Let $F$ be a field. Prove that the polynomial ring $F[x]$ in one variable over $F$ has infinitely many prime ideals. All I can think of if that if we can show there are infinitely many irreducible polynomials in $F[x]$ but I see no reason why that should hold, I have no information about the field $F$.
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Assume there are only a finite number of irreducible monic polynomials, say $p_1(X), p_2(X), . . . , p_n(X)$. Now since $F[X]$ is a UFD, irreducible = prime, so each $p_i(X)$ is prime. Now consider the monic polynomial $q(X) = \prod_i^n p_i(X)+1$. By construction, $q(X)$ is not divisible by any of the $p_i(X)$, hence it is either prime=irreducible itself, or divisible by another prime polynomial greater than $p_n(X)$, contradicting the assumption of a finite number of prime=irreducible monic polynomials.
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You can just duplicate the proof of Euclid, that there are infinitely many prime numbers.
MooS
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There's a simple method if we assume $F$ is infinite.
Clearly $x-a$ is irreducible for any $a \in F$ (note the quotient is $F$ hence prime) and these are always distinct. If they were not, then $x-a$ and $x-b$ differ by a unit, which is an element of $F$ but this is impossible as they are both monic. (Alternatively, evaluate them both at $a$ to contradict them being scalar multiples of each other).
Hence we have an injection of $F$ into the set of prime ideals of $F[x]$. (Indeed these are the only ones when $F$ is algebraically closed, e.g. $\mathbb{C}$).
If $F$ is finite, then the prime ideals are in "bijection with elements" (strictly speaking we should be considering Galois orbits of elements instead) of the algebraic closure $\overline{F}$ again but you probably need the adaptation of Euclid's proof to argue that $\overline{F}$ is infinite.
Matt B
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