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For any positive integer $n$, how many polynomials are there of degree $n$ over $\mathbb{Z}_2$? How many distinct polynomials function from $\mathbb{Z}_2$ to $\mathbb{Z}_2$?

Attempt: first part is clear to me since there are $2$ choices for each coefficient and there are $n$ coefficient so there are $2^n$ such polynomials. I am having problems understanding the second part where I need to find distinct polynomial functions.

If I assume $p(x)$ and $p'(x)$ are two equal polynomial functions over $\mathbb{Z}_2$ such that $p(x)=a_nx^n+\cdots+a_0$ and $p'(x)=a'_nx^n+\cdots+a'_0$, then $p'(x)=p(x)$ for $x=0,1$. So $a'_0=a_0$. And since the degree of these polynomials are $n$ then $a_n=a'_n=1$. So to find distinct polynomial functions we have to consider when $p(x)$ cannot be equal to $p'(x)$ for every value of $x\in\{0,1\}$. From here I cannot proceed. I was looking for solutions. Everywhere I see that they have started the argument with the fact that there are only $4$ such polynomials and then they give the examples of such polynomials. I need help to understand this problem. Thank you

user26857
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Sunit das
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  • You seem to ask different questions in the title and the body of the question. There are only $4$ functions there, but there are (countably) infinitely many polynomials. – tomasz Aug 18 '20 at 09:57
  • I asked about number of distinct polynomial fuctions. – Sunit das Aug 18 '20 at 09:58
  • And you are right there are (countably) infinitely many polynomials. Consider $x^2$+x which is not same as the zero polynomial in $Z_2$[x] but as polynomial functions from $Z_2$ to $Z_2$ they gives equal value at=0 and 1. So i asked about number of such distinct polynomial functions not about polynomials. – Sunit das Aug 18 '20 at 10:05

3 Answers3

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There are only 4 distinct functions $f: \Bbb Z_2 \to \Bbb Z_2$. This is because the cardinality of the set of functions $A \to B$ is $$|B^A|=|B|^{|A|}$$ whenever $A,B$ are finite sets.

It happens that they are polynomial functions. Indeed they are $$f_1(x)=0$$ $$f_2(x)=1$$ $$f_3(x)=x$$ $$f_4(x)=1-x$$ So we have found all of them.

Crostul
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Over $\Bbb{Z}_2$, the polynomial $x(x+1) = x^2 + x$ is identically $0$, which means I can replace $x^2$ with $x$ in any polynomial expression and get the same value. Using this repeatedly, over $\Bbb{Z}_2$, the polynomial $$a_0 + a_1 x + a_2 x^2 + ... + a_n x^n$$ always gives the same value as the polynomial $$a_0 + (a_1 + a_2 + a_3 + ... + a_n)x,$$ and so there are only $4$ distinguishable polynomials over $\Bbb{Z}_2$, depending on whether $a_0 = 0$ or $1$, and whether $a_1 + a_2 + a_3 + ... +a_n = 0$ or $1$.

Rivers McForge
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The answer to your first question should be $2^{n-1}$ rather than $2^{n}$ since the coefficient of $x^n$ is always $1$.

For the second part note that the set of all polynomial functions is the set of all functions in your case.

EDIT: A pointed out in the comment the first part of this answer is incorrect.

Jens Schwaiger
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  • A polynomial of degree n has n+1 coefficient in general form ($a_0$,.......,$a_n$) and one of them is fixed so there are still n places and two choices for each. – Sunit das Aug 18 '20 at 07:10