Definition of $\mathbb{C}$-algebra and $\mathbb{C}$-algebra maps in Invitation to Algebraic Geometry

Question

I am working through Karen Smith's An Invitation to Algebraic Geometry, and I am confused with the following problem from the text.

Let $R$ be a $\mathbb{C}$-algebra, and let $I$ be an ideal of $R$. Prove that the natural surjection $R\rightarrow R/I$ is a $\mathbb{C}$-algebra map.

The reason this question confuses me is that she defines a $\mathbb{C}$-algebra to be a ring $R$ that contains $\mathbb{C}$ as a subring. Furthermore, a $\mathbb{C}$-algebra map is not defined, but a $\mathbb{C}$-algebra homomorphism is defined, so I am assuming the two are the same. She defines a $\mathbb{C}$-algebra homomorphism as follows:

If $R$ and $S$ are $\mathbb{C}$-algebras, then a map $$R\xrightarrow\phi S$$ is said to be a $\mathbb{C}$-algebra homomorphism if it is a ring map (homomorphism) and if it is linear over $\mathbb{C}$, that is, $\phi(\lambda r)=\lambda\phi(r)$ for all $\lambda\in\mathbb{C}$ and $r\in R$.

The reason this is confusing me is that if $I$ is an ideal such that $R/I$ is not a $\mathbb{C}$-algebra, then we can't have a $\mathbb{C}$-algebra map as defined. Thus, should I be making the assumption that $R/I$ is a $\mathbb{C}$ algebra or is one of the definitions incorrect. I am just stuck on trying to even parse where to start on the problem as it seems to me that it might not be true with the given definitions as clearly if we take $I=R$, then $R/R$ is clearly not a $\mathbb{C}$-algebra as defined.

Answer 1

With that definition, the zero ring is not a $\mathbb{C}$-algebra, so $I\ne R$ should be assumed.

Suppose, instead, that $I\ne R$. Then $I\cap\mathbb{C}=\{0\}$, so the induced map $$ \mathbb{C}\to R/I $$ is an injective ring homomorphism and so $R/I$ has a subring isomorphic to $\mathbb{C}$. I guess that the author wants you to identify this with $\mathbb{C}$. But this is a bad way to express things, unless it is proved that once a ring has $\mathbb{C}$ as a subring, then this is the unique subring isomorphic to $\mathbb{C}$. Which is, unfortunately, false.

Extend $\pi$ to a transcendency basis $\{\pi\}\cup B$ of $\mathbb{C}$ over $\mathbb{Q}$. Then the subfield generated by $B$ is a proper subfield of $\mathbb{C}$ and it has an algebraical closure $K$ in $\mathbb{C}$, which is again proper, because it doesn't contain $\pi$. However, the transcendency degree of $K$ over $\mathbb{Q}$ is the same as the one of $\mathbb{C}$, so $K$ and $\mathbb{C}$ are isomorphic.

Then, which subring of $R/I$ isomorphic to $\mathbb{C}$ do we take?

A famous quote (attributed to Albert Einstein, but possibly not his) states

everything should be made as simple as possible, but not simpler.

This is a case where the author is trying to avoid the “complications” with the standard definition, but finds herself in the hole she digged.

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A way out might be to define a $\mathbb{C}$-algebra as a ring with a specified subring isomorphic to $\mathbb{C}$. But this doesn't solve the problem, unless one defines $R/I$ as a $\mathbb{C}$-algebra by specifying the above-mentioned subring built from the specified subring of $R$.

One doesn't gain too much with respect to the standard definition: a $\mathbb{C}$-algebra is a pair $(R,\lambda_R)$, where $\lambda_R\colon \mathbb{C}\to R$ is a ring homomorphism. If $(R,\lambda_R)$ and $(S,\lambda_S)$ are $\mathbb{C}$-algebras, a ring homomorphism $f\colon R\to S$ is a $\mathbb{C}$-algebra homomorphism if $f\circ\lambda_R=\lambda_S$.

Now the exercise becomes a simple observation: if $I$ is an ideal of the $\mathbb{C}$-algebra $(R,\lambda_R$), then the homomorphism $\lambda_{R/I}\colon\mathbb{C}\to R/I$, $\lambda_{R/I}=p\circ\lambda_R$, where $p\colon R\to R/I$ is the canonical projection, is the unique ring homomorphism that makes $R/I$ into a $\mathbb{C}$-algebra so that $p$ is a $\mathbb{C}$-algebra homomorphism.

Then one forgets about $\lambda_R$, because it's normally clear from the context.

Answer 2

The given definition of $\mathbb{C}$-algebra is not the standard definition, and you are right that the problem does not makes sense with the given definition ($R/I$ will pretty much never literally contain $\mathbb{C}$ as a subring; at best you might hope for an isomorphic copy of $\mathbb{C}$, but even that will fail when $I=R$ as you observed). (One formulation of) the standard definition of a (commutative) $\mathbb{C}$-algebra is a commutative ring $A$ together with a ring-homomorphism $f:\mathbb{C}\to A$. Notably, this homomorphism $f$ need not be literally the inclusion of $\mathbb{C}$ as a subring of $A$, and it does not even have to be injective (though if it is not injective, then it is identically $0$ since $\mathbb{C}$ is a field so $A$ must be the zero ring). A $\mathbb{C}$-algebra homomorphism (or $\mathbb{C}$-algebra map) between $\mathbb{C}$-algebras $A,B$ is a ring-homomorphism $h:A\to B$ such that $h\circ f=g$ where $f:\mathbb{C}\to A$ and $g:\mathbb{C}\to B$ are the $\mathbb{C}$-algebra structures on $A$ and $B$. (Equivalently, a $\mathbb{C}$-algebra homomorphism is a ring-homomorphism which is also $\mathbb{C}$-linear, where you make $A$ a $\mathbb{C}$-vector space by defining $\lambda\cdot a=f(\lambda)a$ and similarly for $B$.)

Note in particular that by this definition, a single ring can admit multiple different $\mathbb{C}$-algebra structures, so it still does not make sense to ask whether $R\to R/I$ is a $\mathbb{C}$-algebra homomorphism until you specify the $\mathbb{C}$-algebra structure on $R/I$. The standard way to do so is to let the chosen homomorphism $\mathbb{C}\to R/I$ be the composition of the chosen homomorphism $\mathbb{C}\to R$ with the quotient map $R\to R/I$. This makes the quotient map a $\mathbb{C}$-algebra homomorphism by definition.

Answer 3

Furthermore, a ℂ-algebra map is not defined, but a ℂ-algebra homomorphism is defined, so I am assuming the two are the same.

Yes, they are the same.

The reason this is confusing me is that if 𝐼 is an ideal such that 𝑅/𝐼 is not a ℂ-algebra, then we can't have a ℂ-algebra map as defined. [...] clearly if we take 𝐼=𝑅, then 𝑅/𝑅 is clearly not a ℂ-algebra as defined.

That last line is correct. So technically the problem you're confused about should specify $I\neq R$.

But whenever $I\neq R$ there is no problem: in $R/I$, the subset $\{k+I\mid k\in K\}$ is an isomorphic copy of $k$ inside $R/I$. So $R/I$ is always a $k$-algebra if $R$ is (and $I$ is a proper ideal of $R$.)

So there is not really much to get worked up about here, just an unfortunate incompatibility in the statement of the problem you're referencing to the definition of $k$-algebra you're referencing.