Maximal Ideals - Exercise 5.1 (ii) Rotman AMA

[Math Amateur]

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

Can someone please help me to get a start on this exercise ...

Peter

 

[Deveno]

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

 

[Math Amateur]

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

 

[Deveno]

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:

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

 

[Math Amateur]

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

 

[Deveno]

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).

 

[Math Amateur]

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 ...

Appreciate your help ...

Peter