Splitting Fields

Smallest field holding all roots of f.

The idea

Some polynomials refuse to factor where they live. Over $\mathbb{Q}$ the polynomial $x^{2} - 5$ has no roots; over $\mathbb{R}$ neither does $x^{2} + 1$. The response is not to abandon the polynomial but to enlarge the field until its roots exist: adjoin a symbol for a root, impose the equation that root has to satisfy, and check that what results is again a field.

Once enlarging is allowed you can overshoot. Every rational polynomial has all its roots in $\mathbb{C}$, so $\mathbb{C}$ splits everything and therefore says nothing about any particular $f$. What carries information is the smallest field that does the job. Starting from $F$, adjoin roots of $f$ one at a time until $f$ factors as $c(x - \alpha_{1}) \cdots (x - \alpha_{n})$; the field you land in is the splitting field of $f$ over $F$. Both halves of that matter: every root has to be present, and nothing may be adjoined beyond what the roots generate.

Two facts make it a genuine invariant of $f$. It exists for every polynomial and is unique up to isomorphism, so speaking of the splitting field is legitimate. And it is finite-dimensional as a vector space over $F$ — that dimension is called the degree of the extension, and it is the number the rest of the subject computes with.

Ways to work on it

• Walkthrough. Build the smallest field extension where a polynomial factors completely.
• Practice. Degree of the splitting field of x^2 - D.
• Hardest. Compute the degree of a cubic's splitting field.

Not sure where to start? Take the ten-question placement test.