If I work over an algebraically closed field $K$, and $f(x)\in K[x]$, then $f(x)$ always factorises into a product of linear polynomials, being the roots, is that correct?

I.e. $f(x)=(x-a_1)(x-a_2)\cdots(x-a_n)$?

Is that the right way to phrase it aswell? Linear polynomials, that are the roots?

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    $\begingroup$ When a polynomial factors completely into a product of linear polynomial over a field, we often say it "splits" over that field. But don't confuse the linear polynomials $x-a_i$ and the roots $a_i$. They have different names. $\endgroup$ – hardmath Oct 5 '15 at 11:53

It's true up to a constant that over an algebraically closed field all polynomials factor in the way you describe. The correct statement is that every polynomial factors as

$$f(x) = b(x-a_1)\dots(x-a_n)$$ for $a_i,b \in K$. The point is you can't guarantee the leading coefficient is $1$ (consider the polynomial $2x-1$ for instance). Note also, that we could have $a_i = a_j$ for some $i \neq j$.

The best way to phrase this is probably to say that every polynomial factors as a product of linear polynomials or splits into linear factors.

The roots refer to the elements $a_i \in K$ such that $f(a_i)=0$. While it's true that these (counted with multiplicities) determine the factorization of $f$ up to a constant, these do not refer to the linear factors themselves.

  • $\begingroup$ Thanks I really appreciate this answer. Clears up a few of my worries $\endgroup$ – Understand Oct 5 '15 at 12:00

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