Proving Irreducibility of x^2-3 over Q(sqrt[3]{2}) | Math Proof Solution

• ehrenfest
In summary, if x^2-3 is reducible over \mathbb{Q}(\sqrt[3]{2}), then \mathbb{Q}(\sqrt[3]{2}) contains the roots \pm \alpha of x^2 - 3. This in turn means that \mathbb{Q}(\alpha) \subset \mathbb{Q}(\sqrt[3]{2}).
ehrenfest
[SOLVED] extension fields

Homework Statement

Prove that x^2-3 is irreducible over $$\mathbb{Q}(\sqrt{2})$$.

EDIT: it should be $$\mathbb{Q}(\sqrt[3]{2})$$

The Attempt at a Solution

So, we want to show that $\sqrt{3}$ and $-\sqrt{3}$ are not expressible as $a +b(\sqrt[3]{2})^2 +c\sqrt[3]{2}$. We could take the sixth power of both sides, but then the equation would be just messy. What else can we do?

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Is it $\mathbb{Q}(\sqrt{2})$ or $\mathbb{Q}(\sqrt[3]{2})$?

Sorry. See the EDIT.

If x^2 - 3 is reducible over $\mathbb{Q}(\sqrt[3]{2})$, then $\mathbb{Q}(\sqrt[3]{2})$ contains the roots $\pm \alpha$ of x^2 - 3. This in turn means that $\mathbb{Q}(\alpha) \subset \mathbb{Q}(\sqrt[3]{2})$. Try to think about why this cannot happen.

morphism hints at what is probably the easiest approach.

Another one is the analog of what you would do to show it's irreducible over the rationals -- if x²-3 is reducible, it has a root. It's monic, so the root must be an algebraic integer... in fact, it must be an algebraic integer that divides 2... how many divisors does 2 have in $\mathbb{Q}(\sqrt[3]{2})$?

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morphism said:
If x^2 - 3 is reducible over $\mathbb{Q}(\sqrt[3]{2})$, then $\mathbb{Q}(\sqrt[3]{2})$ contains the roots $\pm \alpha$ of x^2 - 3. This in turn means that $\mathbb{Q}(\alpha) \subset \mathbb{Q}(\sqrt[3]{2})$. Try to think about why this cannot happen.

Why don't you just write $\pm \sqrt{3}$ instead of alpha?

$[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$ and $[\mathbb{Q}(\sqrt[2]{3}):\mathbb{Q}] = 2$ and if $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{Q}(\sqrt[3]{2})$ it must be the case that $\mathbb{Q}(\sqrt[3]{2})$ is a finite extension of $\mathbb{Q}(\sqrt[2]{3})$.

and then we apply the theorem that says

[K:F] = [K:E][E:F]

which in our case is

3 = n*2, where n must be an integer, which is absurd.

Can you help me prove the statement that I underlined?

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I used alpha because it was faster to type. :p

If E is an intermediate field of the finite extension K/F, then [K:E]=[K:F]/[E:F], which is certainly finite because both [K:F] and [E:F] are. (The first is finite by assumption, and the second is finite because a subspace of a finite dimensional space is finite dimensional.)

morphism said:
If E is an intermediate field of the finite extension K/F, then [K:E]=[K:F]/[E:F]

I have not seen a proof of that. My book proves a theorem that says: If E is a finite extension field of a field F, and K is a finite extension of E, then K is a finite extension of F, and [K:F]=[K:E][E:F], but what you said is different.

ehrenfest said:
I have not seen a proof of that. My book proves a theorem that says: If E is a finite extension field of a field F, and K is a finite extension of E, then K is a finite extension of F, and [K:F]=[K:E][E:F], but what you said is different.
Have you thought at all about how this statement might imply what morphism said?

Hurkyl said:
Have you thought at all about how this statement might imply what morphism said?

So, the only thing that you need to prove is that this intermediate field E is a finite extension of F and finitely extends to K. Let's do the former first.

We want to show that E is finite dimensional as a vector space over F. We know that E over F must be a subspace of K over F, which is finite dimensional. So, we need to find a finite set of vectors in E over F that span E over F. Now, we already have a finite number of vectors (call them B) in K over F that span K over F, which includes E over F. If we take the subset of B that is contained in E over F, is it true that those vectors will span E over F? It is definitely not true (consider the standard bases in R^2 and the line x=y; in fact, neither of the two basis vector are contained in that line). What we need to do is this. Let the basis B have size n. Find an arbitrary vector in E over F. Find another vector that is linearly independent with the first. Keep doing this until you get n vectors. If you cannot get n vectors, then you must have a set of less than n vectors that spans E over F. If you get n vectors, then use the fact any set of linearly independent vectors in K over F must be less than or equal to the dimension of the space, which is n. Thus the n vectors we have must span all K over F, which by assumption contains E over F.

Now the latter. Get a maximal set of linearly independent vectors of E over F. We just proved that this set must be finite. Then we can enlarge this set to a basis of K over F just be adding linearly independent vectors until there are none left. We even know how many we need to add because the dimension of K over F is n.

Please confirm this proof.

What is the definition of irreducibility in mathematics?

In mathematics, irreducibility refers to the property of a polynomial or mathematical expression that cannot be factored into smaller, non-constant polynomials with coefficients from the same field or ring.

What does it mean to prove irreducibility of a polynomial?

To prove irreducibility of a polynomial means to show that it cannot be factored into smaller, non-constant polynomials over a specific field or ring. This is often done by using various techniques such as the rational root theorem, Eisenstein's criterion, or the degree and coefficients of the polynomial.

What is the specific polynomial being discussed in "Proving Irreducibility of x^2-3 over Q(sqrt[3]{2}) | Math Proof Solution"?

The polynomial being discussed is x^2-3, which is a quadratic polynomial with a leading coefficient of 1 and a constant term of -3. It is being evaluated over the field Q(sqrt[3]{2}), which is the set of numbers of the form a+b(sqrt[3]{2}), where a and b are rational numbers.

Why is proving irreducibility of x^2-3 over Q(sqrt[3]{2}) significant?

Proving irreducibility of x^2-3 over Q(sqrt[3]{2}) is significant because it shows that this polynomial cannot be factored into smaller, non-constant polynomials with coefficients from the field Q(sqrt[3]{2}). This also means that x^2-3 is a prime element in the ring Q(sqrt[3]{2}), which has implications in various areas of mathematics such as number theory and abstract algebra.

What techniques can be used to prove irreducibility of a polynomial?

Some techniques that can be used to prove irreducibility of a polynomial include the rational root theorem, Eisenstein's criterion, and the degree and coefficients of the polynomial. Other techniques such as the fundamental theorem of algebra and the use of field extensions can also be used in more advanced cases.

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