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My question concerns Kronecker's theorem and Extension fields.

Kronecker's Theorem:

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.

The proof of Kronecker's theorem is given as follows:

Since $F[x]$ is a unique factorization domain, $f (x)$ has an irreducible factor, say, $p(x)$. Clearly, it suffices to construct an extension field $E$ of $F$ in which $p(x)$ has a zero. Our candidate for $E$ is $F[x]/ \langle p(x) \rangle$. Due to a previous theorem, this is a field. Also, since the mapping of $\phi: F \rightarrow E$ given by $\phi(a) = a + \langle p(x) \rangle$ is one-to-one and preserves both operations, $E$ has a subfield isomorphic to $F$. We may think of $E$ as containing $F$ if we simply identify the coset $a + \langle p(x) \rangle$ with its unique coset representative a that belongs to $F$.

Then to show $p(x)$ has a zero in $E$, we substitute in $x + \langle p(x) \rangle$ into $p(x)$ and get $p(x) + \langle p(x) \rangle$ = $0 + \langle p(x) \rangle$.

My question is: This does not show that $F[x] / \langle p(x) \rangle$ is an extension field of $F$, only that there's a homomorphic image of $F$ that's a subfield of $E$. Because by the definition of an extension field:

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 $F$.

But $F$ is not a subset of $F[x] / \langle p(x) \rangle$ because elements of $F$ are just the elements while elements of $F[x] / \langle p(x) \rangle$ are of the form $g(x) + \langle p(x) \rangle$, which are cosets. Can someone explain to me what I'm missing here?

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    $\begingroup$ The $g(x)$ which are constants, i.e. no $x$ terms are the $F$ inside of them. $\endgroup$ – Adam Hughes Sep 20 '16 at 20:23
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    $\begingroup$ Yes, but you asked how $F\subseteq E$, it is embedded via the constants. But the embedding is just an injection, it's not saying everything is a constant. $\endgroup$ – Adam Hughes Sep 20 '16 at 20:38
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    $\begingroup$ that's fair, technically $c$ here is represented by $c+(p(x))$. It's common to consider the cosets just by their representatives, however, and this is likely the intent even if the language is a bit abused. $\endgroup$ – Adam Hughes Sep 20 '16 at 21:16
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    $\begingroup$ I would say $F$ is canonically embedded in $E$, which is usually what one means when they say $F\subseteq E$ in this context. It means there is an obvious way to put $F$ into $E$, in this case $c\mapsto c + (p(x))$ is the embedding. $\endgroup$ – Adam Hughes Sep 20 '16 at 21:32
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    $\begingroup$ I said in the last post "it means there is an obvious way..." $\endgroup$ – Adam Hughes Sep 20 '16 at 21:40
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We have a canonical injective morphism of fields $$F\longrightarrow F[x]/\langle p(x)\rangle$$ given by $f\mapsto f+\langle p(x)\rangle$. This exhibits $F$ as a subfield of $F(x)/\langle p(x)\rangle$.

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  • $\begingroup$ But what element of $F$ is an element of $F[x] / \langle p(x) \rangle$? $\endgroup$ – Oliver G Sep 20 '16 at 20:23
  • $\begingroup$ Pedantically speaking the restriction of the projection displays $F$ as isomorphic to but not equal to a subfield of $F[x]/(p(x))$ and that's the OP"s concern. $\endgroup$ – Rob Arthan Sep 20 '16 at 20:23

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