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Show that $Y^2 + X^2(X+1)^2$ is irreducible over $\mathbf R$.

Are there some general tricks for avoiding barbaric computations in general case?

Lord_Farin
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  • Please ensure that your question is fully understandable without the title. The title's purpose is to attract people to your question (and to make it easily searchable). Once it has done that, it's job is done. The question should be self-contained. I've [edit]ed it for you now. – Lord_Farin Dec 31 '14 at 14:06
  • There are some tricks, but there's no boilerplate solution in general. Here, for instance, you can plug in $ X=1$ and see what you get. What does that tell you about how the polynomial must factor, if it does? – Dustan Levenstein Dec 31 '14 at 14:12

1 Answers1

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If it were reducible it would factor in two non-unit factors. Either both of them are of degree 1 in $Y$ or one of them is of degree zero in $Y$ and the other is of degree 2 in $Y$.

In the second case we see that the factor that is of degree zero in $Y$ must divide $Y^2$. Therefore it must be a unit, which is a contradiction.

Assume then that the two factors are of degree one in $Y$. So

$$Y^2+X^2(X+1)^2=(A(X)Y+B(X))(C(X)Y+D(X))=A(X)C(X)Y^2+(A(X)D(X)+C(X)B(X))Y+B(X)D(X)$$

From this $A(X)$ and $C(X)$ are units. So, we can assume they are $1$. We get

$$Y^2+X^2(X+1)^2=(Y+B(X))(Y+D(X))=Y^2+(B(X)+D(X))Y+B(X)D(X)$$

From this $B(X)+D(X)=0$ and $X^2(X+1)^2=B(X)D(X)$. Therefore

$$X^2(X+1)^2=-B^2(X).$$

Therefore,

$$1=-B^2_{0}$$

where $B_0$ is the leading term of $B(X)$. But there is no such $B_0$ in the reals, who's square is $-1$.

Pp..
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  • For the final step, just observe that $B(1)^2=-4$, without the need of considering the leading coefficient. – egreg Dec 31 '14 at 14:45
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    @egreg I don't think that improves in any way the argument. It rather makes it more complicated. Depending on the definition of polynomial one could have to explain what is $B(1)$. No thanks. I will just look at the leading coefficient. Moreover 1 is prettier, it is in every field and 4 is unlucky for Chinese people. – Pp.. Dec 31 '14 at 14:50
  • From $Y^2+X^2(X+1)^2=(Y+B(X))(Y+D(x))$ you can set $Y=-B(X)$ so that the right hand side is identically zero as a polynomial in $X$ and the left-hand side is the sum of two squares with $X^2(X+1)^2$ having values greater than zero. This method of exploiting a root of the equation is more general. – Mark Bennet Dec 31 '14 at 15:19
  • @MarkBennet This putting $Y=-B(X)$ is equivalent to comparing coefficients. There is no more here, no improvement. I wouldn't argue that $X^2(X+1)^2+B^2(X)\neq0$ by saying that $X^2(X+1)^2>0$. I would compare coefficients. No reference to 'values' needed. – Pp.. Dec 31 '14 at 15:24
  • I am suggesting the method as a short cut in the general case to which the original question refers. It would, for example, work also with a cubic in $Y$, where a linear factor could be tested. I put it in because I think it is useful to know in addition to your suggested answer. – Mark Bennet Dec 31 '14 at 15:31