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I'm currently working through the proof of the weak Nullstellensatz in Atiyah & MacDonald's Introduction to Commutative Algebra, but I'm stuck at the Proof of Proposition 7.9 (Zariski's Lemma):

Let $k$ be a field, $E$ a finitely generated $k$-algebra. If $E$ is a field, then it is a finite algebraic extension of $k$.

Let $E=k[x_1,\ldots,x_n]$, $F=k(x_1,\ldots,x_r)$, where $x_1,\ldots,x_r$ are algebraically independent and $x_{r+1},\ldots,x_r$ are algebraic over $F$.

Then it is shown that $F$ is a finitely generated $k$-algebra. Say $F=[y_1,\ldots,y_k]$, where $y_i = \frac{f_i}{g_i}, \enspace f_i,g_i \in k[x_1,\ldots,x_n]$. Up to this point everything is clear to me.

Then the proof goes on: Since there are infinitely irreducible polynomials in $k[x_1,\ldots,x_n]$ there exists an irreducible polynomial $h$ which is prime to each of the $g_j$ (for example $h=g_1g_2\cdots g_k + 1$).

Here I'm stuck, I can't see why $g_1g_2\cdots g_k + 1$ would necessarily be irreducible (maybe we could take an irreducible factor), and why we need to know that there are infinitely many irreducible polynomials in $k[x_1,\ldots,x_n]$ to say such a polynomial $h$ exists.

Any help would be appreciated.

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  • $\begingroup$ I also can't see why $h$ is irreducible, but an irreducible factor of $h$ does the job. $\endgroup$
    – user26857
    Jan 3, 2018 at 19:09
  • $\begingroup$ The general result is: if $A$ is a UFD, then $A[x]$ is a UFD that has infinitely many non-associate irreducible elements. We mimic Euclid: let $f_1=x$ and define $f_n=1+f_1\cdots f_{n-1}$. Then the $f_i$'s are non-zero and non-units. Moreover, $\operatorname{gcd}(f_k, f_n)=1$ if $k<n$, since any common divisor of $f_k, f_n$ must also divide $1 = f_n − f_1\cdots f_k\cdots f_{n-1}$. Take $g_i$ to be some irreducible polynomial dividing $f_i$. Then the $g_i$'s are pairwise non-associate. $\endgroup$ Apr 2 at 11:44

1 Answer 1

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It is not necessarily true that $h=g_1g_2\cdots g_k + 1$ is irreducible. However, it has no (non-unit) common factors with the $g_j$. This is because any common factor of $h$ and $g_j$ would also be a factor of $h-g_1g_2\cdots g_k=1$. So, if you let $p$ be any irreducible factor of $h$, $p$ must be an irreducible polynomial that does not divide any of the $g_j$. (Such an irreducible factor exists as long as the $g_j$ are not all constant so $h$ is not constant; if they are all constant, you could just take $x_1$ as your irreducible polynomial instead.)

(This is essentially the same as the classic proof that there are infinitely many prime integers: if you have finitely many primes, multiply them together and add $1$, to get a number whose prime factors are not on your list.)

Alternatively, if you already know that there are infinitely many irreducible polynomials in $k[x_1,\dots,x_n]$ (up to units), you can just define $h$ to be any irreducible polynomial that is not a factor of any of the $g_j$. This is possible since each $g_j$ has only finitely many irreducible factors (up to units).

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  • $\begingroup$ It must be noted that having $r\ge 1$ is vital for the existence of $p$. $\endgroup$
    – Atom
    Apr 23, 2023 at 16:23

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