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Let $A$ be the $\Bbb R$-algebra defined by endowing $\Bbb R^2$ with the multiplication $$(a,b)\cdot (c,d)=(ac,ad+bc).$$ Show that $A \cong \Bbb R[x]/(x^2)$.

So I'm planning to use the first isomorphism theorem. For that I need a surjective map $\varphi : \Bbb R[x]\to A$ with kernel $(x^2)$. For any $p = \sum_ka_kx^k \in \Bbb R[x]$ we only need to describe where $x$ maps to, since $$\varphi\left(\sum_ka_kx^k\right)= \sum_ka_k\varphi(x)^k.$$ Since we need to map $x^2$ to $0$ we seek for elements in $A$ that square to $0$. One such element is $(0,1)$ so define $\varphi$ by sending $x$ to $(0,1)$. Extending linearly we get a map $$\varphi\left(\sum_ka_kx^k\right)= \sum_ka_k\varphi(x)^k = \sum_k a_k (0,1)^k.$$

However, I think that there is a problem with this approach. Every image of $\varphi$ has its first component $0$, so it cannot be surjective and so even if I managed to show that $\ker(\varphi) = (x^2)$, I would need to consider the image of $\varphi$ in $A$ only. How should I fix this?

Danlo
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  • Transport the ring structure to the normal form remainders $,a+bx\pmod{!x^2},,$ just like the familiar example $,\Bbb C\cong \Bbb R[x]/(x^2+1),$ here in the linked dupe. – Bill Dubuque Mar 27 '24 at 22:08
  • This is the quotient $\Bbb R[x]/(x^2)$. A priori, I'm defining a map from $\Bbb R[x]$ so I don't have terms as $,a+bx\pmod{!x^2},,$. @BillDubuque – Danlo Mar 27 '24 at 22:44
  • Again, the same method in the linked answer applies here, and immediately yields the sought isomorphism. This is the canonical way to do such in this context. Btw, this ring is known as the algebra of dual numbers over $,\Bbb R.,$ It proves useful as algebraic models for derivatives and tangent / jet spaces, cf. this answer. – Bill Dubuque Mar 27 '24 at 23:20
  • Everything you've written is correct except for your (misplaced) concern at the end. You've forgotten about the constant term of a polynomial (the $k = 0$ term in your sum), which allows the first coordinate to be nonzero. Explicitly, $1 \mapsto (1, 0)$ and $x \mapsto (0, 1)$ provides your surjective homomorphism with kernel $(x^2)$. – Sammy Black Mar 27 '24 at 23:30

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