In the book Contemporary Abstract Algebra by Gallian it defines an extension field as follows:

A field $E$ is an extension field of a field $F$ if $F\subseteq E$ and the operations of $F$ are those of $E$ restricted to $F$.

Question 1) When it says "and the operations of $F$ are those of $E$ restricted to $F$" is this equivalent to saying "and the operations of $F$ are those of $E$ such that $F$ is a subfield"? Is this what is meant by "restricted to $F$"?

Also, Kronecker's threorem states:

Let $F$ be a field and let $f(x)$ be a nonconstant polynomial in $F[x]$. Then there is an extension field $E$ of $F$ in which $f(x)$ has a zero.

Question 2) I know that in the theorem above $E=F[x]/\left<p(x)\right>$ where $p(x)$ is an irreducible factor of $f(x)$ and that the field $E=F[x]/\left<p(x)\right>$ contains an isomorphic copy of $F$. But why doesn't the definition of extension fields take into account up to an isomorphism (since technically $F$ is not a subset of $E$ in this case) when we talk about a field $E$ being an extension of a field $F$?

Can't we define extension fields say as: An extension field of a field $F$ is a pair $(E,\phi)$ such that $\phi$ is a homomorphism from $F$ to $E$ with $\phi(F)\subseteq E$ and the operations of $\phi(F)$ are those of $E$ restricted to $\phi(F)$?


Question 1) Yes, the definition in Gallian is equivalent to "$F$ is a subfield of $E$." When he says that the operations of $F$ are the operations of $E$ restricted to $F$, that does indeed mean that you can add/subtract/multiply elements of $F$ just by viewing them as elements of $E$ and performing the same operations.

Question 2) I like that you are being picky here, and making sure that in the definition is indeed satisfied. Although Gallian does not explicitly say so here, you can identify $F$ with the set of cosets of constant polynomials, and now you really do have a copy of $F$ as a subfield of $E$.

Your alternative definition looks pretty good, but you need specify that $\phi$ is a nonzero homomorphism (unless you require homomorphisms to send 1 to 1, in which case this is already guaranteed). Also, since you have a homomorphism, the operations on $\phi(F)$ are automatically those inhereted from $E$.

  • $\begingroup$ +1:Thank you for you response. For question 2) I guess my dilemma is that even though you really do have a copy of $F$ contained in $E$, the underlying reasoning is still that of an isomorphism and I guess what I want to know is why is the definition not tailored to fit Kronecker's theorem? Also I guess I wasn't very precise in question 1) and I was wondering, does the phrase "restricted to $F$" imply the closure of $F$? because you said that the operations of $\phi(F)$ are automatically inherited from $E$, but does this imply $\phi(F)$ is closed under the operations of (continued below) $\endgroup$ – tcmtan Oct 29 '12 at 23:42
  • $\begingroup$ (continued) $E$ when we say restricted to $\phi(F)$? $\endgroup$ – tcmtan Oct 29 '12 at 23:43
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    $\begingroup$ You're right about question 2, but the isomorphism here is so canonical (ie, there's an obvious choice of what isomorphism to pick and it works well) that we usually just identify $F$ with $\phi(F)$ without really thinking about it. This sort of abuse of language keeps us from getting bogged down in unnecessary semantics. $\endgroup$ – Brett Frankel Oct 30 '12 at 0:23
  • $\begingroup$ As for your comments about being $\phi(F)$ being closed, the phrase "restricted to $F$" does not imply that the operations are closed. But when we assert that $F$ is a field with these operations, then that implies the operations are closed in $F$. $\endgroup$ – Brett Frankel Oct 30 '12 at 0:26

The main thing is that every homomorphism between fields is an embedding (injective). On one hand it means that your definition is correct (assuming that homomorphisms take $1\mapsto 1$, as Brett said), and yes, more precise. On the other hand, it means that if we have a homomorphism $K\to L$, then, basically we can assume that $K\subseteq L$ is a subfield (in this step we identify $K$ with its image), and this viewpoint has some advantages, at least in simplifying the notations.


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