MHB Field Extensions for polynomials

[mathjam0990]

Describe the multiplication in the ring Q[x]/(x²+x+1). Is this a field? What is the multiplicative inverse of [x]?

In describing the multiplication, would I just be describing something in regards to the multiplicative properties of a ring? i.e:
a(bc)=(ab)c
a(b+c)=ab+ac
a*1=1*a=a
ab=ba

Is it true that to show something is a field, the goal to show that every element has a multiplicative inverse?
I started by letting F be a field and E=Q[x]/(p(x)) be an extension field of p(x) over F given p(x)=x²+x+1. Then we need to show every element in E has an inverse?

To find the inverse of [x], do we solve -x²-x=1 for x? If so, is the inverse (-x-1) because x(-x-1)=-x²-x=1 ?

My apologies if I am not on the right track, this material is really hard for me to grasp.

Thanks in advance.

 

[Deveno]

I will describe the steps leading up to your question in "chunks". You can indicate in your reply where you start to feel lost.

Given a ring $R$, a (two-sided) IDEAL of $R$ is a subset $I$ such that:

1. $(I,+)$ is a subgroup of $(R,+)$. Equivalently, $a,b \in I \implies a-b \in I$.
2. For any $r\in R$, and $a \in I$, we have $ra \in I$ and $ar \in I$.

These properties are meant to mimic the following properties of 0:

a) 0+0 = 0
b) 0x = 0

Now $(R,+)$, for any ring, is an abelian group. So $(I,+)$ is a normal subgroup. We can thus define the quotient group:

$(R/I,+)$ where addition is defined by:

$(x + I) + (y + I) = (x+y) + I$.

But that's not all: it turns out that we can define a multiplication on $R/I$ by:

$(x + I)(y + I) = xy + I$

Of course, here is what we need to show that indeed "makes sense":

If $x + I = x' + I$ and $y + I = y' + I$, then $xy + I = x'y' + I$ (so our multiplication depends only on the COSETS, and not the $x$ or $y$).

From group theory, we know that $x + I = x' + I$ if and only if $x - x' \in I$.

So to show that $xy + I = x'y' + I$, we need to show that $xy - x'y' \in I$.

Now $xy - x'y' = xy - x'y + x'y - x'y' = (x - x')y + x'(y - y')$

By property 2 of an ideal, we have $(x-x')y, x'(y-y') \in I$, since both $x-x',y-y' \in I$. By property 1 we have $I$ is closed under addition, and thus the sum of $(x-x')y, x'(y-y')$ is in $I$, and thus $xy - x'y' \in I$.

The above holds for a "general ring". Now we turn our attention to a "specific kind of ring", the ring of polynomials $F[x]$ over a field $F$. This ring is rather special, as it is a Euclidean domain, with the Euclidean function "degree". This, in turn, means that it is a *principal ideal domain* (as all Euclidean domains are), so ANY ideal is generated by a single element.

So, ALL quotient rings of $F[x]$ are of the form $F[x]/\langle f(x)\rangle$, for some polynomial $f(x) \in F[x]$. Furthermore, all these quotients are COMMUTATIVE rings, because $F[x]$ is a commutative ring (in a polynomial ring, we take "$x$" to commute with everything, and the coefficients commute with each other since $F$ is a field, and all fields are commutative rings).

The natural question is: how "nice" can we make $F[x]/\langle f(x)\rangle$? The answer turns out to depend on the generator of the ideal, $f(x)$. If $f$ is PRIME (which in a Euclidean domain is the same as irreducible), then $F[x]/\langle f(x)\rangle$ is an integral domain (no zero-divisors). Which means we're already "pretty close" to it being a field.

But we can say even more:

In a principal ideal domain, we have $\langle a\rangle \subseteq \langle b\rangle$ if and only if $b|a$. But the only thing that divides an irreducible polynomial in $F[x]$ is a unit, or a unit times the polynomial. So, if $f(x)$ is irreducible then $\langle f(x)\rangle$ is a MAXIMAL ideal.

Now we have a correspondence theorem (via the fundamental isomorphism theorem for rings) of ideals of $R/I$ with ideals of $R$ containing $I$. If $I$ is a maximal ideal, the list of ideals of $R$ containing $I$ is very short: $I$, and $R$. So the list of ideals of $R/I$ is also very short:

$\{I\} = \{0_{R/I}\}$
$R/I$

It is a theorem that if an integral domain $D$ has only two ideals, $\langle 0\rangle = \{0\}$ and $D$, then $D$ is a field. So showing $E$ is a field, can be accomplished by showing $x^2 + x + 1$ is irreducible in $\Bbb Q[x]$.

But we can also do this by proving directly, that every coset of $\Bbb Q[x]/\langle x^2 + x +1\rangle$ has a multiplicative inverse (we already know it's a commutative ring).

So let's look at the multiplication of two cosets (we can choose our "representatives" to be of degree < 2, by dint of the division algorithm for polynomials over a field). I will write $I$ for $\langle x^2 + x + 1\rangle$, since it's tiresome to keep writing the same polynomial over and over.

$((a + bx) + I )\ast((c + dx) + I )= ac + (ad + bc)x + bdx^2 + I$.

However, since $x^2 + I = x^2 + x + 1 - x - 1 + I = (-x-1) + I + (x^2 + x + 1) + I = (-x-1) + I + I = (-x - 1) + I$,

we can re-write this as:

$= ac + (ad + bc)x + bd(-x -1) + I = (ac - bd) + (ad + bc - bd)x + I$.

Now the coset of $x$ is: $x + I$ (that is: $a = 0,\ b = 1$). If we are looking for an inverse to this, we need:

$ac - bd = -d = 1$
$ad + bc - bd = c - d = 0$.

This means: $c = -1,d = -1$, so the inverse of $[x] = x + I$ is $[-1 - x] = -1 - x + I$.

 

[mathjam0990]

It is a theorem that if an integral domain $D$ has only two ideals, $\langle 0\rangle = \{0\}$ and $D$, then $D$ is a field. So showing $E$ is a field, can be accomplished by showing $x^2 + x + 1$ is irreducible in $\Bbb Q[x]$.

But we can also do this by proving directly, that every coset of $\Bbb Q[x]/\langle x^2 + x +1\rangle$ has a multiplicative inverse (we already know it's a commutative ring).

So let's look at the multiplication of two cosets (we can choose our "representatives" to be of degree < 2, by dint of the division algorithm for polynomials over a field). I will write $I$ for $\langle x^2 + x + 1\rangle$, since it's tiresome to keep writing the same polynomial over and over.

$(a + bx) + I + (c + dx) + I = ac + (ad + bc)x + bdx^2 + I$.

However, since $x^2 + I = x^2 + x + 1 - x - 1 + I = (-x-1) + I + (x^2 + x + 1) + I = (-x-1) + I + I = (-x - 1) + I$,

we can re-write this as:

$= ac + (ad + bc)x + bd(-x -1) + I = (ac - bd) + (bc - bd)x + I$.

Now the coset of $x$ is: $x + I$ (that is: $a = 0,\ b = 1$). If we are looking for an inverse to this, we need:

$ac - bd = -d = 1$
$bc - bd = c - d = 0$.

This means: $c = -1,d = -1$, so the inverse of $[x] = x + I$ is $[-1 - x] = -1 - x + I$.

Firstly, I cannot thank you enough for taking the time to write out this detailed explanation. We covered majority of what you said in my lecture, but you provided some essential key points that were not mentioned by my professor which kind of helped to tie this altogether much better.

I understood all the background info you provided and it made sense. I only got lost toward the end where you said "let's look at the multiplication of two cosets...I will write I for ⟨x²+x+1⟩."

Q1) Did you mean to write [(a+bx)+I]*[(c+dx)+I] instead of [(a+bx)+I]+[(c+dx)+I]? Your statement in the beginning mentioned that (x+I)(y+I) = xy+I

Q2) I totally understand that x² + I = (-x-1) +I, but when plugging it in you got $= ac + (ad + bc)x + bd(-x -1) + I = (ac - bd) + (bc - bd)x + I. The bold part is what I do not understand I thought it would be...
= (ac - bd) + (ad+ bc - bd)x + I.

Q3) Could you please explain the very last part again where you said: "Now the coset of x is: x+I (that is: a=0, b=1). If we are looking for an inverse to this, we need..." I didn't quite grasp that. How did you get that a=0 and b=1 ?

Thanks again!

 

[Deveno]

You are correct, I had some typos in there.

The polynomial $x$ is the polynomial:

$f(x) = 0 + 1x + 0x^2 + 0x^3 +\dots$

If we write it in the form:

$f(x) = q(x)(x^2 + x + 1) + r(x)$

We have $q(x) = 0$, and $r(x) = x$, so its coset in $F[x]/\langle x^2 + x + 1\rangle$ is:

$r(x) + \langle x^2+x+1\rangle = x + \langle x^2+x+1\rangle$.

This is in the form $a + bx + \langle x^2+x+1\rangle$ with $a = 0$ and $b = 1$.