# Zeroes of irreducible polynomial over splitting field

I'm reading one theorem in the book called Contemporary Abstract Algebra by Gallian, and here's what puzzles me somewhat:

If $f(x) = (x-a)^mg(x) = (x-b)^m\phi(g(x))$ then how does the author deduce that the multiplicity of $b$ is at least $m$? Why not less than $m$ or equal to $m$?

Thank you for clarifying this for me.

• What's the definition of multiplicity? – Swapnil Tripathi Mar 14 '17 at 20:57
• That's how many times a zero of a polynomial appears in the multiset of zeros of the polynomial. – sequence Mar 14 '17 at 21:01
• Yes. So $b$ appears $m$ times as a root already. If $b$ doesn't appear in $\phi(g(x))$ then multiplicity will be exactly $m$, if it appears $n$ times in $\phi(g(x))$, then the multiplicity of $b$ in $f(x)$ would be $m+n$ – Swapnil Tripathi Mar 14 '17 at 21:03
• @sequence In other words, the multiplicity of $b$ is the largest positive integer $k$ such that there exists a polynomial $h(x)$ with the property that $f(x)=(x-b)^kh(x)$ and $x-b$ does not divide $h(x)$. Does this way of formulating the concept of multiplicity help you? – Git Gud Mar 14 '17 at 21:07

Here, $$b$$ appears $$m$$ times as a root of $$f(x)$$ already. If $$b$$ appears as a root of $$\phi(g(x))$$ with multiplicity $$n$$, then the multiplicity of $$b$$ in $$f(x)$$ would be $$m + n$$. Since $$n \geq 0$$, we have $$m + n \geq m$$.
It is useful to view the multiplicity of a root $$b$$ of a polynomial $$f(x)$$ to be the largest non-negative integer $$k$$ such that we can write $$f(x) = (x-b)^k h(x)$$, where $$h(x)$$ is a polynomial such that $$x-b$$ does not divide $$h(x)$$.