I want to show that $f(x)=x^3-9x^2+6x-12$ is reducible over $\mathbb{R}$. Since $\mathbb{R}$ is a field I know that if $f$ has a root in $\mathbb{R}$ then $f$ has a linear factor and hence it is reducible. My way of checking this seems like "cheating" because we don't talk about the notion of continuity in ring theory. One can easily show that $f(0)<0$ and $f(3)>0$, and since polynomials are continuous, the intermediate value theorem guarantees that $f$ has a root in $\mathbb{R}$. How could I show this in another way? Thanks!
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4There's nothing wrong with using continuity to show the required result. – Kenny Lau Jun 02 '17 at 15:31
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The only irreducible polynomial over $\mathbb{R}$ is of degree 1 or 2, since $[\mathbb{C} : \mathbb{R}]=2$ . Also every odd degree polynomial has atleast one real root... – Chinnapparaj R Jun 02 '17 at 15:33
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the real root of the polynomial is $x=8.458372475$! – Amin235 Jun 02 '17 at 15:33
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Actually $f(3)<0$, but $f(9)>0$. Using continuity is fine. – Robert Z Jun 02 '17 at 15:34
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1If you don't talk about the notion of continuity, you aren't talking about $\mathbb{R}$. A defining feature of $\mathbb{R}$ is its completeness, and without it I think there's no way to show that $f(x)$ is reducible over $\mathbb{R}$. – MaudPieTheRocktorate Jun 02 '17 at 15:35
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You are essentially in need of the fundamental theorem of algebra, and there is no way around the real/complex analysis involved since $\mathbb R$ has completeness baked into it. – rschwieb Jun 02 '17 at 19:48
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Irreducible polynomials over the real numbers have degree $1$ or $2$, see here. So any real cubic polynomial is reducible.
Dietrich Burde
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If you don't talk about the notion of continuity, you aren't talking about $\mathbb{R}$. A defining feature of $\mathbb{R}$ is its completeness, and without it I think there's no way to show that $f(x)$ is reducible over $\mathbb{R}$.
Suppose you managed to show $f(x)$ is reducible over $\mathbb{R}$, without using completeness of $\mathbb{R}$, then the same argument would work if you replace each with $\mathbb{R}$ with $\mathbb{Q}$. The reason is that the axioms for the real numbers are the same as those for the rational numbers, other than the axiom of completeness. So any argument about $\mathbb{R}$ that doesn't use completeness must work for $\mathbb{Q}$ as well.
Since $f(x)$ is irreducible over $\mathbb{Q}$, there's no way to prove $f(x)$ is reducible over $\mathbb{R}$ without using its completeness.
MaudPieTheRocktorate
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Cardano's method gives nice numbers for this one. the real root is $$ 3 + \sqrt[3]{24 + \sqrt{233}} + \sqrt[3]{24 - \sqrt{233}} \; \; \approx \; \; 8.458372474 $$
Will Jagy
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The continuity argument is the easiest. However here is another one: Your polynomial $P$ has three roots in $\mathbb{C}$, if $\lambda$ is a root which is not real, then $\bar{lambda}$ too, because $P(\bar{\lambda}) = \bar{P(\lambda)}$. So $P$ = $(X - \lambda)(X- \bar{\lambda})(X-\mu)$. By identifying the degree 2 coefficient, you have $-\mu - lambda - \bar{\lambda} \in \mathbb{R}$ and so $\mu \in \mathbb{R}$.
x4rkz
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