存在任意大的特征为 2 的域

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Sheet II — HT21

II.3. Suppose that $R$ is an infinite integral domain with finitely many units in which every non-unit has an irreducible factor. By emulating Euclid’s proof that there are infinitely many primes show that $R$ contains infinitely many irreducible elements. Apply this to $𝔽_2[X]$ to deduce that there are arbitrarily large fields of characteristic 2.

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前半部分的证明见Commutative Algebra Wiki:

Since \(R\) is not a field, it has at least one irreducible element, say \(p_1\). Suppose now that the set of associate classes of irreducible elements is finite, say with representatives \(p_1,p_2,…,p_r\) for the associate classes. Let \(a=p_1p_2…p_r\). Consider the set $S = \{ 1 + a, 1 + a^2, 1 + a^3, …\}$
Since $a$ is not a unit, \(a^n≠±1\) for any \(n\).
Since \(a≠0,1 - a^{n-m}≠0\) and $R$ is an integral domain,\[(1 + a^m) - (1 + a^n) = a^m(1 - a^{n-m})≠0\]Thus, all the \(1 + a^n\) are distinct. Thus, \(S\) is an infinite set of distinct nonzero elements.
Since there are only finitely many units in \(R\), there exists \(n\) such that \(1 + a^n\) is not a unit.
Since every non-unit has an irreducible factor, \(1 + a^n\) has an irreducible factor, say $p_i$.
But \(p_i | a^n\) and \(p_i | 1 + a^n\), so \(p_i | 1\), a contradiction.

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后半部分如何证明

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Applying this to $𝔽_2[X]$, we deduce that there are infinitely many irreducible elements.
Since irreducible polynomials of degree $<n$ are finite, there is an irreducible polynomial $f$ in $\Bbb F_3[X]$ of degree $>n$, then $𝔽_2[X]/(f)$ has characteristic 2 and cardinality $>2^n$.

Example
a) For $𝔽_4$ we have that $α^3 = 1$ (because $𝔽^∗_4$ is cyclic), but then $α\ne1$ satisfies $α^2 + α + 1 = 0$, since $X^3 − 1 = (X^2 + X + 1)(X − 1)$ and $α\ne1$.
b) For $𝔽_8$ we have that $β^7 = 1$ ($𝔽^∗_8$ is cyclic), hence $β \ne 1$ satisfies $β^3 + β + 1 = 0$ or $β^3 + β^2 + 1 = 0$ since $X^7 − 1 = (X − 1)(X^3 + X^2 + 1)(X^3 + X + 1)$ in $𝔽_2[x]$.