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I have a concern about the first assertion in the proof of proposition 4.3.38 of Qing liu's "Algebraic Geometry and Arithmetic Curves". Referring to smooth morphisms, he says "The stability under base change follows immediately", but I couldn't figure out why.

Let me recall some definitions. An algebraic variety $X$ over $k$ is smooth at $x$ if $X\times_k Spec\ \overline{k}$ is regular at every point lying over $x$, where $\overline{k}$ is the algebraic closure of $k$. $X$ over $k$ is smooth if it is smooth at every point, i.e. $X\times_k Spec\ \overline{k}$ is regular. Let $Y$ be a locally noetherian scheme and $f:X\longrightarrow Y$ a morphism of finite type. We say that $f$ is smooth at $x$ if it is flat at $x$ and if $X_y\rightarrow Spec\ k(y)$ (with $y=f(x)$) is smooth (as an algebraic variety) at $x$. We say that $f$ is smooth is it is smooth at every $x\in X$.

Of course I tried to complete the proof on my own (unsuccessfully). Flatness under base change is easy. Let $Y'$ be an $Y$ scheme, $g$ its structural morphism, $X':=X\times_Y Y'$ and $y=g(y')$. We have to prove that is $X_y\rightarrow Spec\ k(y)$ is geometrically regular (which means "smooth as an algebraic variety") then $X'_{y'}\rightarrow Spec\ k(y')$ is geometrically regular. Calling $\overline{k(y)}$ and $\overline{k(y')}$ their respective algebraic closures, this means to prove that if $X_y\times_{k(y)}Spec\ \overline{k(y)}$ is regular so it is $X'_{y'}\times_{k(y')}Spec\ \overline{k(y')}$. So I reduced to the following proposition.

"Let $X$ be an algebraic variety over $k$ and $K$ any field extension of $k$. If $X$ is geometrically regular then $X\times_k Spec\ K$ is geometrically regular."

Question: if this proposition is true, how can I prove it? Are there easier strategies to show the stability under base change? I would prefer a proof that takes knowledges and definitions from Qing Liu's book.

One last thing. I tryed using Jacobian criterion for smoothness, but it works on closed points. This would be sufficient for regularity, but in order to transfer regularity from $X\times_k Spec\ \overline{k}$ to $X\times_k Spec\ \overline{K}$ I think we should know that the image of a closed point in the induced map $X\times_k Spec\ \overline{K}\longrightarrow X\times_k Spec\ \overline{k}$ is still closed. Is this true or not?

Thank you in advance :)

I have a concern about the first assertion in the proof of proposition 4.3.38 of Qing liu's "Algebraic Geometry and Arithmetic Curves". Referring to smooth morphisms, he says "The stability under base change follows immediately", but I couldn't figure out why.

Let me recall some definitions.$\DeclareMathOperator{\spec}{\textit{Spec}}$ An algebraic variety $X$ over $k$ is smooth at $x$ if $X\times_k \spec \overline{k}$ is regular at every point lying over $x$, where $\overline{k}$ is the algebraic closure of $k$. $X$ over $k$ is smooth if it is smooth at every point, i.e. $X\times_k \spec\ \overline{k}$ is regular. Let $Y$ be a locally noetherian scheme and $f:X\longrightarrow Y$ a morphism of finite type. We say that $f$ is smooth at $x$ if it is flat at $x$ and if $X_y\rightarrow \spec k(y)$ (with $y=f(x)$) is smooth (as an algebraic variety) at $x$. We say that $f$ is smooth is it is smooth at every $x\in X$.

Of course I tried to complete the proof on my own (unsuccessfully). Flatness under base change is easy. Let $Y'$ be an $Y$ scheme, $g$ its structural morphism, $X':=X\times_Y Y'$ and $y=g(y')$. We have to prove that is $X_y\rightarrow \spec k(y)$ is geometrically regular (which means "smooth as an algebraic variety") then $X'_{y'}\rightarrow \spec k(y')$ is geometrically regular. Calling $\overline{k(y)}$ and $\overline{k(y')}$ their respective algebraic closures, this means to prove that if $X_y\times_{k(y)}\spec \overline{k(y)}$ is regular so it is $X'_{y'}\times_{k(y')}\spec \overline{k(y')}$. So I reduced to the following proposition.

"Let $X$ be an algebraic variety over $k$ and $K$ any field extension of $k$. If $X$ is geometrically regular then $X\times_k \spec K$ is geometrically regular."

Question: if this proposition is true, how can I prove it? Are there easier strategies to show the stability under base change? I would prefer a proof that takes knowledges and definitions from Qing Liu's book.

One last thing. I tryed using Jacobian criterion for smoothness, but it works on closed points. This would be sufficient for regularity, but in order to transfer regularity from $X\times_k \spec \overline{k}$ to $X\times_k \spec \overline{K}$ I think we should know that the image of a closed point in the induced map $X\times_k \spec \overline{K}\longrightarrow X\times_k \spec \overline{k}$ is still closed. Is this true or not?

Thank you in advance :)

I have a concern about the first assertion in the proof of proposition 4.3.38 of Qing liu's "Algebraic Geometry and Arithmetic Curves". Referring to smooth morphisms, he says "The stability under base change follows immediately", but I couldn't figure out why.

Let me recall some definitions. An algebraic variety $X$ over $k$ is smooth at $x$ if $X\times_k Spec\ \overline{k}$ is regular at every point lying over $x$, where $\overline{k}$ is the algebraic closure of $k$. $X$ over $k$ is smooth if it is smooth at every point, i.e. $X\times_k Spec\ \overline{k}$ is regular. Let $Y$ be a locally noetherian scheme and $f:X\longrightarrow Y$ a morphism of finite type. We say that $f$ is smooth at $x$ if it is flat at $x$ and if $X_y\rightarrow Spec\ k(y)$ (with $y=f(x)$) is smooth (as an algebraic variety) at $x$. We say that $f$ is smooth is it is smooth at every $x\in X$.

Of course I tried to complete the proof on my own (unsuccessfully). Flatness under base change is easy. Let $Y'$ be an $Y$ scheme, $g$ its structural morphism, $X':=X\times_Y Y'$ and $y=g(y')$. We have to prove that is $X_y\rightarrow Spec\ k(y)$ is geometrically regular (which means "smooth as an algebraic variety") then $X'_{y'}\rightarrow Spec\ k(y')$ is geometrically regular. Calling $\overline{k(y)}$ and $\overline{k(y')}$ their respective algebraic closures, this means to prove that if $X_y\times_{k(y)}Spec\ \overline{k(y)}$ is regular so it is $X'_{y'}\times_{k(y')}Spec\ \overline{k(y')}$. So I reduced to the following proposition.

"Let $X$ be an algebraic variety over $k$ and $K$ any field extension of $k$. If $X$ is geometrically regular then $X\times_k Spec\ K$ is geometrically regular."

Question: if this proposition is true, how can I prove it? Are there easier strategies to show the stability under base change? I would prefer a proof that takes knowledges and definitions from Qing Liu's book.

One last thing. I tryed using Jacobian criterion for smoothness, but it works on closed points. This would be sufficient for regularity, but in order to transfer regularity from $X\times_k \overline{k}$ to $X\times_k \overline{K}$ I think we should know that the image of a closed point in the induced map $X\times_k \overline{K}\longrightarrow X\times_k \overline{k}$ is still closed. Is this true or not?

Thank you in advance :)

I have a concern about the first assertion in the proof of proposition 4.3.38 of Qing liu's "Algebraic Geometry and Arithmetic Curves". Referring to smooth morphisms, he says "The stability under base change follows immediately", but I couldn't figure out why.

Let me recall some definitions. An algebraic variety $X$ over $k$ is smooth at $x$ if $X\times_k Spec\ \overline{k}$ is regular at every point lying over $x$, where $\overline{k}$ is the algebraic closure of $k$. $X$ over $k$ is smooth if it is smooth at every point, i.e. $X\times_k Spec\ \overline{k}$ is regular. Let $Y$ be a locally noetherian scheme and $f:X\longrightarrow Y$ a morphism of finite type. We say that $f$ is smooth at $x$ if it is flat at $x$ and if $X_y\rightarrow Spec\ k(y)$ (with $y=f(x)$) is smooth (as an algebraic variety) at $x$. We say that $f$ is smooth is it is smooth at every $x\in X$.

Of course I tried to complete the proof on my own (unsuccessfully). Flatness under base change is easy. Let $Y'$ be an $Y$ scheme, $g$ its structural morphism, $X':=X\times_Y Y'$ and $y=g(y')$. We have to prove that is $X_y\rightarrow Spec\ k(y)$ is geometrically regular (which means "smooth as an algebraic variety") then $X'_{y'}\rightarrow Spec\ k(y')$ is geometrically regular. Calling $\overline{k(y)}$ and $\overline{k(y')}$ their respective algebraic closures, this means to prove that if $X_y\times_{k(y)}Spec\ \overline{k(y)}$ is regular so it is $X'_{y'}\times_{k(y')}Spec\ \overline{k(y')}$. So I reduced to the following proposition.

"Let $X$ be an algebraic variety over $k$ and $K$ any field extension of $k$. If $X$ is geometrically regular then $X\times_k Spec\ K$ is geometrically regular."

Question: if this proposition is true, how can I prove it? Are there easier strategies to show the stability under base change? I would prefer a proof that takes knowledges and definitions from Qing Liu's book.

One last thing. I tryed using Jacobian criterion for smoothness, but it works on closed points. This would be sufficient for regularity, but in order to transfer regularity from $X\times_k Spec\ \overline{k}$ to $X\times_k Spec\ \overline{K}$ I think we should know that the image of a closed point in the induced map $X\times_k Spec\ \overline{K}\longrightarrow X\times_k Spec\ \overline{k}$ is still closed. Is this true or not?

Thank you in advance :)

I have a concern about the first assertion in the proof of proposition 4.3.38 of Qing liu's "Algebraic Geometry and Arithmetic Curves". Referring to smooth morphisms, he says "The stability under base change follows immediately", but I couldn't figure out why.

Let me recall some definitions. An algebraic variety $X$ over $k$ is smooth at $x$ if $X\times_k Spec\ \overline{k}$ is regular at every point lying over $x$, where $\overline{k}$ is the algebraic closure of $k$. $X$ over $k$ is smooth if it is smooth at every point, i.e. $X\times_k Spec\ \overline{k}$ is regular. Let $Y$ be a locally noetherian scheme and $f:X\longrightarrow Y$ a morphism of finite type. We say that $f$ is smooth at $x$ if it is flat at $x$ and if $X_y\rightarrow Spec\ k(y)$ (with $y=f(x)$) is smooth (as an algebraic variety) at $x$. We say that $f$ is smooth is it is smooth at every $x\in X$.

Of course I tried to complete the proof on my own (unsuccessfully). Flatness under base change is easy. Let $Y'$ be an $Y$ scheme, $g$ its structural morphism, $X':=X\times_Y Y'$ and $y=g(y')$. We have to prove that is $X_y\rightarrow Spec\ k(y)$ is geometrically regular (which means "smooth as an algebraic variety") then $X'_{y'}\rightarrow Spec\ k(y')$ is geometrically regular. Calling $\overline{k(y)}$ and $\overline{k(y')}$ their respective algebraic closures, this means to prove that if $X_y\times_{k(y)}Spec\ \overline{k(y)}$ is regular so it is $X'_{y'}\times_{k(y')}Spec\ \overline{k(y')}$. So I reduced to the following proposition.

"Let $X$ be an algebraic variety over $k$ and $K$ any field extension of $k$. If $X$ is geometrically regular then $X\times_k Spec\ K$ is geometrically regular."

Question: if this proposition is true, how can I prove it? Are there easier strategies to show the stability under base change? I would prefer a proof that takes knowledges and definitions from Qing Liu's book.

One last thing. I tryed using Jacobian criterion for smoothness, but it works on closed points. This would be sufficient for regularity, but in order to transfer regularity from $X\times_k \overline{k}$ to $X\times_k \overline{K}$ I think we should know that the image of a closed point is still closed. Is this true or not?

Thank you in advance :)

I have a concern about the first assertion in the proof of proposition 4.3.38 of Qing liu's "Algebraic Geometry and Arithmetic Curves". Referring to smooth morphisms, he says "The stability under base change follows immediately", but I couldn't figure out why.

Let me recall some definitions. An algebraic variety $X$ over $k$ is smooth at $x$ if $X\times_k Spec\ \overline{k}$ is regular at every point lying over $x$, where $\overline{k}$ is the algebraic closure of $k$. $X$ over $k$ is smooth if it is smooth at every point, i.e. $X\times_k Spec\ \overline{k}$ is regular. Let $Y$ be a locally noetherian scheme and $f:X\longrightarrow Y$ a morphism of finite type. We say that $f$ is smooth at $x$ if it is flat at $x$ and if $X_y\rightarrow Spec\ k(y)$ (with $y=f(x)$) is smooth (as an algebraic variety) at $x$. We say that $f$ is smooth is it is smooth at every $x\in X$.

Of course I tried to complete the proof on my own (unsuccessfully). Flatness under base change is easy. Let $Y'$ be an $Y$ scheme, $g$ its structural morphism, $X':=X\times_Y Y'$ and $y=g(y')$. We have to prove that is $X_y\rightarrow Spec\ k(y)$ is geometrically regular (which means "smooth as an algebraic variety") then $X'_{y'}\rightarrow Spec\ k(y')$ is geometrically regular. Calling $\overline{k(y)}$ and $\overline{k(y')}$ their respective algebraic closures, this means to prove that if $X_y\times_{k(y)}Spec\ \overline{k(y)}$ is regular so it is $X'_{y'}\times_{k(y')}Spec\ \overline{k(y')}$. So I reduced to the following proposition.

"Let $X$ be an algebraic variety over $k$ and $K$ any field extension of $k$. If $X$ is geometrically regular then $X\times_k Spec\ K$ is geometrically regular."

Question: if this proposition is true, how can I prove it? Are there easier strategies to show the stability under base change? I would prefer a proof that takes knowledges and definitions from Qing Liu's book.

One last thing. I tryed using Jacobian criterion for smoothness, but it works on closed points. This would be sufficient for regularity, but in order to transfer regularity from $X\times_k \overline{k}$ to $X\times_k \overline{K}$ I think we should know that the image of a closed point in the induced map $X\times_k \overline{K}\longrightarrow X\times_k \overline{k}$ is still closed. Is this true or not?

Thank you in advance :)