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Context: Here is the problem of my professor give to us:

We all know that a polynomial of degree $n$ in $ \mathbb{C} $ has at most n roots, but is there any ring $A$ such that a polynomial of degree $n$ in $A[x]$ has more than $n$ roots ?

I've tried to find some polynomial on no-integral domain like $Z/nZ$, it's work with some function of specific degree like 2 and 3.

How can i construct a function with an arbitrary degree in $Z/nZ$ that still satisfies the condition ?

meap meap
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    Take for example $A=\mathbb Z/12\mathbb Z$, can you find all the roots of $P(X)=X^2-4$? – Kolakoski54 Oct 11 '24 at 06:47
  • How about a trivial polynomial of degree $n=0$, i.e. with a constant term only, and the term being $0$...? ;) – CiaPan Oct 11 '24 at 06:57
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    @CiaPan The degree of the zero polynomial is problematic – ploosu2 Oct 11 '24 at 07:16
  • @ploosu2 Actually, the wink emoticon shows that my comment was a bit of joke. With problems posed like this one the aim is usually (I suppose) to find some method of finding nontrivial solutions (polynomials in this case) for some reasonably useful ranges of parameters (in this case the degree $n$). Degenerate cases like mine are interesting, but unimportant, because they are either easily excluded by more precise problem statement or can be treated separately as special cases of solutions. – CiaPan Oct 11 '24 at 07:33

2 Answers2

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The fact is that the "factor theorem" doesn't go through for non-integral domains. Use this to provide counterexamples.

For instance, $\Bbb Z_8,$ with $x^2-1.$ The roots are $\pm1,\pm3.$

suckling pig
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  • thanks a lot, i've tried with $f(x) = x(x-1)(x-2)(x-3)$ in $Z_12$ and it still worked, but it feel a bit weird, i couldn't any propositon in my professor text book that involve to this problem, is there any detail explaination for this result ? – meap meap Oct 11 '24 at 09:24
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    Look up the factor theorem. It's sort of the converse of the fundamental theorem of algebra. See if it applied we'd have that the polynomial would equal $(x-1)(x+1)(x-3)(x+3),$ of degree $4.$ btw the Wikipedia page says that this is true in any commutative ring, but it isn't. – suckling pig Oct 11 '24 at 10:15
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When $A$ is a domain the number of roots of a polynomial $F(X)\in A[X]$ of degree $d$ is bounded by $d$.

But when $A$ is not a domain anything can happen. For instance every element in $$ A=\frac{\mathbb{Z}}{2\mathbb{Z}}\times\cdots\times \frac{\mathbb{Z}}{2\mathbb{Z}}\qquad \text{$r$ factors} $$ is a root of $X^2-X$ ($2^r$ roots).

Andrea Mori
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