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I stumbled upon this slick proof of Gauss lemma.

Let $A$ be a commutative ring with identity. Call a polynomial $f\in A[x]$ primitive iff the ideal generated by its coefficients is the entire $A$. Let $f, g\in A$ be primitive. Then $fg$ is primitive too.

I regurgitate the proof:

Proof. Suppose not. Then the ideal generated by the coefficients of $fg$ is proper. Since $A$ has identity, we can apply Zorn's lemma and find a maximal ideal $\mathfrak m$ that contains all the coefficients of $fg$. Since $A$ is commutative too, $\mathfrak m$ is prime as well, and hence, $A/\mathfrak m$ an integral domain. Now, we view the coefficients of polynomials in $A[x]$ as the members of $A/\mathfrak m$. Then, $fg = 0$, but $f, g\ne 0$, which violates the integral domain-ness of $A/\mathfrak m[x]$. (For if $f = 0$, then all the coefficients of $f$ would be in $\mathfrak m$, and since $f$ is primitive, $1\in\mathfrak m$, violating the maximality. Similarly, $g\ne 0$.)

Question: It appears as though the definition of polynomial multiplication is nowhere being used in the proof, which is quite counterintuitive!–for we are making a comment about the coefficients of $fg$, which are evaluated from those of $f$ and $g$ by polynomial multiplication.

One way that the definition of polynomial multiplication is seeping in is due to the usage of the fact that $R[x]$ is an integral domain if $R$ is one. But this seems to be like a "very little" part of the much more mouthful definition of polynomial multiplication.

Can you spot something?

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  • Polynomial multiplication seeps in via use of lead coefficient map is multiplicative $\ell(fg) = \ell(f)\ell(g),,$ see here – Bill Dubuque Jan 08 '23 at 15:11
  • @BillDubuque Yea, I noted that. But as I say, polynomial multiplication seems to be much more than just the multiplicativity of the leading coefficients. Or is it not? – Atom Jan 08 '23 at 15:21
  • Where did you note that? Are you familiar with graded rings? – Bill Dubuque Jan 08 '23 at 15:34
  • @BillDubuque In stating in the last paragraph that polynomial multiplication seeps in due to this fact: $R$ is an integral domain $\implies$ $R[x]$ is an integral domain. And no, I am not familiar with graded rings, why? – Atom Jan 08 '23 at 15:37
  • There are many directions one can pursue toward better understanding Gauss's Lemma. A good place to start chasing links is the survey by D.D. Anderson cited here. – Bill Dubuque Jan 08 '23 at 16:32
  • @BillDubuque Does my answer make sense to you? – Atom Jan 09 '23 at 19:23

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I have got an answer to my question:

In reasoning inside $A/\mathfrak m[x]$, how do we know that the product of the images of $f$, $g$ (which are polynomials in $A[x]$) is actually the image of the product of $fg$? Well this rests on the fact for any rings $R$, $S$, if $\phi\colon R\to S$ is a homomorphism, then the map $R[x]\to S[x]$ given by $$ a_0 + \cdots + a_n x^n\mapsto \phi(a_0) + \cdots + \phi(a_n) x^n $$ is also a homomorphism. And showing this fact requires the full-blown definition of polynomial multiplication!

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