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If we take a polynomial ($x^4+1$, for instance) I do not see the difference in factoring this polynomial into irreducible polynomials in $\mathbb{C}[x]$ and $\mathbb{R}[x]$. Using this concrete example, can you show this difference?

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Factor $X^2+1$ into irreducibles in $\mathbb R[X]$ and in $\mathbb C[X]$. Do you see a difference? Once you have done this, do your example. –  Mariano Suárez-Alvarez Nov 28 '11 at 6:34
    
Isn't $x^2+1$ already irreducible? –  johnnymath Nov 28 '11 at 6:59
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irreducible over what field? –  Mariano Suárez-Alvarez Nov 28 '11 at 7:10

4 Answers 4

Over $\mathbb C[x]$, since $\mathbb C$ is algebraically closed, every polynomial factors as a product of linear factors, i.e. any polynomial $p(x) \in \mathbb C[x]$ of degree greater than $1$ can be written as $$ p(x) = \alpha_p \prod_{i=1}^n (x-\beta_i) $$ where $\alpha_p$ is just any non-zero complex number and the $\beta_i$ are the roots of $p$ (possibly not distinct). Now suppose $p(x) \in \mathbb R[x]$. Since real numbers are also complex, we can factor $p$ as above and get the complex roots of $p$. But you have to notice something really important : if $\beta_i$ is a root of $p$, since $p$'s coefficients are real, they are fixed under complex conjugation, and therefore $\overline{p(\beta_i)} = p \left( \overline{\beta_i} \right) = 0$ (the bar denotes complex conjugation). This means that there are two possibilities : either a root is real, or it is complex (but not real) and its complex conjugate is another distinct root!

If you compute the following : $$ (x-\beta)(x-\bar{\beta}) = x^2 -(\beta+\bar{\beta})x + \beta \bar{\beta}, $$ note that the coefficients of this polynomial are real : thus, if you take the factoring of $p$ up there, you'll notice that any polynomial $p(x)$ of degree greater or equal to $1$ in $\mathbb R[x]$ can be written as $$ p(x) = \alpha_p \left( \prod_{i=1}^n (x-\beta_i) \right) \left( \prod_{j=1}^m (x^2 + \gamma_j x + \delta_j) \right) $$ where $\gamma_j^2 - 4 \delta_j <0$ (i.e. those quadratics are irreducible factors of $p$) and the idea is simple : use the factorization of $p(x)$ over $\mathbb C$ and pair up the non-real roots with their complex conjugates, and expand those factors to get the quadratics.

Hope that helps!

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Note that $$x^4+1=(x^2-\sqrt{2}\,x+1)(x^2+\sqrt{2}\,x+1).$$ The two quadratics are irreducible over $\mathbb{R}$, since they have no real roots.

The two quadratics are not irreducible over $\mathbb{C}$. Indeed no polynomial $P(x)$ of degree $\ge 2$ with coefficients in $\mathbb{C}$ is irreducible over $\mathbb{C}$, since by the Fundamental Theorem of Algebra $P(x)$ has a root $\alpha$ in $\mathbb{C}$, and therefore is divisible by $x-\alpha$.

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HINT $\rm\ x^4+1\ = (x^2+1)^2 - 2x^2 = (x^2+1 -\sqrt{2} x)\ (x^2 + 1 + \sqrt{2}\ x) $ and each factor is irreducible over $\mathbb R$ since discriminant $= -2$; but it splits into linear factors over algebraically closed field $\mathbb C\:.$

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A fourth degree polynomial cannot be irreducible over R ;) –  N. S. Nov 28 '11 at 6:48
    
@N.S. Alas, I originally misread the question as over $\mathbb Q$ vs. $\mathbb R\:.$ –  Bill Dubuque Nov 28 '11 at 7:33

Since $\frac1{\sqrt2}(1+i)$ is a non-real complex root of you polynomial, $x-\frac1{\sqrt2}(1+i)$ is an irreducible factor of it in $\mathbb{C}[x]$; however this factor does not lie in $\mathbb{R}[x]$, so it is not even a candidate for being an irreducible factor of $x^4+1$ in $\mathbb{R}[x]$. So more generally, as soon as a polynomial has a non-real complex root, its factorization into irreducible factors in $\mathbb{R}[x]$ cannot be the same as its factorization in $\mathbb{C}[x]$.

You might wonder if the fact that irreducible factors are unique only up to a nonzero scalar multiple casts any doubt on this conclusion. It does not, since if $a\in\mathbb{C}\setminus\mathbb{R}$ then no complex multiple of $x-a$ lies in $\mathbb{R}[x]$: in order for the leading term to become real the multiplicative factor must be real, but then the constant term $-a$ is transformed into another non-real complex number.

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