1. ## Zero Function

I have had several homeworks lately that refer to the zero function. This time we are asked to prove the set of all p(x) which determine the zero function is an ideal of F[x]. Where F is a field. I dont understand what it means by the zero function. If anyone can help me that would be great thanks

2. ## Re: Zero Function

Originally Posted by spotsymaj
I have had several homeworks lately that refer to the zero function. This time we are asked to prove the set of all p(x) which determine the zero function is an ideal of F[x]. Where F is a field. I dont understand what it means by the zero function. If anyone can help me that would be great thanks
In the case of $F[x]$, the zero function is the function $f(x)=0+0x+0x^2+0x^3+...$

EDIT: I put the solution here in front so people don't have to wade through posts upon posts of confusion:
I just took a look at your book, and it seemed we can just substitute elements of $F$ whenever we want. In this case, our job is quite easy actually. To see that

$\mathcal{S}=\{p(x)\in F[x]|p(a)=0 \quad \forall a\in F\}$

is an ideal, we want to show that it is closed under addition by elements in $\mathcal{S}$ and closed under multiplication by elements of $F[x]$. For the first bit, suppose $p(x),q(x)\in \mathcal{S}$. Then $(p+q)(a)=p(a)+q(a)=0+0=0$ for all $a\in F$. Thus $(p+q)(x)\in \mathcal{S}$. For the second assertion, suppose $p(x)\in \mathcal{S},r(x)\in F[x]$. Then $r(a)p(a)=r(a)\cdot 0 = 0$ for all $a\in F$. Therefore $r(x)p(x)\in \mathcal{S}$.

3. ## Re: Zero Function

So is this a true statement a(c) -d(x) = 0 where a, d are polynomials

4. ## Re: Zero Function

Originally Posted by spotsymaj
So is this a true statement a(c) -d(x) = 0 where a, d are polynomials
I'm not sure what you're asking here. Not in general. Suppose your field is $\mathbb{Q}$. Take $a(x)=x$ and $d(x)=1$. $a(x)-d(x)=x-1$, which is not the zero function.

5. ## Re: Zero Function

Oh. Well then I have no clue where to go to prove that problem. Any hints would be great

6. ## Re: Zero Function

Originally Posted by spotsymaj
Oh. Well then I have no clue where to go to prove that problem. Any hints would be great
In your original question, are the p(x) elements of F[x]?

Yes!

8. ## Re: Zero Function

In that case, the zero function I gave earlier is not supposed to live in $F[x]$. The way I interpret this question is to consider the substitution homomorphism $F[x]\to F$, $p(x)\mapsto p(a)$ for all $a\in F$. The $p(x)$ you want should satisfy $p(a)=0$ for all $a\in F$. For example, in $\mathbb{F}_2[x]$, the polynomial $x^2+x$ evaluates to zero for all elements of $\mathbb{F}_2$ since $0^2+0=0$ and $1^2+1=2=0$

More generally, you want to show that given a field homomorphism $\varphi:F\to F'$, the kernel of the substitution homomorphism $F[x]\to F'$ is an ideal in $F[x]$.

Does this sound vaguely familiar to what you're doing?

9. ## Re: Zero Function

Some not all. So the problem states the set of all p(x) which determine the zero function these polynomials would then have to be always of the form x^m + x^m-1 + .... + x^2 + x^1 + x where m is in F?

10. ## Re: Zero Function

or does m have to deal with deg or modulus??

11. ## Re: Zero Function

Originally Posted by spotsymaj
Some not all. So the problem states the set of all p(x) which determine the zero function these polynomials would then have to be always of the form x^m + x^m-1 + .... + x^2 + x^1 + x where m is in F?
m can be any natural number (and zero). Your restriction to F is on the coefficients of p(x).

Edit: Also, please take note that this reformulation:
More generally, you want to show that given a field homomorphism $\varphi:F\to F'$, the kernel of the substitution homomorphism $F[x]\to F'$ is an ideal in $F[x]$.
makes this question trivial, as the kernel of a ring homomorphism is an ideal.

12. ## Re: Zero Function

so since it is a F[x] is a finite field my generator for the set of all p(x) would be a_mx^m + a_(m-1)x^m-1 + .... + a_2x^2 + a_1x^1 + a_0x ???

13. ## Re: Zero Function

Originally Posted by spotsymaj
so since it is a F[x] is a finite field my generator for the set of all p(x) would be a_mx^m + a_(m-1)x^m-1 + .... + a_2x^2 + a_1x^1 + a_0x ???
Ok, lets start over. Every element of $F[x]$ is of the form $a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ where the $a_i$ are elements of $F$. $n$ can be any natural number (as long as you don't get negative exponents of x). Is that clear?

14. ## Re: Zero Function

Yes! Sorry trying to learn this on my own and it is not working so well

15. ## Re: Zero Function

Originally Posted by spotsymaj
Yes! Sorry trying to learn this on my own and it is not working so well
No problem. Now lets keep going. The substitution principle is a theorem on rings. It says that given a ring homomorphism $R\to R'$ and given an element $b\in R'$, there is a unique homomorphism $\Phi:R[x]\to R'$ which agrees with the map $\varphi$ on constant polynomials $f(x)=a,a\in R$ and sends $x\mapsto b$. Intuitively, this means you're substituting the variable $x$ for the element $b$. For example, $f(x)=3x^2+5x \mapsto f(1)=3(1^2)+5(1)=8$. In grade school terms, we're subbing $b$ for $x$.

Is this one something you know?

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