Let $\mathbb R$ be the reals and $\mathbb R[x]$ be the polynomial ring of one variable with real coefficients. Let $I$ be the principal ideal $(x(x^2+1))$. I want to prove that the ideal of the ideals variety is not the same as its radical, that is, $I(V(I))\not=\text {rad}(I)$. I've reduced this to proving that $I=\text{rad}(I)$. How can I go about that?
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2 Answers
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This $x(x^2 + 1)$ is the factorisation of that polynomial into irreducibles (over $\mathbb{R}$). Once you have such a factorisation, the radical is the factorisation with no powers. So it is radical.
EDIT — I am referencing this:
Let $f \in k[x]$ be a polynomial, and suppose that $f = f_1^{\alpha_1} \cdots f_n^{\alpha_n}$ is the factorisation of $f$ into irreducibles. Then $\sqrt{(f)} = (f_1 \ldots f_n)$.
One inclusion should be clear, and the other can be seen by appealing to the uniqueness of factorisation in $k[x]$.
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$\begingroup$ Your answer makes very little sence to me. Can you expound on it? $\endgroup$ Jun 8, 2013 at 14:06
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$\begingroup$ i have edited my answer to be more explicit, i hope this helps $\endgroup$ Jun 8, 2013 at 14:35
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$\begingroup$ Thank you, this helps, but what do you mean by $\sqrt {(f)}$? $\endgroup$ Jun 8, 2013 at 15:23
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$\begingroup$ it's an alternative notation for the radical of an ideal, some people prefer it $\endgroup$ Jun 8, 2013 at 15:24
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$\begingroup$ Is it true for any ideal of the polynomial ring of n variables over a field, that it is equal to its radical? I feel convinced now that it is true for any principal such ideal. I'm inclined to think it is also true for finitely generated ideals. $\endgroup$ Jun 8, 2013 at 15:52
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Let $f\in{\rm rad}( I)$, that is, $f^n\in I$. You can also think over the complex field: $f=a_n(x-z_1)(x-z_2)\dots (x-z_n)$ for some $z_j\in\Bbb C$. Then $f^n\in I$ means that among the complex roots of $f^n$ we also find the roots of $x(x^2+1)$, i.e., $-i,0,i$. But that means we already had $f\in I$.
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$\begingroup$ Does the same reasoning hold true for any ideal? Which ideals admit to the same argument? (It obviously does for any principal ideal.) $\endgroup$ Jun 8, 2013 at 15:21
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$\begingroup$ No, for example ${\rm rad}\big((x^2-2x+1)\big)=(x-1)$. Check out also the other answer. $\endgroup$– BerciJun 8, 2013 at 21:56