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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,

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    $\begingroup$ As Math Gems recommended: consult the extensive literature on Grobner bases. $\endgroup$ – Bill Dubuque Mar 30 '14 at 15:23
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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.

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  • $\begingroup$ Thank you very much, you start my day. :* I get it now, thank you so much :* $\endgroup$ – user49685 Mar 30 '14 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 '14 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 '14 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 '14 at 15:47

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