[Abstract Algebra] Field and Polynomial Root problem

[RJLiberator]

Homework Statement

Suppose a field F has n elements and $F=(a_1,a_2,...,a_n)$. Show that the polynomial $w(x)=(x-a_1)(x-a_2)...(x-a_n)+1_F$ has no roots in F, where $1_f$ denotes the multiplicative identity in F.

Homework Equations

The Attempt at a Solution

Strategy: We have this polynomial:
$w(x)=(x-a_1)(x-a_2)...(x-a_n)+1_F$
When we set it to 0 we see that the (x-a) terms need to equal -1 for w(x) to have a root.

But here's my problem.
I was thinking of trying to use induction on this proof.
But if n = 0, then we have x+1 = 0 and x = i^2 which is in a field, the complex field.
That has a root.

But ok, let's say n = 1. Then we have x-a+1=0 in which case x = 3, a=4 would be a root.

Must I assume that there is more then n=1 elements?

 

[fresh_42]

The point is that you already know all possible values for $x$. What happens if you try them all one by one?

 

[RJLiberator]

I'm not quite sure what you mean by this. We know all possible values of x? Sure, but couldn't that be from -infinity to +infinity? How does that help?
There must be a deeper understanding to what you are alluding at, but my initial thoughts are that it is confusing. We know all possible values for x?

 

[fresh_42]

I'm not quite sure what you mean by this. We know all possible values of x? Sure, but couldn't that be from -infinity to +infinity? How does that help?
There must be a deeper understanding to what you are alluding at, but my initial thoughts are that it is confusing. We know all possible values for x?

You are looking for a root of $w(x)$, i.e. an element $a ∈ \mathbf{F}$ such that $w(a) = 0$.
Why don't you try all possible $a$ and see what $w(a)$ will get you? There are only finitely many of them, nothing with infinity.

 

[RJLiberator]

Okay, I am starting to see where you are going with this.

w(x) = (x-a)+1 = 0

w(x)= (x-a1)(x-a2)+1 = 0
$x^2-x*a_2-x*a_1+a_1*a_2+1= 0$

Couldn't we have a1 = -1 and a2 = 1 and x = 0 ?

 

[fresh_42]

Okay, I am starting to see where you are going with this.

w(x) = (x-a)+1 = 0

w(x)= (x-a1)(x-a2)+1 = 0
$x^2-x*a_2-x*a_1+a_1*a_2+1= 0$

Couldn't we have a1 = -1 and a2 = 1 and x = 0 ?

I think you don't see the forest for all the trees. (Don't know whether this could be said in English, too.)
Again: What is $w(a)$? (Forget the $x$ for a second.)

 

[RJLiberator]

w(a) = (a-a)+1 = 1.
So clearly that cannot be a root. So if the product of the (x-a) terms are equal to 0, the +1 disallows it from being a root.

This also works for more terms then just one. As (a-a) = 0, we see (a-a)(a-a1)(a-a2)...(a-an)+1 = 1.

 

[fresh_42]

w(a) = (a-a)+1 = 1.
So clearly that cannot be a root. So if the product of the (x-a) terms are equal to 0, the +1 disallows it from being a root.

This also works for more terms then just one. As (a-a) = 0, we see (a-a)(a-a1)(a-a2)...(a-an)+1 = 1.

Yes. $w(a) = (a-a_1) ... (a-a_n) +1_F$ but $a \in \mathbf{F} = \{a_1, ... , a_n\}$. Therefore $w(a) = 1_F$ and $1_F \neq 0_F$ because $\mathbf{F}$ is a field.

 

[RJLiberator]

Ugh, I can't believe I didn't see this obvious result from the start.

I'm so out of it lately.

That is completely obvious :p.

I think you don't see the forest for all the trees. (Don't know whether this could be said in English, too.)

Indeed.