The following Daily Questions were asked this week.

Monday, April 17

For today’s questions, suppose we have a tower of field extensions $F\xrightarrow[]{\phi} E\xrightarrow[]{\psi} D$.

First Question

True or False: If $\psi\circ \phi$ is algebraic, then so are $\phi$ and $\psi$.

Reveal Answer

Answer

True.

Second Question

True or False: If $\psi$ and $\phi$ are both algebraic, then so is $\psi\circ \phi$.

Reveal Answer

Answer

True.

Tuesday, April 18

First Question

By definition, a field extension $\phi:F\to E$ is an algebraic closure of $F$ if…

Reveal Answer

Answer

…it is an algebraic extension and for every nonconstant $f\in F[x]$ the polynomial $\tilde{\phi}(f)$ splits completely in $E[x]$.

Second Question

By definition, a field $E$ is algebraically closed if…

Reveal Answer

Answer

…every nonconstant polynomial $f\in E[x]$ splits completely in $E[x]$.

Equivalent definitions:

– Every nonconstant $f\in E[x]$ has all of its roots in $E$.

– Every nonconstant $f\in E[x]$ has a root in $E$.

– Every nonconstant $F\in E[x]$ has a linear factor in $E[x]$.

Thursday, April 20

First Question

True or False: For every field $F$ there exists an algebraic closure $\phi:F\to \overline{F}$.

Reveal Answer

Answer

True.

Second Question

True or False: If $\phi:F\to E$ is a field extension and $E$ is algebraically closed, then $\phi$ is an algebraic closure of $F$.

Reveal Answer

Answer

False. This is only true if the extension $\phi$ is algebraic. In general, if we let $\overline{F}\subseteq E$ denote the subset of all elements in $E$ that are algebraic over $F$, then $\phi:F\to \overline{F}$ is an algebraic closure of $F$.