I'm reading Lectures on Modules and Rings by T. Y. Lam. It's on page 32 of the book, example 2.19A.

It reads:

(2.19A) Example. Let $k$ be a field. Then in the commutative polynomial ring $R = k[x_1; x_2; ...; x_n]$ with $n \ge 2$ variables, the ideal $\mathfrak{A} = (x_1; x_2; ...; x_n)$ is not invertible, hence not projective. In fact, it turns out that $\mathfrak{A}^{-1} = R$, so $\mathfrak{A} \mathfrak{A}^{-1} = \mathfrak{A} \neq R$. To see this, assume instead that there is some $f/g \in \mathfrak{A}^{-1}$, where $f, g$ are relatively prime to each other (???), and $g \notin k$. Then $(f/g).x_i = h_i \in R$. If $x_1 | g$, then $x_1$ also divides $gh_2 = x_2f$, so $x_1 | f$ (I assume, he means that, since $x_1$, and $x_2$ are relatively prime, so $x_1$ must divides $f$), a contradiction. Therefore $x_1 \not | g$, and $x_1f = gh_1$ implies that $x_1 | h_1$, but then $f/g = h_1/x_1 \in R$, again a contradiction.

I'm fine with most of the things here. I know that a polynomial ring with 1 indeterminate over a field, i.e $k[x]$ is an Euclidean domain, hence a PID, hence a UDF. So, basically two $f, g \in k[x]$ would have some gcd. However, things are somewhat different for the ring $k[x_1; x_2; ...; x_n]$ (say, the Bezout equation doesn't hold) (I'm reading it off here Division algorithm for multivariate polynomials?).

Moreover I'm pretty sure that $k[x_1; x_2; ...; x_n]$ is not a PID, so I'm not even sure if there may (or may not) exist the gcd for any pair of $f, g \in k[x_1; x_2; ...; x_n]$. And how can I define $f, g \in k[x_1; x_2; ...; x_n]$ to be relatively prime?

According to the like above, Math Gems claimed that "But all Euclidean is not lost...", which makes me wonder which properties of $k[x]$ are still true in $k[x_1; x_2; ...; x_n]$.

So it would be great if you guys can guide me through it, or just give me some reference (like a web page, or some textbook) that covers this.

Thank you guys very much in advance,

And have a great day,

  • 1
    $\begingroup$ As Math Gems recommended: consult the extensive literature on Grobner bases. $\endgroup$ Mar 30, 2014 at 15:23

1 Answer 1


In any UFD you can find a greatest common divisor of two (non zero) elements; the greatest common divisor is, of course, only determined up to multiplication by invertible elements.

If any of the two elements $a,b\in R$, where $R$ is a UFD, is invertible, then every common divisor of them is invertible, so $1$ is a greatest common divisor. So, assume $a$ and $b$ are not invertible (and non zero). Factor them into a product of irreducible elements: $$ a=up_1^{\alpha_1}p_2^{\alpha_2}\dots p_r^{\alpha_r}, \qquad b=vp_1^{\beta_1}p_2^{\beta_2}\dots p_r^{\beta_r} $$ where the exponents are non negative (if an irreducible factor doesn't divide either element, take the exponent to be $0$) and $u$ and $v$ are invertible elements. Then $$ d=p_1^{\gamma_1}p_2^{\gamma_2}\dots p_r^{\gamma_r} $$ is a greatest common divisor of $a$ and $b$, where $$ \gamma_i=\min\{\alpha_i,\beta_i\} \quad (i=1,2,\dots,n). $$ The verification is quite easy. The elements are relatively prime if all exponents $\gamma_i$ turn out to be $0$, that is, when no irreducible factor of $a$ divides $b$ and conversely.

Now, if $R$ is a UFD, then also $R[x]$ is a UFD.

  • $\begingroup$ Thank you very much, you start my day. :* I get it now, thank you so much :* $\endgroup$
    – user49685
    Mar 30, 2014 at 15:32
  • $\begingroup$ Wait, can somebody tell me why there is a downvote here? @@ :( I think this answer is great, and straight to the point. Am I missing something? $\endgroup$
    – user49685
    Mar 30, 2014 at 15:37
  • $\begingroup$ @user49685 I'm missing something, too. Of course the answer is not really too detailed; something is left to the reader, for example how to get the expressions of $a$ and $b$ as products, which I'm confident you're willing and able to carry out. $\endgroup$
    – egreg
    Mar 30, 2014 at 15:41
  • $\begingroup$ I'm pretty sure that your answer is logical, and it makes sense. However, to eliminate the fact that there may be some subtle thingy going on around here, I think I'll wait for 1 more day, for the downvoter to jump in, then I'll mark you answer. Thank you very much, :* $\endgroup$
    – user49685
    Mar 30, 2014 at 15:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.