1. ## Fields

Guys I have a couple of questions. All help is appreciated.

1. Is X^3-3 an irreducible polynomial over the real numbers R? over the rational numbers Q? Prove your assertions.

I know the cube root of 3 is a real number so its not irreducible over the reals and it is for the rationals since the cube root of 3 is not a rational number but is that the proof?

2. Q the field of rationals, is

A. Q(sqrt(2)) = {a + b(sqrt(2)) | a,b in Q } a subfield of the real numbers?

B. Q(cube root(3)) = {a + b(cube root(3)) + c(cube root (9)) | a,b,c in Q} a field?

I need some serious help for those two above and this one below.

3. V_R is the set of all differentiable real valued functions of a real variable, is {1, x, x^2,...} a linearly independent subset of V_R? is it a basis of V_R? Prove your assertions.

thanks for the help

2. Originally Posted by Luck of the Irish
1. Is X^3-3 an irreducible polynomial over the real numbers R? over the rational numbers Q? Prove your assertions.
This polynomial is reducible over the real numbers. For example, $\sqrt[3]{3}$ is a root of this polynomial in $\mathbb{R}$. However, it is irreducible over $\mathbb{Q}$ because it has no zeros in $\mathbb{Q}$. In general if $f(x)$ is a any polynomial with real coefficients that has odd degree then the polynomial is reducible over $\mathbb{R}$ unless it is linear, try to prove that!

2. Q the field of rationals, is

A. Q(sqrt(2)) = {a + b(sqrt(2)) | a,b in Q } a subfield of the real numbers?
Yes!

Remember that $\mathbb{Q}(\sqrt{2})$ is the smallest field containing $\sqrt{2}$ and $\mathbb{Q}$. Furthermore, $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ is a vector space with basis $\{1,\sqrt{2}\}$ and so this field consists of elements of the form $a+b\sqrt{2}$.

B. Q(cube root(3)) = {a + b(cube root(3)) + c(cube root (9)) | a,b,c in Q} a field?
Yes!

3. V_R is the set of all differentiable real valued functions of a real variable, is {1, x, x^2,...} a linearly independent subset of V_R? is it a basis of V_R? Prove your assertions.
To show that $\{1,x,x^2,...\}$ is linearly independent it is is required to show that every finite subset is linearly independent. Let $\{ x^{n_1},...,x^{n_k}\}$ be a finite subset where $0\leq n_1 < n_2 < ... . If is is linearly dependent then it means $a_kx^{n_k}+...+a_1x^{n_1} = 0$ for all $x\in \mathbb{R}$ and not all $a_j=0$. But if that is the case then it means that we have a polynomial of degree $n_k$ has only zero values. And so the polynomial must be a zero polynomial. This is a contradiction.

It is not a basis. Hint: consider $e^x$.

3. If I considered the taylor expansion of then {1, x, x^2,...} does not equal 1+x+x^2/2+x^3/6+..+x^n/n! therefore {1, x, x^2,...} is a linearly independent subset since it cannot be expressed as a linear combination of for any scalar lamda. Is that right?

4. Originally Posted by Luck of the Irish
If I considered the taylor expansion of then {1, x, x^2,...} does not equal 1+x+x^2/2+x^3/6+..+x^n/n! therefore {1, x, x^2,...} is a linearly independent subset since it cannot be expressed as a linear combination of for any scalar lamda. Is that right?
If $e^x = p(x)$ where $p$ is a polynomial then taking $n+1$ derivatives of both sides
where $n$ is the degree of the polynomial we get that $e^x = 0$ which is a contratiction.

Therefore, $e^x$ cannot be expressed in a linear combination of elements of that set.