Decide if the polynomial $x^2-7$ is irreducible over $\mathbb Q, \mathbb R, \mathbb C$.

Question

I am trying to understand irreducibility of polynomials and I am trying to solve the following question:

Decide if the polynomial $x^2-7$ is irreducible over $\mathbb Q, \mathbb R, \mathbb C$.

For Q

I tried to use the Eisenstein’s Criterion, 7/7, but 7^2/7 is not, I believe it's irreducible over Q

For R is reducible (x-sqrt{7})(x+sqrt{7})

For C

Not sure

Thanks in advance for your time.

Answer 1

The given polynomial is $$x^2 - 7.$$

Since it is a two degree polynomial having no root in $\mathbb{Q}$, it is irreducible over $\mathbb{Q}$.

But, $$x^2-7=(x+\sqrt7)(x-\sqrt7)$$

So, it is reducible over $\mathbb{R}$.

Now, $\mathbb{R}$ is subset of $\mathbb{C}$ so $\sqrt7$, $-\sqrt7$ are also elements in $\mathbb{C}$ $$\sqrt7=\sqrt7 + 0i$$ Hence, any polynomial which is reducible over $\mathbb{R}$ must be reducible over $\mathbb{C}$, too.

Answer 2

Especially for $\mathbb{C}$ you should read about the fundamental theorem of Algebra which says that a polynomial of degree k has excactly k roots contained in $\mathbb{C}$. Thats why we say that the set of complex numbers is algebraic closed.

Answer 3

The unique factorization is $x^2-7 = (x-\sqrt 7)(x+\sqrt 7)$. By this uniqueness, $x^2-7$ is irreducible over $\Bbb Q$, but factors completely over $\Bbb R,\Bbb C$.