2

This is a question on clearing one of my basic concepts rather than a problem. We know that if $F$ is a field then $F[x]$ is a Euclidean domain and hence an Integral domain. Now in an integral domain say $F[x]$ if $f,g \in F[x]$ if $fg=0$ then either $f=0$ or $g=0$. 1) Now does this $f=0$ imply that $f$ is just the element zero $\in F$ or $f(x)=0$ for all $x \in F$ or something else that is going over my head right now.

This is a basic conception problem that irritates me for some time so that I thought to ask you all. Because 2) in $\Bbb Z_2$ we have $x^2-x$ is zero polynomial but $x^2-x$ isn't the number zero $\in \Bbb Z_2$ but $x^2-x=0$ for all $x \in \Bbb Z_2$. Now $\Bbb Z_2[x]$ is an integral domain so $x^2-x=0 \Rightarrow x=0 $ or $x-1=0$ which is definitely not true $x \in \Bbb Z_2$. I think I am seriously lacking a fundamental concept.

Let me know why this question bugs me a lot. I was trying to prove this to prove this by my own. Then I realized that we have to use Hilbert Nullstelensatz which is standard over an algebraically closed field and hence this is infinite.

But 3) what will happen with finite fields? Why is the infiniteness so important? Probably because of the example I gave in 2). But even in finite field $fg=0$ should imply $f=0$ or $g=0$ then this should be true over finite field that is why I lost my way in a circular arguement.

I don't know whether I am making any sense or not! Probably you have got that I am utterly confused. I am sorry if I am saying something miserable. Please help

Edit:

I read those links still I am confused. Can anyone please answer this:

$x^2-x=x(x-1)=0=f(x)g(x)$ where $f(x)=x$ and $g(x)=x-1$ As $\Bbb Z_2[x]$ is an integral domain we have $f(x)=x=0$ or $g(x)=x-1=0$ what does this mean $f(x)$ is the zero polynomial in $\Bbb Z_2[x]$ or $g(x)$ is the zero polynomial in $\Bbb Z_2[x]$ or what! By the way you can say that $f(x)$ and $g(x)$ both of them are not zero in $\Bbb Z_2[x]$ then where is the fallacy!! you can mark my question as duplicate,downvote but please help me here

Ri-Li
  • 9,038
  • I read it but still could not understand that $x^2-x=x(x-1)=0=f(x)g(x)$ where $f(x)=x$ and $g(x)=x-1$ As $\Bbb Z_2[x]$ is an integral domain we have $f(x)=x=0$ or $g(x)=x-1=0$ what does this mean $f(x)$ is the zero polynomial in $\Bbb Z_2[x]$ or $g(x)$ is the zero polynomial in $\Bbb Z_2[x]$ or what! By the way you can say that $f(x)$ and $g(x)$ both of them are not zero in $\Bbb Z_2[x]$ then where is the fallacy!! – Ri-Li Sep 02 '19 at 23:31
  • In $(\mathbb{Z}/2\mathbb{Z})[x]$, $x^2-x$ is not the zero polynomial. If $K$ is an infinite field, then $0$ is the only polynomial $p(x)\in K[x]$ with the property that $p(a) = 0_K$ for all $a\in K$ (why? because a $p(x)$ has at most $\text{deg}(p)$ roots). But if $K$ is a finite field, there are non-zero polynomials which evaluate to $0_K$ on every element of $K$, e.g. $\prod_{a\in K}(x-a)$. – Alex Kruckman Sep 02 '19 at 23:49
  • Okay, what will happen with $K[x_1,x_2]$ or rather say $K[x_1,....,x_n]$, $K$ infinite, where $n \geq 2$. Here is $0$ the only polynomial $p(x_1,..,x_n)\in K[x_1,....,x_n]$ with with the property that $p(a)=0_K$ for all $a∈K^n$. If so, why? – Ri-Li Sep 02 '19 at 23:59
  • 1
    Induction on $n$. My previous comment settles the base case. Now view a non-zero polynomial $p\in K[x_1,\dots,x_n]$ as a polynomial in the single variable $x_n$ with coefficients in the ring $K[x_1,\dots,x_{n-1}]$. The leading term of this polynomial looks like $q(x_1,\dots,x_{n-1})x_n^d$ for some $d\geq 0$, where $q$ is a non-zero polynomial. By induction, there exist $a_1,\dots,a_{n-1}\in K$ such that $q(a_1,\dots,a_{n-1})\neq 0_K$. Then $p(a_1,\dots,a_{n-1},x_n)\in K[x_n]$ is a non-zero polynomial (it has degree $d$), so by the base case there is $a_n\in K$ s.t. $p(a_1,\dots,a_n)\neq 0$. – Alex Kruckman Sep 03 '19 at 00:14
  • 1
    See also: https://math.stackexchange.com/questions/23241/identically-zero-multivariate-polynomial-function – Alex Kruckman Sep 03 '19 at 00:16
  • You asked the same question yesterday, why did you delete it. The addition multiplication of polynomials in $\Bbb{Z}_2[x]$ is defined only from their coefficients, it is not the same as the polynomials functions $\Bbb{Z}_2\to\Bbb{Z}_2$ – reuns Sep 03 '19 at 00:56
  • @reuns Are you telling me? Neither I asked the same question yesterday and nor I deleted it. I have possibly edited it. If I am getting downvotes because of this please give me upvotes. – Ri-Li Sep 03 '19 at 01:37
  • It is because I wrote the exact same comment yesterday to someone having the same problem as you with polynomials with coefficients in finite fields. Then the ring $k[x]$ is defined abstractly just in term of the coefficients and the operations of $k$, nowhere it is mentionning that we can evaluate $f$ at $f(a),a\in k$ – reuns Sep 03 '19 at 01:40
  • Definitely, not me. You can check my activity. – Ri-Li Sep 03 '19 at 01:41

0 Answers0