Isn't $\mathbb Q[X]/(X^2+1)\cong \mathbb Q[i]$ wrong and should be $\mathbb Q[X]/(X^2+1)\cong \mathbb Q(i)$ ?
Indeed, $\mathbb Q[X]/(X^2+1)$ is a field whereas $\mathbb Q[i]$ is a ring (is the Fraction ring of $\mathbb Q(i)$).
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$\mathbb{Q}[i]$ is in fact a field, since $i^2 = -1$. – Jan 13 '19 at 15:58
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1Actually $\Bbb Q [i]$ happens to be a field, thus it coincides with $\Bbb Q (i)$ – Crostul Jan 13 '19 at 15:58
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4For any $;a\in\Bbb C;$ , we have that $\Bbb Q[a]=\Bbb Q(a)\iff a;$ is algebraic over $;\Bbb Q;$ . You can also change $;\Bbb Q;$ and use your favourite field and its algebraic closure. – DonAntonio Jan 13 '19 at 16:00
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A field is a ring, in case that is worrying you for some reason. – rschwieb Jan 13 '19 at 16:00
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Fields are also rings – Adam Higgins Jan 13 '19 at 16:00
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@Crostul : Really ? but $\mathbb Q[X]/(X^2+1)={aX+b+(X^2+1)\mid a,b\in \mathbb Q}$. But why division is defined ? for example, what is $\frac{X+1}{X+2}$ ? (because division doesn't look defined) – user623855 Jan 13 '19 at 16:01
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@user623855 Your description of that quotient is pretty weak...to say the very least. You should check that carefully – DonAntonio Jan 13 '19 at 16:03
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@PaulK : Why the fact that $i^2=-1$ implies that $\mathbb Q[i]$ is a field ? – user623855 Jan 13 '19 at 16:03
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@DonAntonio: what do you mean ? I'm not correct ? – user623855 Jan 13 '19 at 16:04
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To see why $i^2=-1$ implies $\mathbb{Q}[i]$ is a field, see the first comment by @DonAntonio – saulspatz Jan 13 '19 at 16:08
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@user623855 Whether you're wrong or not depends on what you understand by what you wrote. For example, the elements in $;\Bbb Q[X]/\langle X^2+1\rangle;$ are not polynomials but equivalence classes. If this is clear to you then yes, you're right. And if you do understand this then it is more usual to write those elements in other way. For example, like $;2x+1;$ , where $;x=X+\langle X^2+1\rangle;$ , say., and etc. – DonAntonio Jan 13 '19 at 16:09
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@DonAntonio: Yes I know. I should have written $[X]$ instead of $X$. So what is $\frac{[X+1]}{[X+2]}$ ? – user623855 Jan 13 '19 at 16:14
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@user623855: You have to use the algorithm for rationalizing the denominator. Multiply top and bottom by the conjugate of the denominator. – Cheerful Parsnip Jan 13 '19 at 16:20
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@CheerfulParsnip: What do you mean by "conjugaute of $[X+2]$ ? – user623855 Jan 13 '19 at 16:22
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Go by analogy with what you expect it to be: $-X+2.$ – Cheerful Parsnip Jan 13 '19 at 16:23
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That's the same thing, ring extension equals field extension here. – Wuestenfux Jan 13 '19 at 16:27
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If $\alpha$ is a root of a monic polynomial $\rm\:f(x)\:$ over a domain $\rm\:D\:$ then $\rm\:D[\alpha]\:$ is a field iff $\rm\:D\:$ is a field. Ditto for arbitrary integral extensions of domains. Here is a detailed treament of the quadratic case.
Bill Dubuque
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Addressing your question in the comments: observe that in $\;\Bbb F:=\Bbb Q[X]/\langle X^2+1\rangle\;$ , we have that if $\;x:=Z+\langle x^2+1\rangle\;$ , then $\;x^2+1=0\implies x^2=-1\;$ , so in $\;\Bbb F\;$ we get
$$\frac1{x+2}=\frac{2-x}{4-x^2}=\frac{2-x}{4+1}=\frac15(-x+2)$$
so
$$\frac{x+1}{x+2}=(x+1)\frac1{x+2}=(x+1)\frac15(-x+2)=\frac15(-x^2+x-2)=\frac15(x-1)$$
In similar form you can do other calculation. Observe the is just hte usual operations inthe complex $\;\Bbb C\;$ , just "in disguise".
DonAntonio
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