# I Field Extensions - Lovett, Theorem 7.1.10 ... ...

1. May 1, 2017

### Math Amateur

I am reading Abstract Algebra: Structures and Applications" by Stephen Lovett ...

I am currently focused on Chapter 7: Field Extensions ... ...

I need help with an aspect of the proof of Theorem 7.1.10 ...

Theorem 7.1.10, and the start of its proof, reads as follows:

In the above text from Lovett we read the following ...

" ... ... Let $p(x)$ be a polynomial of least degree such that $p( \alpha ) = 0$ ... ... "

Then Lovett goes on to prove that $p(x)$ is irreducible in $F[x]$ ... ...

... BUT ... I am confused by this since it is my understanding that if $p( \alpha ) = 0$ then $p(x)$ has a linear factor $x - \alpha$ in $F[x]$ and so is not irreducible ... ... ?

Peter

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2. May 1, 2017

### Staff: Mentor

You can use polynomial division here. One can always write $p(x) = q(x) \cdot (x-\alpha) + s(x)$ with $\deg s(x) < \deg p(x)$.
Can you conclude the rest from here?

3. May 1, 2017

### Math Amateur

If $p(x) = q(x) \cdot (x-\alpha) + s(x)$ and $p( \alpha ) = 0$ then we must have $s( \alpha) = 0$

Given this ... I do not know how to square this with D&F Proposition 9, Section 9.4, page 307 which seems to imply that if $p ( \alpha ) = 0$ for $\alpha \in F$ then $p(x)$ has a linear factor ... and is therefore not irreducible ...

Peter

4. May 1, 2017

### Staff: Mentor

$p(x)$ has been chosen to be a (non-zero) polynomial of minimal degree with $p(\alpha)=0$. But now $s(\alpha)=0$ and $\deg s(x) < \deg p(x)$ which can only hold, if $s(x) = 0$. Going back to the equation we have $p(x) = q(x) \cdot (x-\alpha)$ or $(x-\alpha) \mid p(x)$.

5. May 1, 2017

### Math Amateur

Well ... we seem to have shown that $p(x)$ is reducible as a consequence of $p( \alpha ) = 0$ ...

Exactly my problem with this ...

... hmmm ... how are we going to end up with $p(x)$ being irreducible when it has a linear factor ... ???

Peter

6. May 1, 2017

### Staff: Mentor

The equation $p(x)= q(x)\cdot (x-\alpha) + s(x)$ doesn't exist in $F[x]$, because $\alpha \notin F$. It only exists in $F[\alpha][x]$.
Lovett shows that $p(x)\in F[x]$ is irreducible, because $\alpha$ isn't available in $F$. However, it is reducible in $F[\alpha][x]\;\;\;\;$ so the argument with the linear factor (which we have shown is true) only holds over $F[\alpha]$. Irreducibility depends on the ring under consideration, or likewise on the coefficients which are available in the corresponding domain of scalars.

$p(x) \in F[x]$ irreducible (Lovett's proof)
$p(x) \in F[\alpha][x]$ reducible, by your argument with the linear factor $x-\alpha$

Sorry for the confusion, which I think I might have caused, by proving your assertion with the linear factor, instead of recognizing, that there are two different rings involved here: $F[\alpha][x]$ and $F[x]$, which makes the difference.

7. May 1, 2017

### Math Amateur

Thanks so much for the help in clarifying things ... really has been most helpful ...

Peter