# A ring isomorphic to a proper subring of itself

Give an example of a ring which is isomorphic to a proper sub-ring of itself.

HINT: Consider $$\Bbb R^\Bbb N$$.

My try:As given in the hint I considered $$\Bbb R^\Bbb N$$ i.e the set of all sequences from $$\Bbb N\rightarrow \Bbb R$$ with pointwise multiplication and addition.

Consider the set $$X=\{(x_n):x_0=0\}$$ i.e. those sequences from $$\Bbb R^\Bbb N$$ whose first term is zero. This is a proper subset of $$\Bbb R^\Bbb N$$ and also a subring but the isomorphism given by

$$a_0+a_1x+\ldots +a_nx^n+\ldots\mapsto a_1x+a_2x^2+\ldots a_nx^n\ldots$$

is not working .How to find the correct one?

• This is not an isomorphism because it is not injective (you can change $a_0$ but the image will not change). May 14, 2016 at 4:24
• Do your rings need to have a multiplicative identity? If so, does the subring need to have the same multiplicative identity as the original ring? May 14, 2016 at 4:30
• The use of $x$ causes confusion. Map the sequence $a_0,a_1,a_2,\dots$ to $0,a_0,a_1,a_2,\dots$. May 14, 2016 at 4:30

I would advice not to feel obliged to follow hints. The first example that comes to mind is the subring $\mathbf Z[X^2]$ of $\mathbf Z[X]$ of all polynomials in $X^2$ with integer coefficients. More precisely, let $$e\colon \mathbf Z[X]\rightarrow \mathbf Z[X]$$ be the evaluation-at-$X^2$ morphism defined by $e(P)=P(X^2)$. It is injective, hence an isomorphism onto its image, the proper subring $\mathbf Z[X^2]$ of $\mathbf Z[X]$.
Consider the ring of polynomials in variables $x_1$, $x_2$, ... and its subring generated by the variables with even index.