Galois theory (beginner question about normal algebraic extension: Supposedly Z2(X3)⊂Z2(X)Z2(X3)⊂Z2(X) is not a normal alg extension.)

[MathNugget]

Supposedly [imath]Z_2(X^3) \subset Z_2(X)[/imath] is not a normal alg extension. I know by 1 definition that means if an irreducible polynomial from first field has a root in the 2nd, then it has all the roots in the 2nd.
Would [imath]x^3-1 = x^3+1=(x-1)(x^2+x+1)= (x+1)(x^2+x+1)[/imath] solve this? the last polynomial doesn't have any root in [imath]Z_2(X)[/imath], so it would be an example to prove that is not a normal extension...

 

[khansaheb]

Supposedly [imath]Z_2(X^3) \subset Z_2(X)[/imath] is not a normal alg extension. I know by 1 definition that means if an irreducible polynomial from first field has a root in the 2nd, then it has all the roots in the 2nd.
Would [imath]x^3-1 = x^3+1=(x-1)(x^2+x+1)= (x+1)(x^2[SIZE=6]+[/SIZE]x+1)[/imath] solve this? the last polynomial doesn't have any root in [imath]Z_2(X)[/imath], so it would be an example to prove that is not a normal extension...

a^3 - b^3 = (a - b)(a^2 + a*b + b^2) ....... and.........

a^3 + b^3 = (a + b)(a^2 - a*b + b^2)

Check your work carefully....

 

[blamocur]

I am assuming that [imath]\mathbb Z_2[/imath] is the field with two elements (0,1), but what do [imath]\mathbb Z_2(X^3)[/imath] and [imath]\mathbb Z_2(X)[/imath] mean ?

 

[MathNugget]

a^3 - b^3 = (a - b)(a^2 + a*b + b^2) ....... and.........

a^3 + b^3 = (a + b)(a^2 - a*b + b^2)

Check your work carefully....

True, but since it's [imath]\mathbb{Z}_2[/imath], I figured -1 = +1 . Or maybe I misunderstood your comment? I guess I didn't write that properly, it wasn't a random Z, it was [imath]\mathbb{Z}[/imath] (integers).

I am assuming that [imath]\mathbb Z_2[/imath] is the field with two elements (0,1), but what do [imath]\mathbb Z_2(X^3)[/imath] and [imath]\mathbb Z_2(X)[/imath] mean ?

Your guess is as good as mine, since I cannot seem to find this notation somewhere else. Given the inclusion, let's say:
[imath]\mathbb{Z}(X)=a_nX^n+...+a_1X+a_0 \Rightarrow \mathbb{Z}(X^3)=a_nX^{3n}+...+a_1X^3+a_0[/imath] .

 

[blamocur]

True, but since it's [imath]\mathbb{Z}_2[/imath], I figured -1 = +1 . Or maybe I misunderstood your comment? I guess I didn't write that properly, it wasn't a random Z, it was [imath]\mathbb{Z}[/imath] (integers).

Your guess is as good as mine, since I cannot seem to find this notation somewhere else. Given the inclusion, let's say:
[imath]\mathbb{Z}(X)=a_nX^n+...+a_1X+a_0 \Rightarrow \mathbb{Z}(X^3)=a_nX^{3n}+...+a_1X^3+a_0[/imath] .

But then [imath]\mathbb Z_2(X)[/imath] and [imath]\mathbb Z_2(X^3)[/imath] are just rings, not fields.

 

[MathNugget]

But then [imath]\mathbb Z_2(X)[/imath] and [imath]\mathbb Z_2(X^3)[/imath] are just rings, not fields.

You're right. I realize now, nowhere in the material I read did they mention that normal algebraic extensions refer only to fields.
Therefore, I am trying to prove it's a normal algebraic ring extension... with the same definitions and all that.

 

[blamocur]

a^3 - b^3 = (a - b)(a^2 + a*b + b^2) ....... and.........

a^3 + b^3 = (a + b)(a^2 - a*b + b^2)

Check your work carefully....

There is no difference between "+" and "-" in [imath]\mathbb Z_2[/imath].

 

[blamocur]

Supposedly [imath]Z_2(X^3) \subset Z_2(X)[/imath] is not a normal alg extension. I know by 1 definition that means if an irreducible polynomial from first field has a root in the 2nd, then it has all the roots in the 2nd.
Would [imath]x^3-1 = x^3+1=(x-1)(x^2+x+1)= (x+1)(x^2+x+1)[/imath] solve this? the last polynomial doesn't have any root in [imath]Z_2(X)[/imath], so it would be an example to prove that is not a normal extension...

Isn't 1 a root of [imath]x^3\pm 1[/imath] in [imath]\mathbb Z_2[/imath] ?

More general question: what are you trying to solve or prove?

 

[MathNugget]

Isn't 1 a root of [imath]x^3\pm 1[/imath] in [imath]\mathbb Z_2[/imath] ?

More general question: what are you trying to solve or prove?

I am trying to prove that [imath]\mathbb{Z}_2(X^3) \subset \mathbb{Z}_2(X)[/imath] is a normal algebraic extension. One definition would be "all irreducible polynomials from the first set (polynomials from that ring/field) that have at least 1 root in the bigger set (here [imath]\mathbb{Z}_2(X)[/imath] ), have all roots in that bigger set."

So I am on the hunt for a polynomial from [imath]\mathbb{Z}_2(X^3)[/imath] with some roots, but not all, in [imath]\mathbb{Z}_2(X)[/imath].

I think [imath]X^3 + 1[/imath] , as you said, has 1 root in the 2nd ring ( x = 1), but the other 2 roots are obviously not in the ring. I guess I am more interested in whether or not I completely misunderstood what a normal extension is, or the solution was this simple / obvious...

 

[blamocur]

I am trying to prove that [imath]\mathbb{Z}_2(X^3) \subset \mathbb{Z}_2(X)[/imath] is a normal algebraic extension. One definition would be "all irreducible polynomials from the first set (polynomials from that ring/field) that have at least 1 root in the bigger set (here [imath]\mathbb{Z}_2(X)[/imath] ), have all roots in that bigger set."

So I am on the hunt for a polynomial from [imath]\mathbb{Z}_2(X^3)[/imath] with some roots, but not all, in [imath]\mathbb{Z}_2(X)[/imath].

I think [imath]X^3 + 1[/imath] , as you said, has 1 root in the 2nd ring ( x = 1), but the other 2 roots are obviously not in the ring. I guess I am more interested in whether or not I completely misunderstood what a normal extension is, or the solution was this simple / obvious...

But [imath]x^3\pm 1[/imath] is not irreducible since it has a root (=1) in [imath]\mathbb Z_2[/imath].

I am not familiar with normal algebraic extensions of rings, and my online search wasn't too successful. Is this a problem from your homework or a book you study ? Can you post an exact statement, or a snapshot, of the problem?

Thank you.

 

[MathNugget]

But [imath]x^3\pm 1[/imath] is not irreducible since it has a root (=1) in [imath]\mathbb Z_2[/imath].

I am not familiar with normal algebraic extensions of rings, and my online search wasn't too successful. Is this a problem from your homework or a book you study ? Can you post an exact statement, or a snapshot, of the problem?

Thank you.

Sadly it's not from a book. The definitions I got are:
[imath]k \subset K[/imath] is an algebraic extension. We call it normal if [imath]\forall \sigma : K \rightarrow \bar K[/imath] a k - morphism [imath]\rightarrow \sigma[/imath] k-automorphism of K.
[imath]\bar K[/imath] is the algebraic closure of K.

equivalently:
i) [imath]\forall f(x) \in k(x)[/imath] irreducible (in k), if it has a root in K, it has all roots in K.
There's also a 3rd definition, I have trouble translating.

The wikipedia article seems right on it, though, with the 3 equivalent definitions:

Normal extension - Wikipedia

en.wikipedia.org

 

[MathNugget]

I'll write here the definitions from wikipedia:
"Let L/K be an algebraic extension (i.e., L is an algebraic extension of K), such that [imath]L \subseteq \bar K[/imath] (i.e., L is contained in an algebraic closure of K). Then the following conditions, any of which can be regarded as a definition of normal extension, are equivalent:
1) Every embedding of L in [imath]\bar K[/imath] over K induces an automorphism of L.
2) L is the splitting field of a family of polynomials in [imath]K[X][/imath].
3) Every irreducible polynomial of [imath]K[X][/imath] that has a root in L splits into linear factors in L. "

It's a homework, by the way.

 

[blamocur]

I'll write here the definitions from wikipedia:
"Let L/K be an algebraic extension (i.e., L is an algebraic extension of K), such that [imath]L \subseteq \bar K[/imath] (i.e., L is contained in an algebraic closure of K). Then the following conditions, any of which can be regarded as a definition of normal extension, are equivalent:
1) Every embedding of L in [imath]\bar K[/imath] over K induces an automorphism of L.
2) L is the splitting field of a family of polynomials in [imath]K[X][/imath].
3) Every irreducible polynomial of [imath]K[X][/imath] that has a root in L splits into linear factors in L. "

It's a homework, by the way.

All the definitions so far are for fields. As for rings, I've found a couple of mentions of ring extensions on Wikipedia, but none mentions normal extensions:

1. https://en.wikipedia.org/wiki/Algebra_extension
2. https://en.wikipedia.org/wiki/Subring#Ring_extensions

Do you want to post the exact copy of the homework ?

 

[MathNugget]

All the definitions so far are for fields. As for rings, I've found a couple of mentions of ring extensions on Wikipedia, but none mentions normal extensions:

1. https://en.wikipedia.org/wiki/Algebra_extension
2. https://en.wikipedia.org/wiki/Subring#Ring_extensions

Do you want to post the exact copy of the homework ?

Sure, but it's not from a book. The whole thing is:
Exercise: [imath]\mathbb{Z}_2(X^3) \subset \mathbb{Z}_2(X)[/imath] is not normal...
I doubt it's about normal subgroups, given the next exercise is
[imath]\mathbb{Q}(\frac{X^2}{X^3+1}) \subset \mathbb{Q}(X)[/imath] is not normal.

I suppose the notion for fields can be directly applied to rings. All the properties above can be checked for rings...

 

[blamocur]

I doubt it's about normal subgroups, given the next exercise is

I am really rusty on these subjects (not too surprising after not dealing with it for about half-century ): the next exercise reminded me that [imath]F(X)[/imath] is typically used, at least in my textbooks, for fields of rational polynomial functions, as opposed to [imath]F[X][/imath] used to denote polynomial rings. Sorry about a whole bunch of misleading posts

At this point I have no clue how to solve this kind of problems (sorry again), but will appreciate if you post the answers if/when you get them.

 

[MathNugget]

I am really rusty on these subjects (not too surprising after not dealing with it for about half-century ): the next exercise reminded me that [imath]F(X)[/imath] is typically used, at least in my textbooks, for fields of rational polynomial functions, as opposed to [imath]F[X][/imath] used to denote polynomial rings. Sorry about a whole bunch of misleading posts

At this point I have no clue how to solve this kind of problems (sorry again), but will appreciate if you post the answers if/when you get them.

I understand what you're saying. Teachers at this uni have a habit of changing notations often (as 1 of them says, "to further confuse the students").
I'll try to find a solution to the first question, although I think it's the one I posted above.

For the second one though, I am not even convinced
[imath]\mathbb{Q}(\frac{X^2}{X^3+1}) \subset \mathbb{Q}(X)[/imath] is true (the inclusion). But maybe this one too is as simple as taking a polynomial in first set, considering it's degree = t, and amplifying the polynomial with [imath](X^3+1)^t[/imath] ...

Also, don't worry about it. I am messing up the notations and proper words to describe most of the things I have questions about. You've helped me a lot in the past months

 

[blamocur]

Just a thought: polynomial [imath]u^3 - x^3 = 0[/imath] (whose both coefficients are in [imath]\mathbb Z_2(X^3)[/imath]) has a root [imath]u=x[/imath], so [imath]u\in \mathbb Z_2(X)[/imath] but [imath]u\notin \mathbb Z_2(X^3)[/imath]. I believe it can be shown that [imath]u^2 + ux + x^2[/imath] has no roots in [imath]\mathbb Z_2(X)[/imath] -- can you prove this? Do you agree this would prove "abnormality" of this extension?

 

[MathNugget]

Just a thought: polynomial [imath]u^3 - x^3 = 0[/imath] (whose both coefficients are in [imath]\mathbb Z_2(X^3)[/imath]) has a root [imath]u=x[/imath], so [imath]u\in \mathbb Z_2(X)[/imath] but [imath]u\notin \mathbb Z_2(X^3)[/imath]. I believe it can be shown that [imath]u^2 + ux + x^2[/imath] has no roots in [imath]\mathbb Z_2(X)[/imath] -- can you prove this? Do you agree this would prove "abnormality" of this extension?

Absolutely. Given that it's [imath]\mathbb{Z}_2[/imath], though, proving it for a general polynomial [imath]u^3 - x^3 = 0[/imath] seems rather redundant (we're just putting x = 1, x = 0, getting 2 conditions: [imath]u^2 \neq 0 \: and \: u^2 + u + 1 \neq 0[/imath] . it all ends up back to finding u = 1.

For u = 0, we see the example doesn't really work. And I do think the example for u = 1 would fit right into the 3rd definition of normal extension .

 

[blamocur]

Absolutely. Given that it's [imath]\mathbb{Z}_2[/imath], though, proving it for a general polynomial [imath]u^3 - x^3 = 0[/imath] seems rather redundant (we're just putting x = 1, x = 0, getting 2 conditions: [imath]u^2 \neq 0 \: and \: u^2 + u + 1 \neq 0[/imath] . it all ends up back to finding u = 1.

For u = 0, we see the example doesn't really work. And I do think the example for u = 1 would fit right into the 3rd definition of normal extension .

Not sure I follow this, but would be interesting to know what your professor thinks. [imath]u=1 \in \mathbb Z_2(X^3)[/imath], unlike the case of [imath]u=x[/imath].

 

[MathNugget]

Not sure I follow this, but would be interesting to know what your professor thinks. [imath]u=1 \in \mathbb Z_2(X^3)[/imath], unlike the case of [imath]u=x[/imath].

True, but as far as I know, f is reducible if there's g and h in the same ring/field, and f = gh.
Here, 1 is root, but [imath]x^3 - 1 = (x-1) (x^2 + x + 1)[/imath], and [imath]x-1, \: x^2 + x + 1 \notin \mathbb{Z}_2(X^3)[/imath].

I submitted the homework to the teacher (at least what I did from it), but I don't expect an answer (not even a grade), so it would be hard to guess whether it's correct or wrong...