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Apr 14, 2021 at 18:57 history edited user237522 CC BY-SA 4.0
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Apr 14, 2021 at 13:19 comment added user237522 @WillChen, yes, you are right, I meant that every element of $S$ satisfies a polynomial in $R[T]$, bot not every element of $S$ satisfies a monic polynomial in $R[T]$.
Apr 14, 2021 at 13:16 vote accept user237522
Apr 14, 2021 at 3:31 comment added Will Chen Every element of $R$ satisfies a polynomial with coefficients in $R$ with non-invertible leading coefficient. E.g., $r \in R$ satisfies $2T - 2r\in R[T]$. I think what you mean to say is that every element of $S$ satisfies a polynomial in $R[T]$, but not every element in $S$ satisfies a monic polynomial in $R[T]$.
Apr 14, 2021 at 3:28 answer added Will Chen timeline score: 1
Apr 14, 2021 at 2:03 comment added user237522 Well, the rings $\mathbb{Z}$ and $\mathbb{Z}[\frac{1}{2}]$ are not local, but we can consider localizations of them. .
Apr 14, 2021 at 2:00 history edited user237522 CC BY-SA 4.0
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Apr 14, 2021 at 1:52 comment added user237522 @WillChen, thank you for your comment. Actually, I meant 'necessarily, not invertible leading coefficient', since I want the extension to be algebraic but not integral. For an integral extension I have found an example (localizations of polynomial rings). I wish something like $\mathbb{Z} \subset \mathbb{Z}[\frac{1}{2}]$ which is flat, algebraic, non-integral, because $f(T)=2T-1 \in \mathbb{Z}[T]$ is such that $f(\frac{1}{2})=0$. But here the rings are requiered to be $\mathbb{C}$-algebras (I wished to exclude such cases as $\mathbb{Z} \subset \mathbb{Z}[\frac{1}{2}]$).
Apr 14, 2021 at 1:38 comment added Will Chen I assume you meant "not necessarily invertible leading coefficient"?
Apr 14, 2021 at 0:56 history edited user237522 CC BY-SA 4.0
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Apr 14, 2021 at 0:27 history asked user237522 CC BY-SA 4.0