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I want to show there is an isomorphism from $\mathbb{R}[x]/(x^2-1)$ to $\mathbb{R}^2$.

In the solution provided in https://math.stackexchange.com/a/2132888/1092334, it is $(f(1),f(-1))$ for $f\in \mathbb{R}[x]$.

I can show the homomorphic property, but not surjectivity.

I should take $(a,b)\in\mathbb{R}^2$ and show there is a polynomial in $\mathbb{R}[x]$ for which $f(1)=a, f(-1)=b$. Am I right?

Nij
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hirdajarzu
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1 Answers1

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Both rings are 2 dimensional vector spaces over $\mathbb{R}$. So write down what $f$ does as a matrix and check that it has rank 2.

If you identify $\mathbb{R}^2$ with column vectors, then you have: \begin{equation} f(ax+b) = \begin{pmatrix}1 & 1\\-1 & 1\end{pmatrix}\begin{pmatrix}a \\b\end{pmatrix} \end{equation}

You can then trivially perform gaussian elimination, to determine this matrix has full rank, and is therefore surjective.

Lawrence D'Anna
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  • Can you elaborate on "check rank 2"? I don't know this method – hirdajarzu Jan 24 '24 at 18:51
  • Thanks for editing your answer. I will study your method. I have to delete my question because it has been flagged as duplicate, incorrectly, regardless of my multiple attempts to make it clear it was not a duplicate. Nevermind, I can't even delete it – hirdajarzu Jan 24 '24 at 22:06
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Jan 25 '24 at 02:54
  • @BillDubuque I still believe my question is not a duplicate: first, I link in my post the question someone thought this was a duplicate of; also I'm asking "help me prove there is a surjective function such that...". Perhaps the title is wrong; I made it clear the intention was to find an isomorphism and where my problem came from. I couldn't find a way to prove surjectivity as in the linked question, and I could not (due to low reputation) comment there. I did not mean to post any duplicate answer. I am now allowed by the website to delete this question. – hirdajarzu Jan 25 '24 at 20:48