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.

  • $\begingroup$ Kronecker's Theorem seems to be something else. Could you give a link to the result and its proof you are working on? $\endgroup$ – Dietrich Burde May 29 '17 at 14:57
  • $\begingroup$ The latter means the mapping is onto (surjective), but isomorphisms are already surjective. $\endgroup$ – Sean Roberson May 29 '17 at 14:58
  • $\begingroup$ @DietrichBurde:Wait,i'm typing the complete proof. $\endgroup$ – P.Styles May 29 '17 at 14:59
  • $\begingroup$ @DietrichBurde There is a Kronecker's Theorem in field theory. $\endgroup$ – AspiringMathematician May 29 '17 at 14:59
  • $\begingroup$ PK, how do you define an isomorphism? If you grasp that, you've answered your own question. $\endgroup$ – Erik G. May 29 '17 at 15:00

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 $F$.

  • $\begingroup$ :Yeah,the text i'm reading is very old.I've cross checked the proof of this theorem from Gallian's text & Dummit & foote's text.In both texts proof is almost similar both of these texts stated about injectivity and operation preserving property of $\phi$ but the text i'm reading named it as "into isomorphism" specifically. $\endgroup$ – P.Styles May 29 '17 at 16:00
  • $\begingroup$ @PKStyles You should check where the book defines “isomorphism”; most probably it says “injective homomorphism”. This was very common until a few decades ago. $\endgroup$ – egreg May 29 '17 at 16:03
  • $\begingroup$ :There is no mention of injective homomorphism while defining the map but i think in place injective isomorphism it should be injecttive homomorphism. $\endgroup$ – P.Styles May 29 '17 at 16:07
  • $\begingroup$ @PKStyles Can you tell what book it is? $\endgroup$ – egreg May 29 '17 at 16:08
  • $\begingroup$ :Advanced course in Modern Algebra by Gupta & Gupta. $\endgroup$ – P.Styles May 29 '17 at 16:10

It could be just a matter of emphasis (but do see the comments below), so it would be more about writing style than mathematics. If $\phi$ is an isomorphism, then it is already both injective and surjective, but I would imagine that the author is thinking something like

  • Saying "$\phi$ is an isomorphism from $F$ into $F'$" emphasizes to the reader that the target of the map $\phi$ is $F'.$ Using the term into can be important when emphasizing that a map is well defined. Like you can declare that $F'$ is the codomain of $\phi$ all you want, but when you actually define what map $\phi$ does to elements of $F$, the result better be in $F'$. Sometimes there is a little work to showing that $\phi(x) \in F'$ for all $x \in F$, and saying that $\phi$ maps $F$ into $F'$ captures the idea that this was checked.

  • Saying "$\phi$ is an isomorphism from $F$ onto $F′$" emphasizes that the map is surjective, which is possibly the reason the author brought up the fact that $\phi$ is an isomorphism in that part of the argument.

  • 2
    $\begingroup$ Older texts use isomorphism for what we now use injective homomophism. It is not a matter of style, really. $\endgroup$ – Mariano Suárez-Álvarez May 29 '17 at 15:08
  • $\begingroup$ @Mike Pierce:Will you please explain "the target of the map $\phi$ is $F'$"? $\endgroup$ – P.Styles May 29 '17 at 15:32
  • $\begingroup$ @MarianoSuárez-Álvarez Oh, I had no idea. Then it become rather important where OP is reading this proof. Now I'm curious, how old do you mean when you say "older texts"? Like, are there texts that use isomorpism for injective homomorphism that are commonly referenced today? $\endgroup$ – Mike Pierce May 29 '17 at 15:34
  • $\begingroup$ @MikePierce:From first point, you meant that $\phi$ is an isomorphism from $F$ onto "something" inside $F'$? $\endgroup$ – P.Styles May 29 '17 at 15:50
  • 1
    $\begingroup$ @mike, this is quite common. Just google for "isomorphism into" (between quotes) in google, and go to books. Artin used it in his Geometric Algebra, Steenrod in his The topology of fiber bundles, and so on. $\endgroup$ – Mariano Suárez-Álvarez May 29 '17 at 16:19

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