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Let $k$ be a field and let $A$ and $B$ be $k$-algebras.

Assume that $A$ and $B$ are isomorphic as rings.

Can we conclude that $\dim_k(A)=\dim_k(B)$?

Jesko Hüttenhain
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11101
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  • Just to clarify: You do not assume the isomorphism $A\to B$ to be an isomorphism of $k$-algebras, correct? A counterexample would necessarily be one where that is not the case. – Jesko Hüttenhain Jan 27 '20 at 21:21
  • That's correct. We don't assume that they are isomorphic as k-algebras. In this case dimensions would be the same. – 11101 Jan 28 '20 at 15:47

1 Answers1

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Suppose we have a homomorphism of fields $\phi : k \to k$ which is not surjective. (For one example to show this is possible, we can let $k := \mathbb{C}(t)$, and $\phi : f(t) \mapsto f(t^2)$.) Then $\phi$ induces a $k$-algebra structure on $k$ given by $\lambda * x := \phi(\lambda) x$.

Now, let $A := k$ with the usual $k$-algebra structure and $B := k$ with the $k$-algebra structure induced by $\phi$. Then $A$ and $B$ are certainly isomorphic as rings, yet $\dim_k(A) = 1$ and $\dim_k(B) > 1$. (The second follows from the fact that the nonzero vector $1$ has span $\operatorname{im}(\phi) \ne B$ when you consider the corresponding $k$-vector space structure on $B$.)

Daniel Schepler
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  • I guess another way to view the counterexample would be: $A = \mathbb{C}(t)$ and $B = \mathbb{C}(\sqrt{t})$ are isomorphic as rings, via the map $f(t) \mapsto f(\sqrt{t})$. Yet as $\mathbb{C}(t)$-algebras, $\dim(A) = 1$ and $\dim(B) = 2$. – Daniel Schepler Jan 27 '20 at 23:34
  • That trick is useful for producing all sorts of counterexamples: https://math.stackexchange.com/questions/3158953/changing-scalar-product-in-a-vector-space-cause-different-dimensions/3158970#3158970 – Jim Jan 27 '20 at 23:41