Kronecker's theorem: Let $F$ be a field and $f(x)$ a nonconstant polynomial in $F[x]$. Then there is an extension field $E$ of $F$ in which $f(x)$ has zero.
Proof: Since $F[x]$ is a unique factorization domain, $f(x)$ has an
irreducible factor say $p(x)$. Now consider the ideal $\langle p(x)\rangle$ of the
ring of polynomials generated by $p(x)$. Since $\langle p(x)\rangle$ is irreducible
over $F$ then the ideal $\langle p(x)\rangle$ is a maximal ideal of
$F[x]$. Consequently, $F[x]/\langle p(x)\rangle$ is a field.Write $E=F[x]/\langle p(x)\rangle$.
Now we shall show that the field $E$ satisfies every part of the
theorem. Now consider the map$\phi:F\rightarrow E$ defined as $\phi (a)=a+\langle p(x)\rangle$.
This mapping is well defined as any element $a\in F$ can be regarded
as a constant polynomial in $F[x]$.$\phi$ is injective
for $\phi (a)=\phi(b)$
$\implies a+\langle p(x)\rangle=b+\langle p(x)\rangle$
$\implies a-b\in\langle p(x)\rangle$
$a-b=f(x)p(x)$ for some $f(x) \in F[x]$.
Since $\deg(p(x))\geq 1$ then $\deg(f(x)g(x))\geq 1$ while
$\deg(a-b)=0$. Hence, $f(x)=0$. Consequently, $a-b=0 \implies a=b$.Clearly, $\phi$ is an homomorphism.
Thus $\phi$ is an isomorphism from $F$ into $E$.
What is the difference between the phrases “$\phi$ is an isomorphism from $F$ into $F'$” and “$\phi$ is an isomorphism from $F$ onto $F'$”?
Actually, I'm proving Kronecker's theorem and the first phrase arose in the middle of the proof (last line). I got confused as it stated the first phrase just by showing $\phi$ is a homomorphism and injective without showing it is surjective.
Please clarify my doubt, if possible with the help of an example.
Best Answer
It's impossible to know without having a reference to the textbook or lecture notes you're reading.
Some texts use “isomorphism” where the more common terminology, nowadays, is “injective homomorphism” or “monomorphism”. Apparently, your textbook or lecture notes fall in this category. You should check where the book defines “isomorphism”; most probably it says the equivalent of “injective homomorphism”: this terminology was rather common until a few decades ago.
The first part proves that $\phi$ is an injective homomorphism and the second part proves surjectivity.
Actually, proving $\phi$ is injective is not needed: as soon as $\phi$ is a homomorphism that maps $1$ into $1$, it is automatically injective, because its kernel is an ideal and the ideals in a field $F$ are just $\{0\}$ and $F$.