# Is every ideal closed under substitution also a T-ideal?

(I hope this is a research level question suitable for mathoverflow.)

Let $\mathbb F\langle X \rangle$ be the free algebra, namely the ring of polynomials over the indeterminates $X=\{x_1,x_2,...\}$, where multiplication is non-commutative, and $\mathbb F$ is some field.

We say that $I\subseteq \mathbb F\langle X \rangle$ is a T-ideal if $I$ is a two-sided ideal that is closed under endomorphisms.

Let $P \subseteq \mathbb F\langle X \rangle$, and let $S(P)$ be the (two-sided) ideal generated by $P$ and that is closed under all substitutions of polynomials from $\mathbb F\langle X \rangle$ to variables $x_1,x_2,...$.

Question: Is it necessarily true that $S(P)$ is a T-ideal?

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What do you mean by "substitution"? If you mean you replace a variable by any element of the free algebra, then yes, because those are already all endomorphisms. –  anton Jul 16 at 11:35
Endomorphisms of the free algebra are substitutions of polynomials instead of variables... (Each endomorphism is determined completely by what it does to generators etc.) –  Vladimir Dotsenko Jul 16 at 11:40
This is a syntactic terminology. Let p be a polynomial in the variables x1,...,xn. Then a substitution instance of p under the substitution \rho: {x1,...,xn} -> F<X> is the polynomial p(\rho(x1),...,\rho(xn)). –  Dilworth Jul 16 at 11:44
Thanks Vladimir. Sorry for the naive question, But is'nt it possible that an endomorphism will map a constant a to another constant ? –  Dilworth Jul 16 at 11:46
Nope. Endomorphisms of F-algebras preserve F by definition. –  Vladimir Dotsenko Jul 16 at 12:28