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I am having trouble understanding the elements of $\Bbb F_2[x]$. Why is $f_1(x)=x^3+1$ not exactly the same as $f_2(x)=x+1$? $f_1(0)=1=f_2(0),f_1(1)=0=f_2(1)$

I.e. Why can't we reduce $x^n=x$? Then the only irreducible polynomials in $\Bbb F_2[x]$ are $x,x+1,0,1$? What am I misunderstanding?

Galois in the Field
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    A polynomial over a field F is not a function but a formal expression. The distinction between these is only visible when F is finite, as in your case. – KCd Oct 02 '15 at 09:16
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    As functions, they are the same. In other words, $f_1(x) = f_2(x)$ for all $x \in \Bbb{F}_2[x]$. But, they are different elements of the polynomial ring. – Sammy Black Oct 02 '15 at 09:18
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    Ditto the above. Perhaps you would also benefit from http://math.stackexchange.com/q/791845/29335 and it's linked question – rschwieb Oct 02 '15 at 09:23

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The ring $\mathbb{F}_{2}[x]$ does not contain functions; you cannot really argue that objects in this ring are equal because they define the same function. The element $x$ is to be considered as an indeterminate, that is, it is a transcendental element (satisfies no algebraic equations).

If you want to consider the polynomials as functions, you can work in the factor ring $\mathbb{F}_{2}[x]/\langle x^{2} +x\rangle$ (in general, you would use $\mathbb{F}_{q}[x]/\langle x^{q} -x \rangle$); two polynomials in the same equivalence class in this factor ring determine the same mapping from $\mathbb{F}_{2} \to \mathbb{F}_{2}$. Notice that $$x^3 +1 = (x+1)(x^{2} +x) + x + 1,$$ so $x^{3} +1$ and $x+1$ are in the same equivalence class here.

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In the ring $\mathbb{F}_2[x]$ (or even any polynomial ring), much part of the game lies in playing with "coefficients". Thus it is $3x+1=x+1$ because the "coefficients" $3$ and $1$ are same in $\mathbb{F}_2$; but not $x^3+1=x+1$.

Groups
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Why is $f_1(x)=x^3+1$ not exactly the same as $f_2(x)=x+1$? $f_1(0)=1=f_2(0),f_1(1)=0=f_2(1)$[...] Then the only irreducible polynomials in $\Bbb F_2[x]$ are $x,x+1,0,1$?

You have discovered that the two polynomial expressions produce identical functions, but the 'correct' definition of equality of polynomial so is 'the ordered list of coefficients match.'

Using polynomials as functions is most useful for identifying irreducibility of polynomials of degree 3 or less. It's true that a nonzero polynomial of degree 3 or less is reducible iff it has a zero as a function. Beyond that, a polynomial with a root is reducible, but the converse may fail. Perhaps that is where you got the idea for this connection.

I.e. Why can't we reduce $x^n=x$?

You can reduce $x^n-x$ over any field. It at least factors into $(x-1)\sum_{i=0}^{n-1} x^i$. Do you mean reduce in some other way?

If you meant that you believed $x$ is irreducible, then continued that "$x=x^n$" and is therefore reducible as well, then you have run into the 'wrong definition for equality' problem again.

rschwieb
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