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# irreducible polynomial

Let $f(x)=a_{0}\!+\!a_{1}x\!+\cdots+\!a_{n}x^{n}$ be a polynomial with complex coefficients $a_{{\nu}}$ and with the degree $n>0$. If $f(x)$ can not be written as product of two polynomials with positive degrees and with coefficients in the field $\mathbb{Q}(a_{0},\,a_{1},\,\ldots,\,a_{n})$, then the polynomial $f(x)$ is said to be irreducible. Otherwise, $f(x)$ is reducible.

Examples. All linear polynomials are irreducible. The polynomials $x^{2}\!-\!3$, $x^{2}\!+\!1$ and $x^{2}\!-\!i$ are irreducible (although they split in linear factors in the fields $\mathbb{Q}(\sqrt{3})$, $\mathbb{Q}(i)$ and $\mathbb{Q}(\frac{1\!+\!i}{\sqrt{2}})$, respectively). The polynomials $x^{4}\!+\!4$ and $x^{6}\!+\!1$ are not irreducible.

The above definition of irreducible polynomial is special case of the more general setting where $f(x)$ is a non-constant polynomial in the polynomial ring $K[x]$ of a field $K$; if $f(x)$ is not expressible as product of two polynomials with positive degrees in the ring $K[x]$, then $f(x)$ is irreducible (in $K[x]$).

Example. If $K$ is the Galois field with two elements (0 and 1), then the trinomial $x^{2}\!+\!x\!+\!1$ of $K[x]$ is irreducible (because an equation $x^{2}\!+\!x\!+\!1=(x\!+\!a)(x\!+\!b)$ would imply the two conflicting conditions $a\!+\!b=1$ and $ab=1$).

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12D10*no label found*

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## Recent Activity

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

## Comments

## field notation

I have quite a remote idea what the notation

$\mathbb{Q}(a_0, a_1, ..., a_n)$

actually mean. So, could you possibly write that explicitly? I don't know how frequent is its ussage on PM, but if it is frequent then may be it is a good idea to make a separate entry with explanation, and make a link to it here.

Serg.

-------------------------------

knowledge can become a science

only with a help of mathematics

## Re: field notation

$ F = K(a_0, a_1, ..., a_n)$ is the field obtained when one has adjoined to the field K the elements a_0, a_1, ..., a_n (which need not to belong to K but belong to some extension field of K). The procedure is field adjunction. In other words, F consists of all elements which can be obtained of the elements of K and the additional elements a_j via the four fundamental operations. One could also use all rational functions f(X_0, X_1, ..., X_n) with coefficients in K and then substitute

X_j :=a_j for all j (cf. substitution homomorphism). Maybe I shall make an entry.

Regards,

Jussi

## Re: field notation

> The procedure is field adjunction.

Great, thanks a lot! I had a vague feeling that this is what is meant here ;)

> Maybe I shall make an entry.

It would be defenitely very nice! Probably such a field has some specific name, like adjuction field or may be extended field. Thus, having a corresponding entry, one could link to it just by naming Q(...) it correspondingly, e.g. an extended field.

Thanks once more!

Regards

Serg.

-------------------------------

knowledge can become a science

only with a help of mathematics

## Re: field notation

I'm not sure, but there might actually be a "Notation in algebra" or "Notation in number theory" entry floating around here somewhere. That would make a good addition.

Cam

## Re: field notation

> there might actually be a

> "Notation in algebra" or "Notation in number theory"

> entry floating around here somewhere

By making a quick search I have found two entries which could be related to the issue: "composite field" and "extension field". When designing the entry they could be good candidates for "See also" field.

Serg.

-------------------------------

knowledge can become a science

only with a help of mathematics

## Odd browser fault

Hi experts! My Mozilla Firefox (1.5) has behaved for 2 weeks oddly: it does not show any PM entry (except their TeX sources) at all; the same concerns the MS Exploder on my computer. I cannot fix the fault. Please tell what is the name of the fault and how I could fix it.

Quite funny that my Firefox can percectly show the PlanetPhysics entries!

Regards,

Jussi

## Re: Odd browser fault

You really need to give more information. You say it won't show the PlanetMath entries, but what does it show instead? (Is the sidebar still visible? Is there an error message? If so, what does the error message say? Etc.)

It's unlikely to be a browser problem if it affects both Firefox and Internet Explorer.

The images are on port 8080. Perhaps your ISP has started blocking this port (though this would be a stupid thing for it to do).

> Quite funny that my Firefox can percectly

> show the PlanetPhysics entries!

PlanetPhysics doesn't use port 8080.

## Re: Odd browser fault

It may be related to the problem I also have, where the images on PM do not get displayed. It seems that it cannot connect to images.planetmath.org. If everyone else can see them and they are on port 8080, then maybe my ports are being block, ouch!

Ben

## Re: Odd browser fault

> You really need to give more information. You say it won't show the

> PlanetMath entries, but what does it show instead? (Is the sidebar

> still visible? Is there an error message? If so, what does the error

> message say? Etc.)

No error messages.

In the page images mode, Firefox shows the title of e.g. your entry "core of a subgroup" and its _empty_ frame with only the default line "... is owned by yark"; below the frame it shows all usual informations.

In the html mode the frame is'nt empty but contains only the "prose" texts but no symbols and no formulae.

In the TeX source mode, the frame contains all the normal content.

> The images are on port 8080. Perhaps your ISP has started blocking

> this port (though this would be a stupid thing for it to do).

I know of 8080 and ISP equally much as a pig of a windmill.

May the fault have something to do with the Windows "profile"?