# Is a base-change of an integral domain by an extension of its base field without algebraic elements still a domain?

Let $K$ be a field, and let $L/K$ be an algebraically closed field extension (i.e. the only elements of $L$ that are algebraic over $K$ are already in $K$). Let $R$ be a $K$-algebra that is an integral domain. Does it follow that $R \otimes_K L$ is an integral domain? I'm particularly interested in the case where $R$ is a finitely generated $K$-algebra.

My question is closely related to this question, where Will Sawin gives a yes answer when $L$ is purely transcendental over $K$. Also (at least when $R$ is finitely generated as a $K$-algebra), the answer seems to be yes when $K$ is algebraically closed, according to a recent preprint. Furthermore, the answer is typically 'no' if $L$ has algebraic elements over $K$, even when $R$ itself is a field. For instance, let $f$ be an irreducible polynomial in $K[t]$ that has a root in $L$, and let $R := K[t]/(f)$.

• No (but "yes" for $K=\overline{K}$ is classical). An example of MacLane is geometric irreducible over $K$ but not geometrically reduced even with $K$ algebraically closed in the fraction field of $R$ (to avoid lame examples): for $K=\mathbf{F}_p(s,t)$, $R= K[x,y]/(sx^p+ty^p-1)$ is a Dedekind domain and $K$ is algebraically closed in ${\rm{Frac}}(R)$, but $R \otimes_K \overline{K}=\overline{K}[x,y]/(h^p)$ for the linear form $h=s^{1/p}x+t^{1/p}y - 1$. In char. 0, if $K$ is algebraically closed in ${\rm{Frac}}(R)$ then $R\otimes_K K'$ is a domain for any $K'/K$; see 4.3 in EGA IV$_2$ for more. – nfdc23 Jan 12 '17 at 16:15
• I think Neil Epstein's $L$ is @nfdc23's $Frac(R)$ and NE's $R$ is nf's $\overline{K}$ (or $K(s^{1/p},t^{1/p})$ if you want something finitely generated over $K$). – Kevin Buzzard Jan 12 '17 at 16:48
• Why is $K$ algebraically closed in $L := \Frac(R)$? Do you have a reference for MacLane's example? – Neil Epstein Jan 12 '17 at 16:49
• Did you check EGA IV_2 yet? There are explicit references to Bourbaki in there -- lots of material. – Kevin Buzzard Jan 12 '17 at 16:53
• That $K$ is algebraically closed in ${\rm{Frac}}(R)$ is a fun exercise; I don't know a reference (I heard it from somewhere years ago, and worked it out for myself). I mentioned that $R$ is Dedekind as a "hint" on this since that ensures one just has to study elements of $R$ integral over $K$, which one can control by looking in residue fields of $R$ at various maximal ideals. But as Kevin says, the Bourbaki references in the EGA reference dispose of the theoretical aspects on this and other related questions (though I don't know offhand if Bourbaki includes MacLane's example). – nfdc23 Jan 12 '17 at 17:18

No. Let $K = \mathbf{F}_p(s,t)$, $A = K[x,y]/(sx^p + t y^p - 1)$. One checks $A$ is a domain, even Dedekind, so we can define $L = {\rm{Frac}}(A)$. By exploring residue fields of $A$ at maximal ideals and using that $A$ is Dedekind one shows with some thought that $K$ is algebraically closed in $L$. But for the field $R = K[s^{1/p}, t^{1/p}]$ of degree $p^2$ over $K$ we have $R \otimes_K L = L[x,y]/(h^p)$ for $h = s^{1/p}x + t^{1/p}y - 1$, so this is non-reduced. In this case $A$ is a domain of finite type over $K$ that is a geometrically irreducible $K$-algebra that is geometrically everywhere non-reduced. This example is due to MacLane, but I don't know a literature reference for it.
• The property for a $K$-algebra $R$, that $R\otimes_K L$ be an integral domain for every field extension is named “absolutely integral” (absolument intègre) by Bourbaki (A, V, §17, no. 3) and is related to the notion of a regular extension of fields (defined as an absolutely integral extension of fields). A Google search with the terms “maclane example regular extension” led me to an handout by Paul Vojta, which in turn contains the reference: Duke Math. J. <b>5</b> (1939), 372–393. The example is on p. 384, just after the corollary of theorem 14. – ACL Jan 13 '17 at 8:13
• @ACL: Thanks! I see that MacLane doesn't provide a proof that his example has the ground field (which he calls $L$) is algebraically closed in the extension field (though I knew a proof, and the Vojta handout one finds on Google as you indicate gives one). I don't quite follow MacLane's justification of his example (he asserts that certain separable irreducible polynomials relative to a hypothetical separating transcendence basis $\{t\}$ have "exponent" a power of $p$, but I don't know what the "exponent" terminology there means), but it is obvious given the phenomenon above, so no problem. – nfdc23 Jan 13 '17 at 13:22