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Let $Y$ be a regular, integral scheme of dimension $1$. Is it true that $Y$ is locally Noetherian?

To prove local Noetherianity it is enough to show that for any open affine $U=\mathrm{Spec}(A)\subseteq Y$, the ring $A$ is Noetherian. Then the answer to my question would be "yes" if we could prove the following:

Claim: let $A$ be an integral domain of Krull dimension $1$ such that for any nonzero prime ideal $\mathfrak{p}\lhd A$ the local ring $A_{\mathfrak{p}}$ is regular. Then $A$ is Noetherian.

Any help directed towards answering the first question and proving (or disproving) the claim would be greatly appreciated!

Note: I actually know that for a ring $R$, even if every local ring is regular of dimension 1, it could be that $R$ is not Noetherian. However, it seems to me that all the counterexamples to this claim that I know of start from a ring with zero divisors.

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    There are counterexamples: see the section 2 of [this paper](http://www.ams.org/journals/tran/1971-158-02/S0002-9947-1971-0280472-2/S0002-9947-1971-0280472-2.pdf). – 2017-01-08
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    Also here: http://mathoverflow.net/questions/114715/is-a-domain-all-of-whose-localizations-are-noetherian-itself-noetherian – 2017-01-08
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    Thank you very much! But are the rings in these counterexamples of dimension 1? Unfortunately I'm not very well-versed in commutative algebra, so I wasn't able to figure this out for myself. – 2017-01-08
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    Example 2.2 from the linked paper: "*Example of a domain D such that every $D_P$ is a noetherian valuation ring but D is not itself noetherian.*" I suppose this shows clearly that $\dim D=1$. – 2017-01-08
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    See also example 2 [here](http://www.math.ucla.edu/~rse/210c.1.08s/examples.pdf). This seems to be the first example of almost Dedekind domain which is not Dedekind. – 2017-01-08
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    Of course! Thank you very much, you've been really helpful. – 2017-01-08

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