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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.

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3 Answers 3

up vote 3 down vote accepted

Try $3x$ in the ring of integers modulo 6, $\Bbb Z/6\Bbb Z$. The zeroes are 0 and 2.

More generally, if $0\ne a|0$, say so that $ab=0$ with $b\ne 0$ as well, then the polynomial $f(x)=ax$ has at least $2$ zeroes: $0,b\in R$.

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Thanks . I was trying to prove this for an $nth$ degree equation. How can we construct an equation over $R$ so that it has more than $n$ zeroes? –  VHP Jul 20 at 6:12
1  
Oh I got it. Thanks :) –  VHP Jul 20 at 6:14

Yes, it is true. Consider $x^2=1\mod 8$.

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Thank you for your answer :) –  VHP Jul 20 at 6:14
  1. Let $K$ be an algebraically closed field and $P$ be a polynomial of degree $n$ with coefficients in $K$. This polynomial has $n$ roots (with multiplicity) $(a_i)_i$ in $K$. Now $K$ can be considered as a subring of $K\times K$ and then $P\in (K\times K)[x]$ admits $n^2$ roots (with multiplicity) $((a_i,a_j))_{i,j}$ in the ring $K\times K$.

  2. The equation $x^2=1$ is fundamental in cryptography. Let $p,q$ be great distinct randomly chosen PRIME integers (for instance of length $1000$ bits). Then the equation $x^2=1$ admits exactly $4$ roots in $\mathbb{Z}_p\times \mathbb{Z}_q$: $(1,-1),(1,1),(-1,-1),(-1,1)$. Yet, if we know $pq$ and not $p,q$, then nobody in the world can explicitly find these $4$ solutions in $\mathbb{Z}_{pq}$. Otherwise the cryptographic system RSA would collapse.

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