# Why is $\mathbb{Q}(\sqrt{2})$ not a stem field for the polynomial $X^3-2 \in \mathbb{Q}[X]$?

I'm trying to learn Galois theory on Coursera. The lecturer gave the following definition of a stem field:

Let $P$ be an irreducible monic polynomial in $K[X]$ with a root $\alpha$. A stem field is an extension E such that $\alpha \in E$ and $E=K(\alpha)$.

A textbook I have says $\mathbb{Q}(\sqrt{2})$ is not a stem field for the polynomial $X^3-2 \in \mathbb{Q}[X]$, but to me it seems to match the definition. Can someone help me understand why it isn't?

• According to your definition, I agree that $\mathbb{Q}(\sqrt{2})$ is a stem field for $X^3-2$. Are you sure that the textbook doesn't say that $\mathbb{Q}(\sqrt{2})$ isn't a splitting field for $X^3-2$? Mar 5 '16 at 22:33
• Something must be lacking in the definition, I think. You're right, imo, to say it matches perfectly what you say is a "stem field". By the way, from where or what book is that definition with that name? Perhaps they meant "splitting". Mar 5 '16 at 22:33
• I have never heard about the term "stem field", and Google does not give me any reference. What is your book? Are you translating from some other language than English? Mar 5 '16 at 22:40
• Thanks! My book (in Finnish) was indeed talking about splitting fields. The word used for splitting fields in Finnish is almost a literal translation of the word stem field for some reason... Btw the term stem field seems to come from A. Albert, Modern Higher Algebra, 1937. Mar 5 '16 at 22:51
• A splitting field is an extension in which the polynomial splits completely, that is, splits into linear factors; it's an extension comtaining not just one of, but all of the zeros of the polynomial. ${\bf Q}(\root3\of2)$ is not a splitting field for $x^3-2$. Mar 5 '16 at 23:05

Since the polynomial $$~P = X^3 - 2~$$ is generated using the root $$~\alpha~$$ where $$~\alpha = \sqrt{2} ~$$,$$~Q(\sqrt2)~$$ is indeed a stem field, however when we attempt to split it over $$~Q~$$ we fail to do so since $$~2~$$ out of the $$~3~$$ roots $$~\epsilon$$ $$C~$$.
Hence your stated $$~Q(\sqrt2)~$$ is in fact not a splitting field but definitely a stem field for $$~P~$$.