# How to get the splitting field of a polynomial?

I'm sorry if I sound too ignorant. I don't have a high level of knowledge in math.

While reading an article about Galois theory, I've become confused over the concept of a splitting field. The author defines the splitting Field of the polynomial $p(x)$ as the smallest field extension of $Q$ that contains all the roots of $p(x)$. He/She then gives the examples of $p(x)=x^2−2$, which splitting Field is $\mathbb{Q}[\sqrt2]$ since "it contains all the roots of $p(x)$ and if it had fewer elements it either wouldn't contain all the roots or wouldn't be a field" and that of $p(x)=x^4−5x^2+6$ which splitting field is $\mathbb{Q}[\sqrt 2,\sqrt 3]$, then the author decides to finish the paragraph by asking the question "Can you see why?".

And that is the issue, I cannot see why :/

I struggled for a while trying to think of a general method or explanation to why those field extensions correspond to those polynomials, but I couldn't come up with anything.

Any thoughts/ideas would be really appreciated!

• Hint : $x^4-5x^2+6=(x^2-2)(x^2-3)$ , hence $\sqrt{2}$ and $\sqrt{3}$ are the positive roots. – Peter Jul 11 '17 at 22:40
• The phrase "splitting field" of a polynomial refers to the factoring of the polynomial into first-degree polynomials, so "splitting" the polynomial in that sense. – hardmath Jul 11 '17 at 22:42
• Yup. But how can one prove that there exist no other field extension of Q smaller than Q[√2,√3] that contains the roots of $x^4−5x^2+6$? – Leo Jul 11 '17 at 22:43
• There are other fields that contain those roots, but the splitting field is the smallest one in a suitable sense. If that is the point you are confused by, I'll try to find a previous Question that hits the point squarely or (in the alternative) post a new Answer here. – hardmath Jul 11 '17 at 22:45
• @Leo $Q[\sqrt 2]$ is by definition the subfield generated by $Q$ and $\sqrt 2$. Prove that $\{a+b \sqrt 2\}$ is a subfield containing $Q$ and $\sqrt 2$ and contained in every subfield containing $Q$ and $\sqrt 2$. – Cauchy Jul 11 '17 at 23:11