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According to Theorem 4.12 of Linear Algebra Done Right by Sheldon Axler, for every polynomial p $\in \mathcal{P}(\mathbb{F})$ with degree $m\geq0$, the polynomial p has at most m distinct zeros in $\mathbb{F}$.

He further defines $\lambda \in\mathbb{F}$ to be a zero of a polynomial p $\in \mathcal{P} (\mathbb{F})$ if p($\lambda) = 0$

However, for $p(z) = 0$ all $z \in \mathbb{F}$ are zeros of p. This would make the number of distinct zeros greater than the degree and break the $m=0$ case of Theorem 4.12?

Is my reasoning valid? I know this is a small distinction, but I want to know if I my conclusions are right or if I'm missing anything.

Aidan McNabb
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    While the non-zero constant polynomial has degree 0, the zero polynomial is a special case, and considered to have infinite degree, or undefined degree for precisely this reason. Also because when you multiply two polynomials, their degrees should add up, so again the zero polynomial is special. – Jaap Scherphuis May 13 '22 at 09:32
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    For $p=0$, one way to define the degree is $\deg (p)=-\infty$. For a discussion, see this post. – Dietrich Burde May 13 '22 at 09:39
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    I was too late to edit my comment - it should be minus infinity so that adding two polynomials gives a polynomial of degree max(d1,d2). – Jaap Scherphuis May 13 '22 at 09:39
  • You are right , the zero-polynomial does not satisfy this rule, but it is no counterexample since it has no degree being a non-negative integer. – Peter May 13 '22 at 09:41
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    See the answer by Surb at the duplicate: "Indeed, let $P$ a polynomial of degree $\geq 1$. Then, you have that $$\deg(PQ)=\deg(P)+\deg(Q),$$ for every polynomial $Q$. Now, if you define $\deg(0)=0$, you'll get $$\deg(0\cdot P)=0+\deg(P)>0,$$ which is not compatible with the degree formula. The only way to give a sense to this formula is to define $\deg(0)=-\infty $." – Dietrich Burde May 13 '22 at 09:41

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