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I need to identify $\mathbb{R}[x]/(x^2-k)$, where $k>0$ (if $k<0$ I believe it's isomorphic to $\mathbb{C}$).

If we let $f(x) = x^2-k$, then according to Artin, since $\sqrt{k}$ satisfies $f(x)=0$ the set $(1, \sqrt{k})$ is a basis for $\mathbb{R}[\sqrt{k}]$ = $\mathbb{R}[x]/(f)$, i.e. every element of the quotient ring can be written uniquely as $a + b\sqrt{k}$. But this isn't true, because the representation isn't unique.

If we assume $\mathbb{R}[x]/(f) = \mathbb{R}[\sqrt{k}] = \{a + b\sqrt{k} | a,b\in\mathbb{R}\}$ then it looks isomorphic to $\mathbb{R}$, which I don't think it is.

Where have I gone wrong?

Flash
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4 Answers4

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For $k > 0$ we have $\mathbb R[x]/(x^2-k) \cong \mathbb R[x]/(x-\sqrt{k}) \times \mathbb R[x]/(x+\sqrt{k}) \cong \mathbb R \times \mathbb R$ by the Chinese Remainder Theorem.

MooS
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More generally, consider the quotient $\,R[x]/f(x)\,$ where $\,R\,$ is any commutative ring, and $\,f(x) = (x\!-\!a)(x\!-\!b)\,$ for $\,a,b\in R\,$ and $\,a\!-\!b\,$ is invertible in $ R\,$ (e.g. $\,a\ne b\,$ and $ R\,$ is a field).

Define $\ h\,:\, R[x]\to R^2\ $ via $\ g(x)\, \to\, (g(a),g(b)),\ $ a bi-evaluation hom.

Notice $\,\color{#0a0}{h\rm\ \ is\ \ onto}\ $ by $\ r + (s\!-\!r)\,\dfrac{a-x}{a-b\, }\,\to\, (r,s)\in R^2\ \ $ [Lagrange interpolation $\cong$ CRT]

and $\,\ker h\, =\, f(x)R[x]\,$ by the Bifactor Theorem, $ $ since $\,a\!-\!b\,$ is cancellable.

Thus $\ R[x]/f(x)\, =\, R[x]/\ker h\, \cong\, \color{#0a0}{{\rm im}\ h\, =\, R^2}\ $ by the First Isomorphism Theorem.

Community
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Bill Dubuque
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The mistake you made is in the assertion that $\{1,\surd k\}$ is a basis. This is true only in case $\surd k$ is not in the field you started with. But the field of real numbers has square roots (two) for every positive real number. So you get a ring that is a direct product of two copies of R. In particular it has zero divisors.

P Vanchinathan
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More generally, if you quotient by a reducible square-free quadratic polynomial, you get an instance of a ring that I was (once) surprised to find having a name, the "split complex numbers". One you have determined the idempotents in the ring, you find that it is also decomposes as product ring $\Bbb R\times\Bbb R$.

Marc van Leeuwen
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