A polynomial of degree $n$ over a commutative ring with zero divisors may have more than $n$ zeroes.
Attempt: Let $R$ be the commutatve ring which has a zero divisor $a \neq 0$. Then $\exists~~b \in R ,. b \neq 0$ such that $ab=0$.
We need to prove that $\exists~ f \in R[x]$ of degree $n$ such that $f$ has more than $n$ zeroes.
If we take the polynomial $ab x + ab^2 x^2 + ….. + ab^n x^n$ , this is essentially the zero polynomial as $ab=0$ and not a polynomial of degree $n$. So, I am not sure if this polynomial will work or not.
I am a bit confused. How do I move forward. Honestly, I am not sure if the problem statement is right itself.
Please note that my book has just introduced polynomial rings but none of reducability, irreducability, factorization etc.
Best Answer
Yes, it is true. Consider $x^2=1\mod 8$.