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I am a bit confused on the proof that Gallian delivers of the theorem mentioned in the title. Although I have seen other proofs of this fact and have been able to parse them, his specifically trips me up as I cannot understand the purpose of a lot of it. Word for word:

A polynomial of degree $n$ over a field has at most $n$ zeros, counting multiplicity.

Proof: We proceed by induction on $n$. Clearly, a polynomial of degree $0$ over a field has no zeros. Now suppose that $f(x)$ is a polynomial of degree $n$ over a field and $a$ is a zero of $f(x)$ of multiplicity $k$. Then, $f(x) = (x-a)^kq(x)$ and $q(a) \ne 0$; and, since $n = \text{deg} f(x) = \text{deg} (x-a)^kq(x) = k + \text{deg} q(x)$, we have $k \le n$ (see Exercise 19) If $f(x)$ has no zeros other than $a$, we are done. On the other hand, if $b \ne a$ and $b$ is a zero of $f(x)$ then $0 = f(b) = (b-a)^k q(b)$ so that $b$ is also a zero of $q(x)$ with the same multiplicity as it has for $f(x)$ (see Exercise 21). By the Second Principle of Mathematical Induction, we know that $q(x)$ has at most deg$q(x) = n-k$ zeros, counting multiplicity. Thus $f(x)$ has at most $k + n - k = n$ zeros, counting multiplicity.

I'm wondering why couldn't of it been possible to say something along the lines of:

Proof Assuming that the theorem is true for a polynomial of degree $n$, we show that it is true for a polynomial $f_{n+1}$ of degree $n+1$. The base case of a degree $0$ polynomial is trivial.

Let $f_{n+1}$ have a zero $a$ of multiplicity $k \ge 1$. Then we know that

$$f_{n+1} = (x-a)^k q(x)$$

Since there are no zero divisors in a field, we know that the 'degree rule' holds: that is $\text{deg} (f(x)g(x)) = \text{deg} f(x) + \text{deg} g(x)$. And so we know that $\text{deg} q(x) = n+1-k$. Since $q(x)$ has degree $\le n$, we can utilize our induction hypothesis and come to the conclusion that $q(x)$ has at most $n + 1 - k$ zeros. And so, $f_{n+1}$ has at most $n + 1 - k + (k) = n + 1$ zeros, and the theorem is proven. $\blacksquare$

This doesn't utilize any information that wasn't used in Gallian's proof - proving the "Degree Rule" was what "Exercise 19" was. I both don't really understand or see why the whole second part of Gallian's proof introducing $b$ is necessary. On top of that, it's kind of hard to understand.

I was hoping if someone could help me in both explaining why my proof is incorrect (if it is) and how the second half of Gallian's proof operates.

Thank you!

adam dhalla
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  • @PaulFrost, you are right!!! Thanks for reminding that. I will delete the comment – dmtri Mar 17 '23 at 13:54
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    The proof can be viewed more conceptually as $,(x!-!a_1)^{e_1},\ldots, (x-a_j)^{e_j}!\mid f\Rightarrow \prod_i (x!-!a_i)^{e_i}\mid f,$ because LCM = product for pair-coprimes, where the inductive step amounts to Euclid's Lemma (the key point your argument omits). If this interests you then let me know and I can explain further in an answer. – Bill Dubuque Mar 17 '23 at 20:45
  • @BillDubuque Bill, that would actually be wonderful - I'm still having a bit of trouble understanding the connections between the theorem and Euclid's Lemma. – adam dhalla Mar 18 '23 at 07:25

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I would have used the same proof as you suggested, but I realized that it has a little gap: We have to show that all zeros of $f(x)$ are zeros of one of its factors. This seems to be trivial; nevertheless it deserves a proof. The sentence

On the other hand, if $b \ne a$ and $b$ is a zero of $f(x)$ then $0 = f(b) = (b-a)^k q(b)$ so that $b$ is also a zero of $q(x)$ with the same multiplicity as it has for $f(x)$

gives a proof of this fact.

Paul Frost
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    The proof is a special case of the standard proof that LCM = product for coprimes, and that sentence (inductive step) is merely an application of Euclid's Lemma (cf. my comment on the answer). – Bill Dubuque Mar 17 '23 at 20:51