4

Are the roots of polynomial $\lambda^2-\lambda=0$ always the addition identity $0$ and multiplication identity $1$ for any field $F$? If not, anyone can help give a counterexample?

Tuyet
  • 847

2 Answers2

5

Yes.

In a field, you have the distributative laws: $$a(b+c) = ab+ac$$ So, we can apply that to your polynomial: $$\lambda^2-\lambda = \lambda(\lambda-1)$$ Now, for $\lambda(\lambda-1) = 0$, we have one of the two productands must be zero (fields don't have zero divisors, so there can't be $a,b\neq 0$ such that $ab = 0$).

Now, this means that either $\lambda = 0$ or $\lambda-1 = 0$. In the first case we have that $\lambda = 0$, the additive identity. In the second case, we can add one ot both sides to get that $\lambda -1+1 = 1$, then use that inverses cancel to get that $\lambda = 1$.

Mark Schultz-Wu
  • 6,986
4

$\lambda^2-\lambda=0$ is the same as $\lambda(\lambda-1)=0$ where $1$ is the multiplicative identity. Fields have no zero-divisors, so $\lambda =0$ or $\lambda-1=0$. So yes, the only solutions are $\lambda=0$ or $\lambda=1$.

kccu
  • 20,808
  • 1
  • 22
  • 41
  • This is because a field demands that multiplication must distribute over addition? – mathreadler Mar 30 '17 at 02:47
  • That is one property that was used (to say $\lambda^2-\lambda=\lambda(\lambda-1)$. The really important one is that fields have no zero-divisors besides zero itself. – kccu Mar 30 '17 at 03:44
  • For instance, in the ring $\mathbb{Z}_6$, $3^2-3=9-3=3-3=0$, yet $3$ is neither $0$ nor $1$. We see that $3^2-3=3(3-1)=3\cdot 2$, which is $0$ in $\mathbb{Z}_6$ even though neither $2$ nor $3$ is $0$. So the result fails because $\mathbb{Z}_6$ has zero divisors, namely $2$ and $3$. – kccu Mar 30 '17 at 03:48