dfrac{1}{4} -2i\dfrac{\sqrt{3}}{4} - \dfrac{3}{4} - \dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} + 1$

$= 0$

By the discussion above, $\overline{\omega} = -\dfrac{1}{2} -i\dfrac{\sqrt{3}}{2}$ is *also* a root of $x^2 + x + 1$.

These two roots of this REAL polynomial $x^2 + x + 1$ are both non-real, and any possible factorization of a quadratic would be into the two linear factors $(x - r_1)(x - r_2)$, so since both roots are non-real, we see that $x^2 + x + 1$ is an irreducible quadratic in $\Bbb R[x]$ (there AREN'T any irreducible complex polynomials over $\Bbb C$ except those of degree 0 or 1).
Thanks for the post, Deveno ...

That certainly clarified the matter ...

Peter

## 1. What is a maximal ideal?

A maximal ideal is a proper subset of a ring that is not contained in any other proper ideal. In other words, it is an ideal that cannot be enlarged while still remaining a proper subset of the ring.

## 2. How do you determine if an ideal is maximal?

An ideal is maximal if and only if it is not contained in any other proper ideal. In other words, if there is no ideal that properly contains the given ideal, then it is maximal.

## 3. What is the significance of maximal ideals in ring theory?

Maximal ideals play a crucial role in the structure of rings. They are used to define important concepts such as prime ideals and quotient rings. They also help to classify rings into different types based on their maximal ideals, such as local rings and integral domains.

## 4. Can a maximal ideal be a prime ideal as well?

Yes, a maximal ideal can also be a prime ideal. In fact, in a commutative ring, a maximal ideal is always a prime ideal.

## 5. How do you prove that an ideal is maximal?

To prove that an ideal is maximal, you must show that it is not contained in any other proper ideal. This can be done by assuming the opposite and arriving at a contradiction, or by using the fact that the quotient ring by a maximal ideal is a field.

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