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Suppose we define an irreducible polynomial in the following way:

Let $P(X) \in \{K[x] \text{ of degree at least 1}\}$. $P(x)$ is irreducible over the ring $K$ iif: $$ \forall A(x), B(x)\in K[x]:A(x)B(x) = P(x) \implies (\text{deg}(A)=0 \text{ } \lor \text{deg(B)}=0) $$

Now consider the following test question:

Every polynomial of degree 1 from $K[x]$ is irreducible. Is this statement true or false?

G. Chiusole
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  • $K$ is a ring, not a field (as the letter would suggest)? – Hagen von Eitzen Dec 15 '17 at 11:13
  • @HagenvonEitzen I don't know what the $K$ stands for, but it is indeed a ring. –  Dec 15 '17 at 11:53
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    I would have $R$ or $A$ for a ring, whereas $K$ (and $F$, of course) is often used for fields. While $F$ for fields is obvious, the use of $K$ seems to origin form the German word "Körper". – Hagen von Eitzen Dec 15 '17 at 22:05
  • @HagenvonEitzen Thank you, "Körper" explains a lot. I just realized that since I translated the definition from Czech, where the word "kruh" means "circle", and "okruh" means "ring", the $K$ is for the Czech okruh. –  Dec 16 '17 at 09:12
  • @Josef That may be a coincidence. "Körper" does not mean "field" or "ring" in German, it means "body". It just happens that they chose a different name for the same algebraic structure. These names are fairly arbitrary in most languages, the algebraic structure does not much resemble the day to day meaning of field or body. – badjohn Apr 12 '19 at 09:39

1 Answers1

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The claim is true if $K$ is a field or at least an integral domain. For general rings $K$, however, the claim is false.

By a suitable choice of $K$, we can find a degree 1 polynomial $P(x)$ that is the product of degree 1 polynomials $A(x)=3x+1$ and $B(x)=2x+1$, for example. You may object that this would make $A(x)B(x)=6x^2+5x+1$ of degree $>1$. But not so if $K=\Bbb Z/6\Bbb Z$!

Hagen von Eitzen
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  • Thank you for your explanation! By the way, your the $6x$ in your equation is missing an exponent. –  Dec 15 '17 at 11:59
  • See also https://math.stackexchange.com/a/2539517/589 for a linear polynomial that decomposes as a product of two quadratic polynomials – lhf Dec 15 '17 at 12:53
  • @Josef Thanks, fixed – Hagen von Eitzen Dec 15 '17 at 22:02
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    I think it is worth mentioning that the definition in the question is (to my knowledge) not the standard definition of an irreducible polynomial. The usual definition requires the stronger condition that $A$ or $B$ is a unit in $K[X]$ and hence a unit in $K$ (which implies that the degree is $0$). However, when using this definition, the result only holds when every non-zero element in $K$ is a unit i.e. when $K$ is a field – G. Chiusole Apr 12 '19 at 09:19