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I'm reading the article

M. Hochster, Prime ideal structure in commutative rings, Trans. Amer. Math. Soc. 142 (1969), 43--60. Freely available here on the journal's website.

But, I can not find the definition of the symbol "$[b_1 , b_2]$" in the proof of Theorem 4 (page 49, which is 7th page of the pdf).

Could someone explain what this notation means there? Possibly it is explained somewhere in the paper, but I could not locate it.

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    $\begingroup$ Not sure how this is off-topic...I just took another look at the paper myself and am puzzling over what the notation is supposed to mean. I can't tell for sure, but I think it is supposed to denote the subring generated by $b_1$ and $b_2$. $\endgroup$ Mar 28, 2016 at 23:05
  • $\begingroup$ I agree with @EricWofsey it is not very clear. Normally I'd assume it is some kind of LCM but this does not quite seem to fit. I added a way to locate the document easily. It might have been still better to include the relevant part in the post, but then perhaps it is sufficient like this. $\endgroup$
    – user9072
    Mar 28, 2016 at 23:53
  • $\begingroup$ I am grateful for your support. $\endgroup$
    – user89541
    Mar 29, 2016 at 0:36

1 Answer 1

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My best guess for what $[b_1,b_2]$ is supposed to mean is the subring of $A$ generated by $b_1$ and $b_2$. However, even with this interpretation, there are some statements that aren't quite right (though as I recall, this particular paper has a lot of minor errors of this sort, so this shouldn't be too surprising). Here is how I would rewrite the final three sentences of the proof of Theorem 4:

Let $f:\mathbb{Z}[x_1,x_2]\to A$ be the unique homomorphism sending $x_i$ to $b_i$ and let $I=(x_1,x_2)\subset\mathbb{Z}[x_1,x_2]$. For any $z\in I$, $f(z)$ vanishes off of $d(b_1)\cup d(b_2)$, and its image restricted to $d(b_1)\cup d(b_2)$ is equal to its image restricted to the finite set $Y$. Thus we need only find $z\in I$ such that $z\not\in f^{-1}(\phi(y))$ for each $y\in Y$, and then $c=f(z)$ will be an element of $(b_1,b_2)$ such that $d(c)=d(b_1)\cup d(b_2)$. But each $f^{-1}(\phi(y))$ is a prime ideal in $\mathbb{Z}[x_1,x_2]$ which does not contain $I$ (since every point of $Y$ is in $d(b_1)\cup d(b_2)$), and there are only finitely many of them, so by prime avoidance such a $z$ must exist.

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