# I Kronecker's Theorem - Anderson and Feil, Theorem 42.1

1. Apr 26, 2017

### Math Amateur

I am reading Anderson and Feil - A First Course in Abstract Algebra.

I am currently focused on Ch. 42: Field Extensions and Kronecker's Theorem ...

I need some help with an aspect of the proof of Theorem 42.1 ( Kronecker's Theorem) ...

Theorem 42.1 and its proof read as follows:

In the above text by Anderson and Feil we read the following:

" ... ... We show that there is an isomorphism from $F$ into $F[x] / <p>$ by considering the function $\psi \ : \ F \longrightarrow F[x] / <p>$ defined by $\psi (a) = <p> + a$, where $a \in F$. ... ... "

Te authors show that $\psi$ is one-to-one or injective but do not show that $\psi$ is onto or surjective ...

My question is ... how do we know that $\psi$ is surjective ...

... for example if a polynomial in $F[x]$, say $f$, is degree 5, and $p$ is degree 3 then dividing $f$ by $p$ gives a polynomial remainder $r$ of degree 2 ... then $r + <p>$ will not be of the form $<p> + a$ where $a \in F$ ... ... and so it seems that $\psi$ is not surjective ... since the coset of $f$ is not of the form $<p> + a$ where $a \in F$ ...

... ???

Obviously my thinking is somehow mistaken ...

... can anyone help by demonstrating that $\psi$ is surjective ... and hence (given that Anderson and Feil have demonstrated it is injective) an isomorphism ...

Help will be appreciated ...

Peter

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2. Apr 26, 2017

### andrewkirk

$\psi$ is onto because an arbitrary element of $F[x]\ /\ \langle p\rangle$ has the form $q+\langle p\rangle$ for some polynomial $q$ in $F[x]$. We note that, by definition, $\psi(q)=q+\langle p\rangle$ so that $q+\langle p\rangle\in Im\ \psi$.

I don't quite see why you are dividing $f$ by $p$. Are you mixing up the quotient sign $/$ with a division sign? In your example $r+\langle p\rangle$ is of the form $a+\langle p\rangle$ because we can use $r$ as $a$ since $r\in F[x]$. Similarly, $f+\langle p\rangle$ is of the form $a+ \langle p\rangle$ by setting $a=f$.

3. Apr 26, 2017

### Stephen Tashi

As I read the text, $\psi$ is a isomorphism from $F$ into $E = F[x]/<p>$, not onto $E$. We don't want $F$ to be isomorphic to all of $E$ because $E$ is supposed to be an extension field of $F$ that contains an element not in $F$.

4. Apr 26, 2017

### Staff: Mentor

$\psi$ is not surjective. $F \hookrightarrow F[x]/\langle p(x) \rangle$ is a proper embedding for otherwise $p(x)$ would have to be linear.
The notation "$\psi$ maps $F$ isomorphic into $F[x]/\langle p(x) \rangle$" is a bit sloppy for "$\psi$ maps $F$ isomorphic onto its image in $F[x]/\langle p(x) \rangle$", i.e. is a monomorphism, an injective homomorphism.

5. Apr 26, 2017

### Math Amateur

Thanks Andrew ...

You write ...

" ... ... I don't quite see why you are dividing $f$ by $p$. ..."

Well ... we have that $f, p \in F[x]$ with $f$ being of degree 5 and $p$ of degree 3 ...

Suppose after polynomial division we have $f = qp + r$ ... then I'm assuming that the coset of $f$ in $F[x]$ is $r + <p>$ .. ... and $r$ may well be a degree 2 polynomial (depending on the division) and hence not equal to any $a \in F$ (that is a zero degree polynomial in F[x] ...)

Hope my meaning is now clear ... let me know what you think ...

I am now thinking that $\psi$ is not an isomorphism ... see Stephen's post ... I read the theorem rather carelessly ...

Peter

6. Apr 26, 2017

### Math Amateur

Thanks to fresh_42 and to Stephen for clarifying the matter ... I appreciate your help ...

Peter

7. Apr 26, 2017

### andrewkirk

Sorry. I misread the text as saying that the domain of $\psi$ was $F[x]$ when in fact it said it was $F$.

I really must get my eyes checked.

8. Apr 27, 2017

### Math Amateur

No problems Andrew ... you have helped me with many problems ...

Peter

9. Apr 27, 2017

### Staff: Mentor

By the way, this is the reason I like the (old fashioned) words "monomorphism" for injective homomorphisms and "epimorphism" for surjective homomorphisms. It reserves the word "isomorphism" for bijective homomorphisms and keeps the road clear.

10. Apr 27, 2017

### andrewkirk

I wouldn't call that old-fashioned. If it is, I missed the update memo. To me, the use you describe is simply 'correct' and the way the authors have used the word 'isomorphism' is simply incorrect. The correct term was homomorphism. To convey injectivity one can say 'injective homomorphism' or 'monomorphism' or say that it is 'one-to-one'. But to call it an isomorphism is wrong.

11. Apr 27, 2017

### Staff: Mentor

Yes. The entire information with this - in your eyes wrong - notation is condensed in the words "into" and "onto". I think as well, that it is confusing if paired with otherwise reserved names. I called it sloppy, but that's a semantic discussion. But I find "one-to-one" not less confusing for injectivity, as it might be understood to hit all elements, as in the definition of sets with equal cardinality, a one-to-one correspondence so to say, which it is not if used for injectivity.

Another helpful way to express the situation is by the usage of certain arrows:
• $\hookrightarrow$ : canonical embedding (injective homomorphism)
• $\rightarrowtail$ : monomorphism (injective homomorphism)
• $\twoheadrightarrow$ : epimorphism (surjective homomorphism)
• $\cong$ or an overlay, resp. a combination of the two previous ones for which I haven't found a LaTex command (like \twoheadrightarrowtail): isomorphism (bijective homomorphism)
• $\leftrightarrow$ or $\stackrel{1:1}{\leftrightarrow}$ : bijective mapping (not necessarily a homomorphism)
One advantage of this is, that a short exact sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ can be written really short as $A \rightarrowtail B \twoheadrightarrow C$.