# I Extension fields

1. May 5, 2017

### Mr Davis 97

I am a little confused about terminology when it comes to extension fields. In my textbook, E is a field extension of F if F is a subfield of E. This is understandable. However, in proving that all polynomials have a zero in an extension field, $F[x] / \langle p(x) \rangle$, where $p(x)$ is irreducible, is identified as an extension field of $F$. But how does that match the definition of extension field given above? $F$ isn't a subfield of $F[x] / \langle p(x) \rangle$ at all, but rather isomorphic to a subfield by the isomorphism $\mu (a) = a + \langle p(x) \rangle$, right?

2. May 5, 2017

### Staff: Mentor

Yes.

Now $F[x]/\langle p(x) \rangle$ is an $F-$vector space with a basis vector $1$, so the embedding $\mu$ is a canonical one (maybe even natural) and $F$ can be considered as a subfield. Usually we also write the elements of $E = F[x]/ \langle p(x) \rangle$ by Latin letters like $v \in E$ instead of $[v]_{\langle p(x) \rangle }$ and elements $\mu(a)$ as $a$ instead of $a\cdot 1$. It is nothing gained by a rigorous notation here, so it's easier to drop all the extra $\mu \, , \, [\,.\,]_{\langle p(x) \rangle} \, , \, \cdot 1 , \cdot x , \ldots , \cdot x^{\deg p -1}$.

It's similar to what we do with complex numbers: $a + i\cdot b$ is far more convenient than $a\cdot 1 + b \cdot x$ or $\mu(a) + \mu(b)x$ or even $a \cdot [1]_{\langle x^2+1 \rangle} + b \cdot [x]_{\langle x^2+1 \rangle}$. As can be seen here, $F=\mathbb{R}$ would have be to mentioned additionally anyhow. Imagine we would always have to speak of representatives of cosets in $\mathbb{R}[x]/{\langle x^2+1 \rangle}$ instead of real numbers in $\mathbb{C}$.